refraction of light,from lens and slabs

$: Refraction of light,from lens and slabs$
Refraction at a Boundary

Boundary Behavior

A wave doesn't just stop when it reaches the end of the medium. Rather, a wave will undergo

certain behaviors when it encounters the end of the medium. Specifically, there will be some

reflection off the boundary and some transmission into the new medium. The transmitted wave

undergoes refraction (or bending) if it approaches the boundary at an angle. If the boundary is

merely an obstacle implanted within the medium, and if the dimensions of the obstacle are

smaller than the wavelength of the wave, then there will be very noticeable diffraction of the

wave around the object. The reflection, refraction, and diffraction of waves were first

introduced in Unit 10 of The Physics Classroom tutorial. In Unit 11 of The Physics Classroom

Tutorial, the reflection, refraction, and diffraction of sound waves was discussed. Since light is

a wave, it too will undergo these same behaviors when it reaches a boundary between two

medium. The boundary behavior of light waves has already been introduced in Unit 12 of The

Physics Classroom Tutorial. In this unit, we will focus on the refraction of light in great detail.

We will explore the conceptual and mathematical principles governing the bending of waves as

they cross the boundary between two media. To understand light refraction, we will need to

back up a few steps and investigate the behavior of waves when they reach the end of a

medium.

Boundary Behavior for Waves on a Rope

Suppose that there is a thin rope attached to a thick rope, with each rope held at opposite

ends by people. And suppose that a pulse is introduced by the person holding the end of the

thin rope. If this is the case, there will be an incident pulse traveling in the less dense medium

(thin rope) towards the boundary with a more dense medium (thick rope).

Upon reaching the boundary, two behaviors will occur.

A portion of the energy carried by the incident pulse is reflected and returns towards the left end of the thin rope. The disturbance which returns to the left after bouncing off the boundary is known as the reflected pulse.

A portion of the energy carried by the incident pulse is transmitted into the thick rope. The disturbance which continues moving to the right is known as thetransmitted pulse.

These two behaviors - reflection and transmission - were first introduced in Unit 10 of The

Physics Classroom. In that unit, it was mentioned that the passage of the energy from the

incident medium into the transmitted medium was accompanied by a change in speed and

http://www.physicsclassroom.com/Class/waves/u10l3a.cfm

http://www.physicsclassroom.com/Class/waves/u10l3a.cfm

http://www.physicsclassroom.com/Class/light/u12l1a.cfm


http://www.physicsclassroom.com/Class/sound/u11l3d.cfm

http://www.physicsclassroom.com/Class/sound/u11l3d.cfm

http://www.physicsclassroom.com/Class/waves/u10l3b.cfm

wavelength. In the case of a pulse crossing the boundary from a less dense medium into a

more dense medium, the speed and the wavelength are both decreased. On the other hand,

if a pulse crosses the boundary from a more dense medium into a less dense medium, the

speed and the wavelength are both increased.

Refraction of Light Waves

The above discussion was limited to the behavior of a wave on a rope. But what if the wave is

a light wave traveling in a three-dimensional medium? For example, what would happen if a

light wave is traveling through air and reaches the boundary with a glass surface? How can the

reflection and transmission behavior of a light wave be described? First, the light wave

behaves like the wave on the rope: a portion of the wave is transmitted into the new medium

(glass) and a portion of the wave reflects off the air-glass boundary. Second, the same wave

property changes which were observed for the wave on the rope are also observed for the light

wave passing from air into glass; there is a change in speed and wavelength of the wave as it

crosses the air-glass boundary. When passing from air into glass, both the speed and the

wavelength decrease. Finally, and most importantly, the light is observed to change directions

as it crosses the boundary separating the air and the glass. This bending of the path of light is

known as refraction. A one-word synonym for refraction is bending. The transmitted wave

experiences this refraction at the boundary. As seen in the diagram at the right, each

individual wavefront is bent only along the boundary. Once the wavefront has passed across

the boundary, it travels in a straight line. For this reason, refraction is called a boundary

behavior. A ray is drawn perpendicular to the wavefronts; this ray represents the direction

which the light wave is traveling. Observe that the ray is a straight line inside of each of the

two media, but bends at the boundary. Again, refraction is a boundary behavior.

The Ray Model of Light

In this unit, we will rely heavily on the use of rays to represent the direction in which light is

moving. While we often think of light behaving as a wave, we will still find it useful to represent

its movement through a medium using a line segment with an arrowhead (i.e., a ray) to depict

the refraction of light. The ray is constructed in a direction perpendicular to the wavefronts of

the light wave; this accurately depicts the light wave's direction. In this sense, we are viewing

light as behaving as a stream of particles which head in the direction of the ray. The idea that

the path of light can be represented by a ray is known as the ray model of light. The same ray


http://www.physicsclassroom.com/mmedia/waves/ltm.cfm

model was utilized in Unit 13 of The Physics Classroom Tutorial to discuss the reflection of light

waves.


Refraction and Sight

In Unit 13 of The Physics Classroom Tutorial, it was emphasized that we are able to see

because light from an object can travel to our eyes. Every object that can be seen is seen only

because light from that object travels to our eyes. As you look at Mary in class, you are able to

see Mary because she is illuminated with light and that light reflects off of her and travels to

your eye. In the process of viewing Mary, you are directing your sight along a line in the

direction of Mary. If you wish to view the top of Mary's head, then you direct your sight along a

line towards the top of her head. If you wish to view Mary's feet, then you direct your sight

along a line towards Mary's feet. And if you wish to view the image of Mary in a mirror, then

you must direct your sight along a line towards the location of Mary's image. This directing of

our sight in a specific direction is sometimes referred to as the line of sight.

As light travels through a given medium, it travels in a straight line. However, when light

passes from one medium into a second medium, the light path bends. Refraction takes place.

The refraction occurs only at the boundary. Once the light has crossed the boundary between

the two media, it continues to travel in a straight line. Only now, the direction of that line is

different than it was in the former medium. If when sighting at an object, light from that object

changes media on the way to your eye, a visual distortion is likely to occur. This visual

distortion is witnessed if you look at a pencil submerged in a glass half-filled with water. As you

sight through the side of the glass at the portion of the pencil located above the water's

surface, light travels directly from the pencil to your eye. Since this light does not change

medium, it will not refract. (Actually, there is a change of medium from air to glass and back

into air. Because the glass is so thin and because the light starts and finished in air, the

refraction into and out of the glass causes little deviation in the light's original direction.) As

you sight at the portion of the pencil which was submerged in the water, light travels from

water to air (or from water to glass to air). This light ray changes medium and subsequently

undergoes refraction. As a result, the image of the pencil appears to be broken. Furthermore,

the portion of the pencil which is submerged in water appears to be wider than the portion of

http://www.physicsclassroom.com/Class/refrn/u14l1a.cfm#refrn

http://www.physicsclassroom.com/Class/refln/u13l1a.cfm

http://www.physicsclassroom.com/Class/refln/

the pencil which is not submerged. These visual distortions are explained by the

refraction of light.

In this case, the light rays which undergo a deviation from their original path are

those which travel from the submerged portion of the pencil, through the water,

across the boundary, into the air, and ultimately to the eye. At the boundary, this ray refracts.

The eye-brain interaction cannot account for the refraction of light. As was emphasized in Unit

13, the brain judges the image location to be the location where light rays appear to originate

from. This image location is the location where either reflected or refracted rays intersect. The

eye and brain assume that light travels in a straight line and then extends all incoming rays of

light backwards until they intersect. Light rays from the submerged portion of the pencil will

intersect in a different location than light rays from the portion of the pencil which extends

above the surface of the water. For this reason, the submerged portion of the pencil appears to

be in a different location than the portion of the pencil which extends above the water. The

diagram at the right shows a God's-eye view of the light path from the submerged portion of

the pencil to each of your two eyes. Only the left and right extremities (edges) of the pencil are

considered. The blue lines depict the path of light to your right eye and the red lines depict the

path of light to your left eye. Observe that the light path has bent at the boundary. Dashed

lines represent the extensions of the lines of sight backwards into the water. Observe that the

these extension lines intersect at a given point; the point represents the image of the left and

the right edge of the pencil. Finally, observe that the image of the pencil is wider than the

actual pencil. A ray model of light which considers the refraction of light at boundaries

adequately explains the broken pencil observations.

The broken pencil phenomenon occurs during your everyday spear-fishing outing. Fortunately

for the fish, light refracts as it travels from the fish in the water to the eyes of the hunter. The

refraction occurs at the water-air boundary. Due to this bending of the path of light, a fish

appears to be at a location where it isn't. A visual distortion occurs. Subsequently, the hunter

launches the spear at the location where the fish is thought to be and misses the fish. Of

course, the fish are never concerned about such hunters; they know that light refracts at the

boundary and that the location where the hunter is sighting is not the same location as the

actual fish. How did the fish get so smart and learn all this? They live in schools.



http://www.physicsclassroom.com/mmedia/optics/bp.cfm

Now any fish who has done his/her physics homework knows that the amount of refraction

which occurs is dependent upon the angle at which the light approaches the boundary. We will

investigate this aspect of refraction in great detail in Lesson 2. For now, it is sufficient to say

that as the hunter with the spear sights more perpendicular to the water, the amount of

refraction decreases. The most successful hunters are those who sight perpendicular to the

water. And the smartest fish are those who head for the deep when they spot hunters who

sight in this direction.

Since refraction of light occurs when it crosses the boundary, visual distortions often occur.

These distortions occur when light changes medium as it travels from the object to our eyes.


The Cause of Refraction

We have learned that refraction occurs as light passes across the boundary between two

medium. Refraction is merely one of several possible boundary behaviors by which a light

wave could behave when it encounters a new medium or an obstacle in its path. The

transmission of light across a boundary between two media is accompanied by a change in

both the speed and wavelength of the wave. The light wave not only changes directions at the

boundary, it also speeds up or slows down and transforms into a wave with a larger or a

shorter wavelength. The only time that a wave can be transmitted across a boundary, change

its speed, and still not refract is when the light wave approaches the boundary in a direction

which is perpendicular to it. As long as the light wave changes speed and approaches the

boundary at an angle, refraction is observed.

But why does light refract? What is the cause of such behavior? And why is there this one

exception to the refraction of light? An analogy of marching soldiers is often used to address

this question. In fact, it is not uncommon that the analogy be illustrated in a Physics class with

http://www.physicsclassroom.com/Class/refrn/u14l1a.cfm#bb

http://www.physicsclassroom.com/Class/refrn/u14l1a.cfm#refrn

http://www.physicsclassroom.com/Class/refrn/u14l2a.cfm

a student demonstration. A group of students forms a straight line (shoulder to shoulder)

and connect themselves to their nearest neighbor using meter sticks. A strip of masking tape

divides the room into two media. In one of the media (on one side of the tape), students walk

at a normal pace. In the other media (or on the other side of the tape), students walk very

slowly using baby steps. The group of students walk forward together in a straight line towards

the diagonal strip of masking tape. The students maintain the line as they approach the

masking tape. When an individual student reaches the tape, that student abruptly changes the

pace of her/his walk. The group of students continue walking until all students in the line have

entered into the second medium. The diagram below represents the line of students

approaching the boundary (the masking tape) between the two medium. On the diagram, an

arrow is used to show the general direction of travel for the group of students in both medium.

Observe that the direction of the students changes at the "boundary."

The fundamental feature of the students' motion which leads to this change in direction is the

change in speed. Upon reaching the masking tape, each individual student abruptly changes

speed. Because the students approach the masking tape at an angle, each individual student

reaches the tape at a different time. The student who reaches the tape first, slows down while

the rest of the line of students marches ahead. This occurs for every student in the line of

students. Once a student reaches the boundary, that student slows down while his/her nearest

neighbor marches ahead at the original pace. The result is that the direction that the line of

students is heading is altered at the boundary. The change in speed of the line of students

causes a change in direction.

Conditions of Refraction

Will this refractive behavior always occur? No! There are two conditions which are required in

order to observe the change in direction of the path of the students:

The students must change speed when crossing the boundary.

The students must approach the boundary at an angle; refraction will not occur when they approach the boundary head-on (i.e., heading perpendicular to it).

These are both reasonable enough conditions if you consider the previous paragraph. If the

students do not change speed, then there is no cause factor. Recall that it was the change in

speed of the students which caused the change in direction. The second condition is also

reasonable. If the students approach the masking tape in a direction which is perpendicular to

it , then each student will reach the tape at the exact same time. Recall that the line of student

changed their direction because they had reached the masking tape at different times. The

first student reached the tape, slowed down, and observed the rest of the students marching

ahead at the original speed. The change in direction of the line of students only occurs at the

boundary when the students change speed and approach at an angle.

The Marching Soldiers analogy provides an excellent analogy to understanding the cause of

light refraction. The line of students approaching the masking tape are analogous to a

wavefront of light. The masking tape is analogous to a boundary between two medium. The

change in speed which occurred for the line of students would also occur for a wave of light.

And like the marching students, a light wave will not undergo refraction if it approaches the

boundary in a direction which is perpendicular to it.

The same two conditions which are necessary for bending the path of the line of students are

also necessary for bending the direction of a light ray. Light refracts at a boundary because of

a change in speed. Their is a distinct cause-effect relationship. The change in speed is the

cause and the change in direction (refraction) is the effect.


Optical Density and Light Speed

Refraction is the bending of the path of a light wave as it passes from one material to another

material. The refraction occurs at the boundary and is caused by a change in the speed of the

light wave upon crossing the boundary. The tendency of a ray of light to bend one direction or

another is dependent upon whether the light wave speeds up or slows down upon crossing the

boundary. Because a major focus of our study will be upon the direction of bending, it will be

important to understand the factors which effect the speed at which a light wave is

transported through a medium.

The mechanism by which a light wave is transported through a medium occurs in a manner

which is similar to the way that any other wave is transported - by particle-to-particle

interaction. In Unit 10 of The Physics Classroom Tutorial, the particle-to-particle interaction

mechanism by which a mechanical wave transports energy was discussed in detail. In Unit 12

of The Physics Classroom Tutorial, the mechanism of energy transport by an electromagnetic

wave was briefly discussed. Here we will look at this method in more detail.

An electromagnetic wave (i.e., a light wave) is produced by a vibrating electric charge. As the

wave moves through the vacuum of empty space, it travels at a speed of c (3 x 108m/s). This

value is the speed of light in a vacuum. When the wave impinges upon a particle of matter, the

energy is absorbed and sets electrons within the atoms into vibrational motion. If the

frequency of the electromagnetic wave does not match theresonant frequency of vibration of

the electron, then the energy is reemitted in the form of an electromagnetic wave. This new

electromagnetic wave has the same frequency as the original wave and it too will travel at a

speed of c through the empty space between atoms. The newly emitted light wave continues

to move through the interatomic space until it impinges upon a neighboring particle. The

energy is absorbed by this new particle and sets the electrons of its atoms into vibration

motion. And once more, if there is no match between the frequency of the electromagnetic

wave and the resonant frequency of the electron, the energy is reemitted in the form of a new

electromagnetic wave. The cycle of absorption and reemission continues as the energy is

transported from particle to particle through the bulk of a medium. Every photon (bundle of

electromagnetic energy) travels between the interatomic void at a speed of c; yet time delay

involved in the process of being absorbed and reemitted by the atoms of the material lowers

the net speed of transport from one end of the medium to the other. Subsequently, the net

http://www.physicsclassroom.com/Class/light/u12l2c.cfm

http://www.physicsclassroom.com/Class/light/u12l2c.cfm

http://www.physicsclassroom.com/Class/light/u12l1e.cfm#scat

http://www.physicsclassroom.com/Class/light/u12l1e.cfm#scat

http://www.physicsclassroom.com/Class/waves/u10l1b.cfm

http://www.physicsclassroom.com/Class/refrn/u14l1c.cfm


http://www.physicsclassroom.com/mmedia/optics/bp.cfm

speed of an electromagnetic wave in any medium is somewhat less than its speed in a vacuum

- c (3 x 108 m/s).

Optical Density and the Index of Refraction

Like any wave, the speed of a light wave is dependent upon the properties of the medium. In

the case of an electromagnetic wave, the speed of the wave depends upon the optical

density of that material. The optical density of a medium is not the same as its physical

density. The physical density of a material refers to the mass/volume ratio. The optical density

of a material relates to the sluggish tendency of the atoms of a material to maintain the

absorbed energy of an electromagnetic wave in the form of vibrating electrons before

reemitting it as a new electromagnetic disturbance. The more optically dense which a material

is, the slower that a wave will move through the material.

One indicator of the optical density of a material is the index of refraction value of the

material. Index of refraction values (represented by the symbol n) are numerical index values

which are expressed relative to the speed of light in a vacuum. The index of refraction value of

a material is a number which indicates the number of times slower that a light wave would be

in that material than it is in a vacuum. A vacuum is given an n value of 1.0000. The n values of

other materials are found from the following equation:

The table below lists index of refraction values for a variety of medium. The materials listed at the top of the table are those through which light travels fastest; these are the least optically dense materials. The materials listed at the bottom of the table are those through which light travels slowest; these are the most optically dense materials. So as the index of refraction value increases, the optical density increases, and the speed of light in that material decreases.

Material Index of Refraction

Vacuum 1.0000<--lowest optical density

Air 1.0003

Ice 1.31

Water 1.333

Ethyl Alcohol 1.36

Plexiglas 1.51

Crown Glass 1.52

Light Flint Glass 1.58

Dense Flint Glass 1.66

Zircon 1.923

Diamond 2.417

Rutile 2.907

Gallium phosphide 3.50<--highest optical density

The index of refraction values thus provide a measure of the relative speed of a light wave in a

particular medium. Knowledge of such relative speeds allows a student of physics to predict

which way a light ray would bend when passing from one medium to the another. In the next

part of Lesson 1, the rules for the direction of bending will be discussed in detail.


The Direction of Bending

Refraction is the bending of the path of a light wave as it passes from one material into

another material. The refraction occurs at the boundary and is caused by a change in the

speed of the light wave upon crossing the boundary. The tendency of a ray of light to bend one

direction or another is dependent upon whether the light wave speeds up or slows down upon

crossing the boundary. The speed of a light wave is dependent upon the optical density of the

material through which it moves. For this reason, the direction which the path of a light wave

bends depends on whether the light wave is traveling from a more dense (slow) medium to a

less dense (fast) medium or from a less dense medium to a more dense medium. In this part of

Lesson 1, we will investigate this topic of the direction of bending of a light wave.

Recall the Marching Soldiers analogy discussed earlier in this lesson. The analogy served as a

model for understanding the boundary behavior of light waves. As discussed, the analogy is

often illustrated in a Physics classroom by a student demonstration. In the demonstration, a

line of students (representing a light wave) march towards a masking tape (representing the

boundary) and slow down upon crossing the boundary (representative of entering a new

medium). The direction of the line of students changes upon crossing the boundary. The

http://www.physicsclassroom.com/Class/refrn/u14l1c.cfm#march

http://www.physicsclassroom.com/Class/refrn/u14l1d.cfm




http://www.physicsclassroom.com/Class/refrn/u14l1e.cfm


diagram below depicts this change in direction for a line of students who slow down upon

crossing the boundary.

On the diagram, the direction of the students is represented by two arrows known asrays. The

direction of the students as they approach the boundary is represented by anincident

ray (drawn in blue). And the direction of the students after they cross the boundary is

represented by a refracted ray (drawn in red). Since the students change direction (i.e.,

refract), the incident ray and the refracted ray do not point in the same direction. Also, note

that a perpendicular line is drawn to the boundary at the point where the incident ray strikes

the boundary (i.e., masking tape). A line drawn perpendicular to the boundary at the point of

incidence is known as a normal line. Observe that the refracted ray lies closer to the normal

line than the incident ray does. In such an instance as this, we would say that the path of the

students has benttowards the normal. We can extend this analogy to light and conclude that:

Light Traveling from a Fast to a Slow MediumIf a ray of light passes across the boundary from a material in which it travels fast into a material in which

travels slower, then the light ray will bend towards the normal line.

The above principle applies to light passing from a material in which it travels fast across a

boundary and into a material in which it travels slow. But what if light wave does the opposite?

What if a light wave passes from a material in which it travels slow across a boundary and into

a material in which it travels fast? The answer to this question can be answered if we

reconsider the Marching Soldier analogy. Now suppose that the each individual student in the

train of students speeds up once they cross the masking tape. The first student to reach the

boundary will speed up and pull ahead of the other students. When the second student

reaches the boundary, he/she will also speed up and pull ahead of the other students who

have not yet reached the boundary. This continues for each consecutive student, causing the

line of students to now be traveling in a direction further from the normal. This is depicted in

the diagram below.

From this analogy and the diagram above, we see that the refracted ray (in red)

is further away from the normal then the incident ray (in blue). In such an

instance as this, we would say that the path of the students has bent away

from the normal. We can once more extend this analogy to light and conclude that:

Light Traveling from a Slow to a Fast MediumIf a ray of light passes across the boundary from a material in which it travels slow into a material in which

travels faster, then the light ray will bend away from the normal line.

The Tractor Analogy

Now lets consider another analogy to assist in our understanding of these two important

principles. Suppose that a tractor is moving across an asphalt surface towards a rectangular

plot of grass (as shown in the diagram at the right). Upon entering the grass, the tractors'

wheels will sink into the surface and slow down. Upon exiting the plot of grass on the opposite

side, the tractor wheels will speed up and achieve their original speed. In effect, this analogy

would be representative of a light wave crossing two boundaries. At the first boundary (the

asphalt to grass boundary), the light wave (or the tractor) would be slowing down; and at the

second boundary (the grass to asphalt boundary), the light wave (or the tractor) would be

speeding up. We can apply our two important principles listed above and predict the direction

of bending and the path of the tractor as it travels through the rectangular plot of grass. As

indicated on the diagram, upon entering the grass, the wheels slow down and the path of the

tractor bends towards the normal (perpendicular line drawn to the surface). Upon exiting the

plot of grass, the wheels speed up and the path of the tractor bends away from the normal.

The path of the tractor is closer to the normal in the slower medium and farther away from the

normal in the faster medium.

This analogy can be extended to the path of a light wave as it passes from air into and out of a

rectangular block of glass. Since air is less optically dense than glass, the light wave will slow

down upon entering the glass and speed up when exiting the glass. In other words, the light

wave will be undergoing the same change in speed as the tractor in the above diagram. For

this reason, the direction of bending for the light wave upon entering and exiting the glass will

be the same as in the diagram above. The light ray refracts towards the normal upon entering

the glass (crossing from a fast to a slow medium) and refracts away from the normal upon

exiting the glass (crossing from a slow to a fast medium). This is shown in the diagram at the

right.

There is an important point to be noted in these diagrams of the rectangular plot of grass and

rectangular block of glass. Notice that the direction of the original incident ray is the same as

the direction of the final refracted ray. Put another way, the direction at which the light is

traveling when entering the rectangular block of glass is the same as the direction which the

light travels after exiting the rectangular block of glass. There is no ultimate change in the

direction which the light is traveling. This small detail will only be the case under two

conditions:

the two sides of the glass through which the light enters and exits are parallel to each other

the medium surrounding the glass on the side through which the light enters and exits are the same

These two conditions are met in the case of a rectangular block of glass surrounded by air.

The diagrams below provides a contrast to the rectangular plot of grass and the rectangular

block of glass. Both diagrams involve the refraction of a tractor or a light wave as it passes into

and out of a triangular plot of grass and a triangular block of glass.

Copy this diagram onto a sheet of paper and apply your understanding of refraction principles

to predict the path of the tractor and the light wave as it travels through the triangle-shaped

obstacle. Draw the path on your separate sheet of paper and then click on the button below to

check your answer

Least Time Principle

Another means of approaching the subject of the direction which light bends when crossing a

boundary between two medium is through the Least Time Principle. This Least Time Principle is sometimes stated as follows:

Least Time PrincipleOf all the possible paths that light might take to get from one point to another, it always takes the path that

requires the least amount of time.

A useful analogy to understanding the principle involves a life guard who has become aware of

a drowning swimmer in the water. In order to save the drowning swimmer, the life guard must

run through the sand, cross the boundary between the sand and the water, and then swim

through the water to the drowning swimmer. Of course, the guard must reach the swimmer in

as little time as possible. Since the guard can run faster on sand than she can swim in water, it

would make sense that the guard cover more distance in the sand than she does in the water.

In other words, she will not run directly at the drowning swimmer. The optimal entry point into

the water is the point which would allow the life guard to reach the drowning swimmer in the

least amount of time. Obviously, this point would be at a location closer to the swimmer than

to the guard. The diagram below depicts such an entry point.

Observe in the diagram, that minimizing the time to reach the swimmer means that the life

guard will approach the boundary at a steep angle to the normal and then will bend towards

the normal upon crossing the boundary. This analogy demonstrates that the Least Time

Principle would predict the following direction of bending:

A ray of light will bend towards the normal when crossing the boundary from a medium in which it travels fast into a medium in which it travels slow.

This is the very generalization which was made earlier on this page.

http://www.physicsclassroom.com/Class/refrn/u14l1e.cfm#fst%23fst


Some Useful Mnemonics

Using the above principles and logic to explain and predict the direction which light refracts when crossing a boundary will be a major objective of this unit. Rather than merely restating the principle, you will be asked to apply it to a variety of situations (such as those in the Check Your Understanding section below). Part of accomplishing this task will involve remembering the principles. For this reason, the following useful mnemonics are offered.

FST = Fast to Slow, Towards NormalIf a ray of light passes across the boundary from a material in which it travelsfast into a material in which

travels slower, then the light ray will bend towards the normal line.

SFA = Slow to Fast, Away From NormalIf a ray of light passes across the boundary from a material in which it travelsslow into a material in which

travels faster, then the light ray will bend away from the normal line.

A mnemonic is a tool used to help one remember and difficult-to-remember idea. Of course,

there is always the risk that the mnemonic will be forgotten. And since FST and SFA might not

be the most easily remembered mnemonics, perhaps the following oddity will help. You can

remember FST (fast to slow; towards) by simply thinking about thoseFreaky Science

Teachers which you have had through the years. And you can remember SFA by thinking

about the disgusting habit of your friend Sara (or Susan or Sammy or Samir or Somebody ...)

- Sarah Farts Alot.

Check Your Understanding

Test your ability to apply these principles by answering the following questions.

1. When light passes from a more optically dense medium into a less optically densemedium, it

will bend _______ (towards, away from) the normal.

2. When light passes from a less optically dense medium into a more optically densemedium, it

will bend _______ (towards, away from) the normal.






3. When light passes from a medium with a high index of refraction value into a medium with a

low index of refraction value, it will bend _______ (towards, away from) the normal.

4. When light passes from a medium with a low index of refraction value into a medium with a

high index of refraction value, it will bend _______ (towards, away from) the normal.

5. In each diagram, draw the "missing" ray (either incident or refracted) in order to

appropriately show that the direction of bending is towards or away from the normal.

6. Arthur Podd's method of fishing involves spearing the fish while standing on the shore. The

actual location of a fish is shown in the diagram below. Because of the refraction of light, the

observed location of the fish is different than its actual location. Indicate on the diagram the

approximate location where Arthur observes the fish to be. Must Arthur aim above or below

where the fish appears to be in order to strike the fish?

http://www.physicsclassroom.com/Class/refrn/u14l1d.cfm#index





7. For the following two cases, state whether the ray of light will bend towards or away from

the normal upon crossing the boundary.


If I Were an Archer Fish

In the quiet waters of the Orient, there is an unusual fish known as the Archer fish. The Archer

fish is unlike any other fish in that the Archer fish finds its prey living outside the water. An

insect, butterfly, spider or similar creature is the target of the Archer fish's powerful spray of

water. The Archer fish will search for prey that are resting upon a branch or twig above the

water. With pinpoint accuracy, the fish knocks the prey off the branch using a powerful jet of

water. The prey falls to the water as the Archer fish simultaneously swims directly to the

location on the surface where the prey strikes the water, wasting no time to retrieve its meal.

The feat of shooting a stream of water to knock the prey off a branch is remarkable. The fact

that the Archer fish can do this time and again with pinpoint accuracy is even more

remarkable. The fact that the fish can determine the exact location of incidence at which the

prey will subsequently strike the water is incredibly remarkable. But most remarkable of all is

that the Archer fish can accomplish this trick despite the fact that light from the prey to

its eye undergoes refraction at the air-water boundary. Such refraction would cause

avisual distortion, making the prey appear to be in a location where it isn't. Yet the

Archer fish is hardly ever fooled. What is the secret of the Archer fish? How is it able to

overcome the visual distortion caused by refraction in order to accomplish these

remarkable hunting tasks? Biologists are not quite sure, though as of this writing it is a topic

which is under considerable experimental investigation.

Now we will entertain the question: What if I were an Archer fish? What could I do to perform

such a remarkable aquatic magic trick? Unlike a fish, I don't live in schools; nonetheless, there

is a principle taught in schools that might assist me in such a trick. That principle is usually

taught in a Physics classroom.

There is only one condition in which light can pass from one medium to another, change its

speed, and still not refract. If the light is traveling in a direction which is perpendicular to the

boundary, no refraction occurs. As the light wave crosses over the boundary, its speed and

wavelength still change. Yet, since the light wave is approaching the boundary in a

perpendicular direction, each point on the wavefront will reach the boundary at the same time.

For this reason, there is no refraction of the light. Such a ray of light is said to be approaching

the boundary while traveling along the normal. (The normal is a line drawn perpendicular to

the surface.)

The only means by which I could remotely match the marvels of the Archer fish would be to

line up my sight with the prey from a position directly underneath the prey. From this vantage

point, light from the prey travels directly to my eye without undergoing a change in direction.

Since the light is traveling along the normal to the surface, it does not refract. The light passes

straight through the water to my eyes. Normally, when light from an object changes medium

on the way to the eye, there is a visual distortion of the image. But if I sight along the normal,

there is no refraction and no visual distortion of the image. From this ideal line of sight, I would

be able to hit my prey time after time (assuming I could master the task of spraying a jet of

water in the desired direction). Using my physics understanding of the refraction of light, I

could pretend to mimic the Archer fish.

http://www.physicsclassroom.com/Class/refrn/u14l1b.cfm



And now for the rest of the story. From this discussion, one might conclude that the secret of

the Archer fish is to aim at its prey from directly below. Refraction is less when sighting along

the normal. However, the Archer fish's accomplishments are more remarkable than that. It has

been found that Archer fish are able to strike their prey when sighting upwards at angles of 40

degrees with the normal. In fact, it has been found that hit probabilities do not show significant

variance with the angle of sighting, meaning that an Archer fish is just as likely to strike its

prey whether the amount of refraction is great or minimal. Now that's remarkable!

The Mathematics of Refraction

The Angle of Refraction

Refraction is the bending of the path of a light wave as it passes across the boundary

separating two media. Refraction is caused by the change in speed experienced by a wave

when it changes medium. In Lesson 1, we learned that if a light wave passes from a medium in

which it travels slow (relatively speaking) into a medium in which it travels fast, then the light

wave will refract away from the normal. In such a case, the refracted ray will be farther from

the normal line than the incident ray; this is the SFA rule of refraction. On the other hand, if a

light wave passes from a medium in which it travels fast (relatively speaking) into a medium in

which it travels slow, then the light wave will refract towards the normal. In such a case, the

refracted ray will be closer to the normal line than the incident ray is; this is the FST rule of

refraction. These two rules regarding the refraction of light only indicate the direction which a

light ray bends; they do not indicate how much bending occurs. Lesson 1 focused on the topics

of "What causes refraction?" and "Which direction does light refract?" Lesson 2 will focus on

the question of "By how much does light refract when it crosses a boundary?"

http://www.physicsclassroom.com/Class/refrn/u14l1e.cfm#fst


http://www.physicsclassroom.com/Class/refrn/u14l1e.cfm#sfa




The question is: "By how much does light refract when it crosses a boundary?"

Perhaps there are numerous answers to such a question. (For example, " a lot,"

"a little," "like wow! quite a bit dude," etc.) The concern of this lesson is to

express the amount of refraction of a light ray in terms of a measurable

quantity that has a mathematical value. The diagram to the right shows a light

ray undergoing refraction as it passes from air into water. As mentioned in Lesson 1,

the incident rayis a ray (drawn perpendicular to the wavefronts) which shows the direction

which light travels as it approaches the boundary. (The meaning of an incident ray was first

introduced in the discussion of Reflection of Light in Unit 13 of The Physics Classroom Tutorial.)

Similarly, the refracted ray is a ray (drawn perpendicular to the wavefronts) which shows the

direction which light travels after it has crossed over the boundary. In the diagram, a normal

line is drawn to the surface at the point of incidence, This line is always drawn perpendicular to

the boundary. The angle which the incident ray makes with the normal line is referred to as

the angle of incidence. Similarly, the angle which the refracted ray makes with the normal

line is referred to as the angle of refraction. The angle of incidence and angle of refraction are denoted by the following symbols:

= angle of incidence

= angle of refraction

The amount of bending which a light ray experiences can be expressed in terms of the angle of

refraction (more accurately, by the difference between the angle of refraction and the angle of

incidence). A ray of light may approach the boundary at an angle of incidence of 45-degrees

and bend towards the normal. If the medium into which it enters causes a small amount of

refraction, then the angle of refraction might be a value of about 42-degrees. On the other

hand if the medium into which the light enters causes a large amount of refraction, the angle

of refraction might be 22-degrees. (These values are merely arbitrarily chosen values to

illustrate a point.) The diagram below depicts a ray of light approaching three different

boundaries at an angle of incidence of 45-degrees. The refractive medium is different in each

case, causing different amounts of refraction. The angles of refraction are shown on the

diagram.


http://www.physicsclassroom.com/Class/refrn/u14l1e.cfm#terms

Of the three boundaries in the diagram, the light ray refracts the most at the air-

diamond boundary. This is evident by the fact that the difference between the angle of

incidence and the angle of refraction is greatest for the air-diamond boundary. But how

can this be explained? The cause of refraction is a change in light speed; and wherever

the light speed changes most, the refraction is greatest. We have already learned that the

speed is related to the optical density of a material which is related to the index of refraction of

a material. Of the four materials present in the above diagram, air is the least dense material

(lowest index of refraction value) and diamond is the most dense material (largest index of

refraction value). Thus, it would be reasonable that the most refraction occurs for the

transmission of light across an air-diamond boundary.

In this example, the angle of refraction is the measurable quantity which indicates the amount

of refraction taking place at any boundary. A comparison of the angle of refraction to the angle

of incidence provides a good measure of the refractive ability of any given boundary. For any

given angle of incidence, the angle of refraction is dependent upon the speeds of light in each

of the two materials. The speed is in turn dependent upon the optical density and the index of

refraction values of the two materials. There is a mathematical equation relating the angles

which the light rays make with the normal to the indices (plural for index) of refraction of the

two materials on each side of the boundary. This mathematical equation is known as Snell's

Law and is the topic of the next section of Lesson 2.


Snell's Law

Refraction is the bending of the path of a light wave as it passes across the boundary

separating two media. Refraction is caused by the change in speed experienced by a wave

when it changes medium. Lesson 1, focused on the topics of "What causes refraction?" and

"Which direction does light refract?" In that lesson, we learned that light can either

refract towards the normal (when slowing down while crossing the boundary) or away from the

normal (when speeding up while crossing the boundary). The focus of Lesson 2 is upon the

question of "By how much does light refract when it crosses a boundary?" In the first part of

Lesson 2, we learned that a comparison of the angle of refraction to the angle of incidence

provides a good measure of the refractive ability of any given boundary. The more that light

refracts, the bigger the difference between these two angles. In this part of Lesson 2, we will

learn about a mathematical equation relating these two angles and the indices of refraction of

the two materials on each side of the boundary.

To begin, consider a hemi-cylindrical dish filled with water. Suppose that a laser beam is

directed towards the flat side of the dish at the exact center of the dish. The angle of incidence












can be measured at the point of incidence. This ray will refract, bending towards the normal

(since the light is passing from a medium in which it travels fast into one in which it travels

slow - FST). Once the light ray enters the water, it travels in a straight line until it reaches the

second boundary. At the second boundary, the light ray is approaching along the normal to the

curved surface (this stems from the geometry of circles). The ray does not refract upon exiting

since the angle of incidence is 0-degrees (recall the If I Were An Archer Fishpage). The ray of laser light therefore exits at the same angle as the refracted ray of light made at the first boundary. These two angles can be measured and recorded. The angle of incidence of the laser beam can be changed to 5-degrees and new measurements can be made and recorded. This process can be repeated until a complete data set of accurate values has been collected. The data below show a representative set of data for such an experiment.

Angle of Incidence (degrees) Angle of Refraction (degrees)0.00 0.00

5.00 3.8

10.0 7.5

15.0 11.2

20.0 14.9

25.0 18.5

30.0 22.1

35.0 25.5

40.0 28.9

45.0 32.1

50.0 35.2

55.0 38.0

60.0 40.6

65.0 43.0

70.0 45.0

75.0 46.6

80.0 47.8

85.0 48.5

An inspection of the data above reveal that there is no clear linear relationship between the

angle of incidence and the angle of refraction. For example, a doubling of the angle of

incidence from 40 degrees to 80 degrees does not result in a doubling of the angle of

refraction. Thus, a plot of this data would not yield a straight line. If however, the sine of the

angle of incidence and the sine of the angle of refraction were plotted, the plot would be a

straight line, indicating a linear relationship between the sines of the important angles. If two

quantities form a straight line on a graph, then a mathematical relationship can be written in y

= m*x + b form. A plot of the sine of the angle of incidence vs. the sine of the angle of

refraction is shown below.

http://www.physicsclassroom.com/Class/refrn/u14l1f.cfm


The equation relating the angles of incidence ("theta i") and the angle of refraction ("theta r")

for light passing from air into water is given as

Observe that the constant of proportionality in this equation is 1.33 - the index of refraction

value of water. Perhaps it's just a coincidence. But if the semi-cylindrical dish full of water was

replaced by a semi-cylindrical disk of Plexiglas, the constant of proportionality would be 1.51

- the index of refraction value of Plexiglas. This is not just a coincidence. The same pattern

would result for light traveling from air into any material. Experimentally, it is found that for a

ray of light traveling from air into some material, the following equation can be written.

where nmaterial = index of refraction of the material

This study of the refraction of light as it crosses from one material into a second material yields

a general relationship between the sines of the angle of incidence and the angle of refraction.

This general relationship is expressed by the following equation:

where ("theta i") = angle of incidence

("theta r") = angle of refraction

ni = index of refraction of the incident medium

nr = index of refraction of the refractive medium

This relationship between the angles of incidence and refraction and the indices of refraction of

the two medium is known as Snell's Law. Snell's law applies to the refraction of light in any

situation, regardless of what the two media are.

http://www.physicsclassroom.com/Class/refrn/u14l1d.cfm#table




Using Snell's Law to Predict An Angle Value

As with any equation in physics, the Snell's Law equation is valued for its predictive ability. If any three of the four variables in the equation are known, the fourth variable can be predicted if appropriate problem-solving skills are employed. This is illustrated in the two examples below.

Example ProblemsIn the following two examples, use Snell's law, the sine button on your calculator, a protractor, and the index of

refraction values to complete the following diagrams. Measure , calculate , and draw in the refracted ray with the calculated angle of refraction.

In each of these two example problems, the angle of refraction is the variable to be

determined. The indices of refraction (ni and nr) are given and the angle of incidence can be

measured. With three of the four variable known, substitution into Snell's law followed by

algebraic manipulation will lead to the answer.

Solution to Example A

First, use a protractor to measure the angle of incidence. An appropriate measurement would

be some angle close to 45-degrees.

Second, list all known values and the unknown value for which you wish to solve:

Given:

ni = 1.00 nr = 1.33 = 45 degrees

Find:

= ???

Third, list the relevant equation:

Fourth, substitute known values into the equation and algebraically manipulate the equation in

order to solve for the unknown variable - "theta r."

1.00 * sine (45 degrees) = 1.33 * sine (theta r)

0.7071 = 1.33 * sine (theta r)

0.532 = sine (theta r)

sine-1 (0.532) = sine-1 ( sine (theta r))

32.1 degrees = theta r

Proper algebra yields to the answer of 32.1 degrees for the angle of refraction. The diagram

showing the refracted ray can be viewed by clicking the button below.

The solution to Example A is given as an example. Try Example B on your own and click on the

button to check your answer.

Snell's Law provides the quantitative means of answering the question of "By how much does

the light ray refract?" The task of answering this question involves using indices of refraction

and the angle of incidence values in order to determine the angle of refraction. This problem-

solving process is discussed in more detail on the remaining pages of Lesson 2.

Next Section: Ray Tracing and Problem-SolvingJump To Lesson 3: Total Internal Reflection


Ray Tracing and Problem-Solving

In a previous part of Lesson 2, we learned about a mathematical equation relating the two angles (angles of incidence and refraction) and the indices of refraction of the two


http://www.physicsclassroom.com/Class/refrn/U14l3a.cfm

http://www.physicsclassroom.com/Class/refrn/U14l2c.cfm

materials on each side of the boundary. The equation is known as the Snell's Law equation and is expressed as follows.





As with any equation in physics, the Snell's Law equation is valued for its predictive ability. If any three of the four variables in the equation are known, the fourth variable can be predicted if appropriate problem-solving skills are employed. In this part of Lesson 2, we will investigate several of the types of problems that you will have to solve, and learn the task of tracing the refracted ray if given the incident ray and the indices of refraction.

Example Problem AA ray of light in air is approaching the boundary with water at an angle of 52 degrees. Determine the angle of refraction

of the light ray. Refer to the table of indices of refraction if necessary.

Solution to Problem A

The solution to this problem begins like any problem: a diagram is constructed to assist in the visualization of the physical situation, the known values are listed, and the unknown value (desired quantity) is identified. This is shown below:

Diagram: Given:

ni = 1.00 (from table)

nr = 1.333 (from table)

= 52 degrees

Find:

=??

Now list the relevant equation (Snell's Law), substitute known values into the equation, and

perform the proper algebraic steps to solve for the unknown.

1.00 * sine (52 degrees) = 1.333 * sine (theta r)

0.7880 = 1.333 * sine (theta r)







Proper algebra yields the answer of 36.2 degrees for the angle of refraction. When finished, it

is always a wise idea to apply the FST and SFA principles as a check of your numerical answer.

In this problem, the light ray is traveling from a less optically dense or fast medium (air) into

a more optically dense or slow medium (water), and so the light ray should refract towards the

normal - FST. Thus, the angle of refraction should be smaller than the angle of refraction. And

indeed it is - 36.2 degrees (theta r) is smaller than 52.0 degrees (theta i). Using this conceptual

criteria as a check of your answer can often identify incorrect solutions to problems.

Example Problem BA ray of light in air is approaching the boundary with a layer of crown glass at an angle of 42.0 degrees. Determine the angle of refraction of the light ray upon entering the crown glass and upon leaving the crown glass. Refer to the table of

indices of refraction if necessary.

Solution to Problem B

This problem is slightly more complicated than Problem A since refraction is taking place at

two boundaries. This is an example of a layer problem where the light refracts upon entering the layer (boundary #1: air to crown glass) and again upon leaving the layer (boundary #2: crown glass to air). Despite this complication, the solution begins like the above problem: a diagram is constructed to assist in the visualization of the physical situation, the known values are listed, and the unknown value (desired quantity) is identified. This is shown below:

Diagram:

Note that the angle of

refraction at boundary #1

is the same as the angle of

Given:

Boundary #1



= 42.0 degrees

Boundary #2

Find:

at

boundary #1

and

at

boundary #2









incidence at boundary #2. ni = 1.52 (from table)



perform the proper algebraic steps to solve for the unknown. Begin the process at boundary

#1 and then repeat for boundary #2 until the final answer is found.

Boundary #1:

1.00 * sine (42.0 degrees) = 1.52 * sine (theta r)

0.669 = 1.52 * sine (theta r)




The value of 26.1 degrees corresponds to the angle of refraction at boundary #1. Since

boundary #1 is parallel to boundary #2, the angle of refraction at boundary #1 will be the

same as the angle of incidence at boundary #2 (see diagram above). So now repeat the

process in order to solve for the angle of refraction at boundary #2.

Boundary #2:


1.52 * (0.4402) = 1.00 * sine (theta r)




http://www.physicsclassroom.com/Class/refrn/U14l2c.cfm#diagramB%23diagramB




The answers to this problem are 26.1 degrees and 42.0 degrees.

There is an important conceptual idea which is found from an inspection of the above answer.

The ray of light approached the top surface of the layer at 42 degrees and exited through the

bottom surface of the layer with the same angle of 42 degrees. The light ray refracted one

direction upon entering and the other direction upon exiting; the two individual effects have

balanced each other and the ray is moving in the same direction. The important concept is

this:

When light approaches a layer which has the shape of a parallelogram that is bounded on both sides by the same material, then the angle at which the light enters the material is equal to the angle at which light exits the layer.

If the layer is not a parallelogram or is not bound on both sides by the same material, then this will not be the case. Knowing this concept will allow you to conduct a quick check of an answer in a situation like this.

Example Problem C

A ray of light in air is approaches a triangular piece of crown glass at an angle of 0.00 degrees (as

shown in the diagram at the right). Perform the necessary calculations in order to trace the path of

the light ray as it enters and exits the crown glass. Refer to the table of indices of refraction if

necessary.

Solution to Problem C

This problem is even more complicated than Practice Problem B. Like Practice Problem B, there

are two boundaries; but unlike Problem B, the two boundaries are not parallel to each other.

The problem can be treated like a layer problem in which the light refracts upon entering the

glass (boundary #1: air to crown glass) and upon leaving the glass (boundary #2: crown glass

to air).

Despite the complication of there being nonparallel boundaries, the solution begins like the above problem: a diagram is constructed to assist in the visualization of the physical situation, the known values are listed, and the unknown value (desired quantity) is identified. This is shown below:

Diagram:

Given:

Boundary #1

Find:

Trace path of light.




= 0.0 degrees

Boundary #2



That is, find at

boundary #1

and

and at

boundary #2


perform the proper algebraic steps to solve for the unknown. Begin the process at boundary

#1 and then repeat for boundary #2 until the final answer is found.

Boundary #1:


0.000 = 1.52 * sine (theta r)




This problem is made easier if you draw upon your conceptual knowledge of what occurs when

a light ray approaches at an angle of incidence of 0-degrees (recall the If I Were an Archer

Fish page). When approaching along the normal, the light ray passes across the boundary

without refracting. If you did not know this, then you would merely recognize it upon

performing your first calculation of the angle of refraction at the first boundary. The fact that

the answer is 0 degrees - the same as the incident angle - means that light did not refract at

this boundary.







The next step demands that the light ray be traced through the triangular block until it

reaches the second boundary. Draw the refracted ray at 0 degrees (i.e., trace the incident ray

straight through the first boundary). At the second boundary, the normal line must be drawn

(labeled N) and the angle of incidence (between the incident ray and the normal) must be

measured. This is shown on the diagram at the right. The measured value of the angle of

incidence at the second boundary is 30.0 degrees. This angle measurement now provides

knowledge of three of the four variables in the Snell's Law equation and allows for the

determination of the fourth variable (the angle of refraction) at the second boundary.

(Note: the given angle measures for the 30-60-90 degree triangle can be used along with the

fact that any three angles of a triangle add to 180 degrees in order to geometrically determine

this angle measure.)

Boundary #2:


1.52 * (0.5000) = 1.00 * sine (theta r)




The refracted ray at the second boundary will exit at an angle of 49.5 degrees from the

normal. This can be measured on the diagram and drawn with a straight edge as shown in the

diagram at the right.

The above three practice problems demonstrate a sampling of the variety of problems which

could be encountered. In thenext part of Lesson 2, we will see one more type of problem.



1. Determine the angle of refraction for the following two refraction problems.

2. Perform the necessary calculations at each boundary in order to trace the path of the light

ray through the following series of layers. Use a protractor and a ruler and show all your work.


3. A ray of light in crown glass exits into air at an angle of 25.0 degrees. Determine the angle

at which the light approached the glass-air boundary. Refer to the table of indices of

refraction if necessary.

4. A ray of light is traveling through air (n = 1.00) towards a lucite block (n = 1.40) in the

shape of a 30-60-90 triangle. Trace the path of the light ray through the lucite block shown in

the diagram below.

Next Section: Determination of n ValuesJump To Lesson 3: Total Internal Reflection


Determination of n Values

In a previous part of Lesson 2, we learned about a mathematical equation relating the two angles (angles of incidence and refraction) and the indices of refraction of the two materials on each side of the boundary. The equation is known as the Snell's Law equation and is expressed as follows.



http://www.physicsclassroom.com/Class/refrn/U14l2d.cfm







As with any equation in physics, the Snell's Law equation is valued for its predictive ability. If

any three of the four variables in the equation are known, the fourth variable can be predicted

if appropriate problem-solving skills are employed. The task of using the equation to solve

several varieties of problems was thoroughly discussed in a previous part of this Lesson 2. In a

similar manner, the equation can be used to determine the index of refraction of a material if

the path of light through the material is known. In this part of Lesson 2, we will investigate the

details of this task.

A common lab performed in a Physics class involves the determination of the index of

refraction of an unknown material. Typically, a series of transparent objects (glass and lucite

squares, rectangles and triangles) made of an unknown material are distributed and the task is

assigned to determine the index of refraction of each unknown material. Of course, the task of

determining the index of refraction value of an unknown material would be easy if the other

three quantities in the Snell's Law equation are known. Thus, a line of sight method or a laser

beam is used to determine the path of the light into the material, through the material and out

of the material to the students' eyes. The diagram depicts a typical path of one such ray

through a square block. Observe that the light encounters two boundaries - the boundary upon

entering the glass and the boundary upon exiting the glass. The ray of light refracts at each

boundary. By measuring the angles of incidence and refraction and using the index of

refraction of air, the index of refraction of the unknown material can be found. Calculations can

be performed at each boundary and the results can be averaged. The data table below

represent sample data for the Index of Refraction Lab; the listed values correspond to the

diagram at the right.

Sample Datasine sine Calculated n

value

Light Ray 45 deg. 24 deg. 0.7071 0.4067 1.74


Entering

Material

Light Ray

Exiting

Material

24 deg. 45 deg. 0.4067 0.7071 1.74

Ave. n

Value --->

1.74

The accompanying work for this lab are as follows:

First Boundary: Light Ray Entering MaterialDiagram: Known:


= 45 degrees

= 24 degrees

Find:nr = ???

1.00 * sine(45 deg.) = nr * sine(24 deg.)

0.7071 = 0.4067 * nr

(0.7071)/(0.4067) = nr

1.74 = nr

Second Boundary: Light Ray Exiting Material

Diagram: Known:


Find:ni = ???



= 45 degrees

= 24 degrees

ni * sine(24 deg.) = 1.000 * sine(45 deg.)

ni * 0.4067 = 0.7071

ni = (0.7071)/(0.4067)

ni = 1.74

Observe in the data table above the two calculated n values are the same. This should not be a

surprise since the unknown material only has one index of refraction value. If different values

were obtained, it would be the result of some form of measurement error in the retrieval of

data.

This method of determining the index of refraction of an unknown material is one more

application of the Snell's Law equation. Test your ability to use the equation in this manner by

answering the following questions.


1. Cal Culator is performing an experiment to determine the index of refraction of an unknown

material (in the shape of a 45-45-90 triangle). Cal determines that the light follows the path as

shown on the diagram below. Use this path, a protractor, a calculator and Snell's Law to

determine the index of refraction of the unknown material.


2. The path of a light ray through an unknown material is shown in the diagram below. Make

some measurements and determine the index of refraction of the material.

3. Light traveling through air (n = 1.00) is incident upon a triangular block made of an

unknown material. The path of the light through the material is shown in the diagram below.

Using a protractor and a calculator, determine the index of refraction of the unknown material.


4. Light traveling through air (n = 1.00) is incident upon a 60-60-60 triangular block (the

triangle is equilateral; the sides make 60-degree angles with each other) made of an unknown

material. The path of the light through the material is shown in the diagram below. Using a

protractor and a calculator, determine the index of refraction of the unknown material.

Total Internal Reflection

Boundary Behavior Revisited

Earlier in this unit, the boundary behavior of light waves was discussed. It was

mentioned that a light wave doesn't just stop when it reaches the end of the

medium. Rather, the light wave undergoes certain behaviors when it encounters

the end of the medium - such behaviors include reflection,

transmission/refraction, and diffraction. InUnit 13 of The Physics Classroom Tutorial, the

primary focus was the reflective behavior of light waves at the boundary. In this unit, our

primary interest has been the refractive behavior of light waves at the boundary. In Lesson 3,

we will investigate the connection between light reflection and light refraction.

A light wave, like any wave, is an energy-transport phenomenon. A light wave transports

energy from one location to another. When a light wave strikes a boundary between two

distinct media, a portion of the energy will be transmitted into the new medium and a portion

of the energy will be reflected off the boundary and stay within the original medium. The

actual percentage of energy which is transmitted and reflected is dependent upon a number of

variables; these will be discussed as we proceed through Lesson 3. For now, our concern is to

review and internalize the basic concepts and terminology associated with boundary behavior.

Reflection of a light wave involves the bouncing of a light wave off the boundary, while

refraction of a light wave involves the bending of the path of a light wave upon crossing a

boundary and entering a new medium. Both reflection and refraction involve a change in

direction of a wave, but only refraction involves a change in medium.

The diagram at the right shows several wavefronts approaching a boundary between two

media. These wavefronts are referred to as the incident waves and the ray which points in the

direction which they are traveling is referred to as the incident ray. The incident ray is drawn

in blue on the diagram at the right. Notice on the diagram that the incident ray leads into two

other rays at the point of incidence with the boundary. The reflected waves are the waves

which bounce off the boundary and head back upwards and the reflected ray is the ray which

points in the direction which the reflected waves are traveling. The reflected ray is drawn in

green on the diagram at the right. The refracted waves are the waves which are transmitted

across the boundary and continues moving downwards, only at a different angle than before.

The refracted ray is the ray which points in the direction which the refracted waves are

traveling. The refracted ray is drawn in red on the diagram at the right. At the point of

incidence (the point where the incident ray strikes the boundary), a normal line is drawn.

The normal line is always drawn perpendicular to the surface at the point of incidence. The

normal line creates a variety of angles with the light rays; these angles are important and are

given special names. The angle between the incident ray and the normal is the angle of

incidence. The angle between the reflected ray and the normal is the angle of reflection.

And the angle between the refracted ray and the normal is the angle of refraction.



The fundamental law which governs the reflection of light is called the law of reflection.

Whether the light is reflecting off a rough surface or a smooth surface, a curved surface or a

planar surface, the light ray follows the law of reflection. The law of reflection states that

When a light ray reflects off a surface, the angle of incidence is equal to the angle of reflection.

The fundamental law which governs the refraction of light is Snell's Law. Snell's Lawstates

that

When a light ray is transmitted into a new medium, the relationship between the angle of incidence and the angle of refraction is given by the following equation

where the ni and nr values represent the indices of refraction of the incident and the refractive

medium respectively.

As we proceed through this Lesson, we will see that there is a connection between the

reflection and the refraction of light. Each of these two behaviors usually occur together. But

as we will see, there are two conditions, which when both met, will cause the light waves to

undergo reflection without any accompanying refraction.



A common Physics lab is to sight through the long side of an isosceles triangle at a pin or other

object held behind the opposite face. When done so, an unusual observation - a discrepant

event - is observed. The diagram on the left below depicts the physical situation. A ray of light

entered the face of the triangular block at a right angle to the boundary. This ray of light

passes across the boundary without refraction since it was incident along the normal (recall

the If I Were An Archer Fish page). The ray of light then travels in a straight line through the

glass until it reaches the second boundary. Now instead of transmitting across this boundary,

all of the light seems to reflect off the boundary and transmit out the opposite face of the

isosceles triangle. This discrepant event bothers many as they spend several minutes looking

for the light to refract through the second boundary. Then finally, to their amazement, they

looked through the third face of the block and clearly see the ray. What happened? Why did

light not refract through the second face?



http://www.physicsclassroom.com/Class/refln/u13l1c.cfm

The phenomenon observed in this part of the lab is known as total internal reflection.Total

internal reflection, or TIR as it is intimately called, is the reflection of the total amount of

incident light at the boundary between two medium. TIR is the topic of focus in Lesson 3.

To understand total internal reflection, we will begin with a thought experiment. Suppose that

a laser beam is submerged in a tank of water (don't do this at home) and pointed upwards

towards water-air boundary. Then suppose that the angle at which the beam is directed

upwards is slowly altered, beginning with small angles of incidence and proceeding towards

larger and larger angles of incidence. What would be observed in such an experiment? If we

understand the principles of boundary behavior, we would expect that we would observe both

reflection and refraction. And indeed, that is what is observed (mostly). But that's not the only

observation which we could make. We would also observe that the intensity of the reflected

and refracted rays do not remain constant. At angle of incidence close to 0 degrees, most of

the light energy is transmitted across the boundary and very little of it is reflected. As the

angle is increased to greater and greater angles, we would begin to observe less refraction and

more reflection. That is, as the angle of incidence is increased, the brightness of the refracted

ray decreases and the brightness of the reflected ray increases. Finally, we would observe that

the angles of the reflection and refraction are not equal. Since the light waves would refract

away from the normal (a case of the SFA principle of refraction), the angle of refraction would

be greater than the angle of incidence. And if this is the case, the angle of refraction would

also be greater than the angle of reflection (since the angles of reflection and incidence are the

same). As the angle of incidence is increased, the angle of refraction would eventually reach a

90-degree angle. These principles are depicted in the diagram below.


The maximum possible angle of refraction is 90-degrees. If you think about it (a practice which

always helps), you recognize that if the angle of refraction were greater than 90 degrees, then

the refracted ray would lie on the incident side of the medium - that's just not possible. So in

the case of the laser beam in the water, there is some specific value for the angle of incidence

(we'll call it the critical angle) which yields an angle of refraction of 90-degrees. This particular

value for the angle of incidence could be calculated usingSnell's Law (ni = 1.33, nr = 1.000,

= 90 degrees, = ???) and would be found to be 48.6 degrees. Any angle of incidence which

is greater than 48.6 degrees would not result in refraction. Instead, when the angles of

incidence is greater than 48.6 degrees (the critical angle), all of the energy (the total energy)

carried by the incident wave to the boundary stays within the water (internal to the original

medium) and undergoesreflection off the boundary. When this happens, total internal

reflection occurs.

Two Requirements for Total Internal Reflection

Total internal reflection (TIR) is the phenomenon which involves the reflection of all the

incident light off the boundary. TIR only takes place when both of the following two conditions

are met:

the light is in the more dense medium and approaching the less dense medium.

the angle of incidence is greater than the so-called critical angle.

Total internal reflection will not take place unless the incident light is traveling within the

more optically dense medium towards the less optically dense medium. TIR will happen for

light traveling from water towards air, but it will not happen for light traveling from air towards

water. TIR would happen for light traveling from water towards air, but it will not happen for

light traveling from water (n=1.333) towards crown glass (n=1.52). TIR occurs because the




angle of refraction reaches a 90-degree angle before the angle of incidence reaches a 90-

degree angle. The only way for the angle of refraction to be greater than the angle of

incidence, is for light to bend away from the normal. Since light only bends away from the

normal when passing from a more dense medium into a less dense medium, then this would

be a necessary condition for total internal reflection.

Total internal reflection only occurs with large angles of incidence. Question: How large is

large? Answer: larger than the critical angle. As mentioned above, the critical angle for the

water-air boundary is 48.6 degrees. So for angles of incidence greater than 48.6-degrees, TIR

occurs. But 48.6 degrees is the critical angle only for the water-air boundary. The actual value

of the critical angle is dependent upon the two materials on either side of the boundary. For

the crown glass-air boundary, the critical angle is 41.1 degrees. For the diamond-air boundary,

the critical angle is 24.4 degrees. For the diamond-water boundary, the critical angle is 33.4

degrees. The critical angle is different for different media. In the next part of Lesson 3, we will

investigate how to determine the critical angle for any two materials. For now, let's internalize

the idea that TIR can only occur if the angle of incidence is greater than the critical angle for

the particular combination of materials.

Light Piping and Optical Fibers

Total internal reflection is often demonstrated in a Physics class through a variety of

demonstrations. In one such demonstration, a beam of laser light is directed into a coiled

plastic thing-a-ma jig. The plastic served as a light pipe, directing the light through the coils

until it finally exits out the opposite end. Once the light entered the plastic, it was in the more

dense medium. Every time the light approached the plastic-air boundary, it is approaching at

angles greater than the critical angle. The two conditions necessary for TIR are met, and all of

the incident light at the plastic-air boundary stays internal to the plastic and undergoes

reflection. And with the room lights off, every student becomes quickly aware of the ancient

truth that Physics is better than drugs.


This demonstration helps to illustrate the principle by which optical fibers work.

The use of a long strand of plastic (or other material such as glass) to pipe light

from one end of the medium to the other is the basis for modern day use

of optical fibers. Optical fibers are are used in communication systems and micro-

surgeries. Since total internal reflection takes place within the fibers, no incident

energy is ever lost due to the transmission of light across the boundary. The

intensity of the signal remains constant.

Another common Physics demonstration involves the use of a large jug filled with water and a

laser beam. The jug has a pea-sized hole drilled in its side such that when the cork is removed

from the top of the jug, water begins to stream out the jug's side. The beam of laser light is

then directed into the jug from the opposite side of the hole, through the water and into the

falling stream. The laser light exits the jug through the hole but is still in the water. As the

stream of water begins to fall as a projectile along a parabolic path to the ground, the laser

light becomes trapped within the water due to total internal reflection. Being in the more

dense medium (water) and heading towards a boundary with a less dense medium (air), and

being at angles of incidence greater than the critical angle, the light never leaves the stream

of water. In fact, the stream of water acts as a light pipe to pipe the laser beam along its

trajectory. Once more, students viewing the demonstration are convinced of the fact that

Physics is better than drugs.


1. For each combination of media, which light ray (A or B) will undergo total internal reflection

if the incident angle is gradually increased?



The Critical Angle

In the previous part of Lesson 3, the phenomenon of total internal reflection was introduced.

Total internal reflection (TIR) is the phenomenon which involves the reflection of all the

incident light off the boundary. TIR only takes place when both of the following two conditions

are met:

a light ray is in the more dense medium and approaching the less dense medium.

the angle of incidence for the light ray is greater than the so-called critical angle.

In our introduction to TIR, we used the example of light traveling through water towards the

boundary with a less dense material such as air. When the angle of incidence in water reaches

a certain critical value, the refracted ray lies along the boundary, having an angle of refraction

of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of

incidence for which refraction can still occur. For any angle of incidence greater than the

critical angle, light will undergo total internal reflection.

So the critical angle is defined as the angle of incidence which provides an angle of refraction

of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For

the water-air boundary, the critical angle is 48.6-degrees. For the crown glass-water boundary,

the critical angle is 61.0-degrees. The actual value of the critical angle is dependent upon the

combination of materials present on each side of the boundary.

Let's consider two different media - creatively named medium i (incident medium) and medium

r (refractive medium). The critical angle is the which gives a value of 90-degrees. If this

information is substituted into Snell's Law equation, a generic equation for predicting the

critical angle can be derived. The derivation is shown below.

ni *• sine( ) = nr • sine ( )

ni • sine( ) = nr • sine (90 degrees)


ni • sine( ) = nr

sine( ) = nr/ni

= sine-1 (nr/ni) = invsine (nr/ni)

The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of

refraction. The ratio of nr/ni is a value less than 1.0. In fact, for the equation to even give a

correct answer, the ratio of nr/ni must be less than 1.0. Since TIR only occurs if the refractive

medium is less dense than the incident medium, the value of ni must be greater than the value

of nr. If at any time the values for the numerator and denominator become accidentally

switched, the critical angle value cannot be calculated. Mathematically, this would involve

finding the inverse-sine of a number greater than 1.00 - which is not possible. Physically, this

would involve finding the critical angle for a situation in which the light is traveling from the

less dense medium into the more dense medium - which again, is not possible.

This equation for the critical angle can be used to predict the critical angle for any boundary,

provided that the indices of refraction of the two materials on each side of the boundary are

known. Examples of its use are shown below:

Example ACalculate the critical angle for the crown glass-air boundary. Refer to the table of indices of refraction if necessary.

The solution to the problem involves the use of the above equation for the critical angle.

= sin-1 (nr/ni) = invsine (nr/ni)

= sin-1 (1.000/1.52) = 41.1 degrees

Example BCalculate the critical angle for the diamond-air boundary. Refer to the table of indices of refraction if necessary.

The solution to the problem involves the use of the above equation for the critical angle.

= sin-1 (nr/ni) = invsine (nr/ni)

= sin-1 (1.000/2.42) = 24.4 degrees

TIR and the Sparkle of Diamonds



Relatively speaking, the critical angle for the diamond-air boundary is an extremely small

number. Of all the possible combinations of materials which could interface to form a

boundary, the combination of diamond and air provides one of the largest difference in the

index of refraction values. This means that there will be a very small nr/ni ratio and

subsequently a small critical angle. This peculiarity about the diamond-air boundary plays an

important role in the brilliance of a diamond gemstone. Having a small critical angle, light has

the tendency to become "trapped" inside of a diamond once it enters. A light ray will typically

undergo TIR several times before finally refracting out of the diamond. Because the diamond-

air boundary has such a small critical angle (due to diamond's large index of refraction), most

rays approach the diamond at angles of incidence greater than the critical angle. This gives

diamond a tendency to sparkle. The effect can be enhanced by the cutting of a diamond

gemstone with a strategically planned shape. The diagram below depicts the total internal

reflection within a diamond gemstone with a strategic and a non-strategic cut.


1. Suppose that the angle of incidence of a laser beam in water and heading towards air is

adjusted to 50-degrees. Use Snell's law to calculate the angle of refraction? Explain your result

(or lack of result).

2. Aaron Agin is trying to determine the critical angle of the diamond-glass

surface. He looks up the index of refraction values of diamond (2.42) and crown

glass (1.52) and then tries to compute the critical angle by taking the

invsine(2.42/1.52).

Unfortunately, Aaron's calculator keeps telling him he has an ERROR! Aaron hits

the calculator and throws it own the ground a few times; he then repeats the calculation with

the same result. He then utters something strange about the pizza he had slopped on it the

evening before and runs out of the library with a disappointed disposition. What is Aaron's

problem? (That is, what is the problem with his method of calculating the critical angle?)

3. Calculate the critical angle for an ethanol-air boundary. Refer to the table of indices of

refraction if necessary.

4. Calculate the critical angle for an flint glass(light)-air boundary. Refer to the table of indices

of refraction if necessary.

5. Calculate the critical angle for a diamond-crown glass boundary. Refer to the table of indices

of refraction if necessary.

6. Some optical instruments, such as periscopes and binoculars use trigonal prisms instead of

mirrors to reflect light around corners. Light typically enters perpendicular to the face of the

prism, undergoes TIR off the opposite face and then exits out the third face. Why do you

suppose the manufacturer prefers the use of prisms instead of mirrors?








Interesting Refraction Phenomena

Dispersion of Light by Prisms

In the Light and Color unit of The Physics Classroom Tutorial, the visible light spectrum was

introduced and discussed. Visible light, also known as white light, consists of a collection of

component colors. These colors are often observed as light passes through a triangular prism.

Upon passage through the prism, the white light is separated into its component colors - red,

orange, yellow, green, blue and violet. The separation of visible light into its different colors is

known as dispersion. It was mentioned in the Light and Color unit that each color is

characteristic of a distinct wave frequency; and different frequencies of light waves will bend

varying amounts upon passage through a prism. In this unit, we will investigate the dispersion

of light in more detail, pondering the reasons why different frequencies of light bend or refract

different amounts when passing through the prism.

Earlier in this unit, the concept of optical density was introduced. Different materials are

distinguished from each other by their different optical densities. The optical density is simply

a measure of the tendency of a material to slow down light as it travels through it. As

mentioned earlier, a light wave traveling through a transparent material interacts with the

atoms of that material. When a light wave impinges upon an atom of the material, it is

absorbed by that atom. The absorbed energy causes the electrons in the atom to vibrate. If the

frequency of the light wave does not match the resonance frequency of the vibrating

electrons, then the light will be reemitted by the atom at the same frequency at which it

impinged upon it. The light wave then travels through the interatomic vacuum towards the

next atom of the material. Once it impinges upon the next atom, the process of absorption and

reemission is repeated.

The optical density of a material is the result of the tendency of the atoms of a material to

maintain the absorbed energy of the light wave in the form of vibrating electrons before

reemitting it as a new electromagnetic disturbance. Thus, while a light wave travels through a

vacuum at a speed of c (3.00 x 108 m/s), it travels through a transparent material at speeds

less than c. The index of refraction value (n) provides a quantitative expression of the optical

density of a given medium. Materials with higher index of refraction values have a tendency

to hold onto the absorbed light energy for greater lengths of time before reemitting it to the

interatomic void. The more closely that the frequency of the light wave matches the resonant



http://www.physicsclassroom.com/Class/light/

frequency of the electrons of the atoms of a material, the greater the optical density and the

greater the index of refraction. A light wave would be slowed down to a greater extent when

passing through such a material

What was not mentioned earlier in this unit is that the index of refraction values are dependent

upon the frequency of light. For visible light, the n value does not show a large variation with

frequency, but nonetheless it shows a variation. For instance, the nvalue for frequencies of

violet light is 1.53; and the n value for frequencies of red light is 1.51. The absorption and

reemission process causes the higher frequency (lower wavelength) violet light to travel slower

through crown glass than the lower frequency (higher wavelength) red light. It is this

difference in n value for the varying frequencies(and wavelengths) which causes the dispersion

of light by a triangular prism. Violet light, being slowed down to a greater extent by the

absorption and reemission process, refracts more than red light. Upon entry of white light at

the first boundary of a triangular prism, there will be a slight separation of the white light into

the component colors of the spectrum. Upon exiting the triangular prism at the second

boundary, the separation becomes even greater andROYGBIV is observed in all its splendor.

The Angle of Deviation

The amount of overall refraction caused by the passage of a light ray through a prism is often

expressed in terms of the angle of deviation ( ). The angle of deviation is the angle made

between the incident ray of light entering the first face of the prism and the refracted ray

which emerges from the second face of the prism. Because of the different indices of refraction

for the different wavelengths of visible light, the angle of deviation varies with wavelength.

Colors of the visible light spectrum which have shorter wavelengths (BIV) will deviated more

from their original path than the colors with longer wavelengths (ROY). The emergence of

different colors of light from a triangular prism at different angles leads an observer to see the

component colors of visible light separated from each other.

Of course the discussion of the dispersion of light by triangular prisms begs the following

question: Why doesn't a square or rectangular prism cause the dispersion of a narrow beam of

white light? The short answer is that it does. The long answer is provided in the following

discussion and illustrated by the diagram below.

Suppose that a flashlight could be covered with black paper with a slit across it so as to create

a beam of white light. And suppose that the beam of white light with its component colors

unseparated were directed at an angle towards the surface of a rectangular glass prism. As

would be expected, the light would refract towards the normal upon entering the glass and

away from the normal upon exiting the glass. But since the violet light has a shorter

wavelength, it would refract more than the longer wavelength red light. The refraction of light

at the entry location into the rectangular glass prism would cause a little separation of the

white light. However, upon exiting the glass prism, the refraction takes place in the opposite

direction. The light refracts away from the normal, with the violet light bending a bit more than

the red light. Unlike the passage through the triangular prism with non-parallel sides, there is

no overall angle of deviation for the various colors of white light. Both the red and the violet

components of light are traveling in the same direction as they were traveling before entry into

the prism. There is however a thin red fringe present on one end of the beam and thin violet

fringe present on the opposite side of the beam. This fringe is evidence of dispersion. Because

there is a different angle of deviation of the various components of white light after

transmission across the first boundary, the violet is separated ever so slightly from the red.

Upon transmission across the second boundary, the direction of refraction is reversed; yet

because the violet light has traveled further downward when passing through the rectangle it

is the primary color present in the lower edge of the beam. The same can be said for red light

on the upper edge of the beam.

Dispersion of light provides evidence for the existence of a spectrum of wavelengths present in

visible light. It is also the basis for understanding the formation of rainbows.Rainbow

formation is the next topic of discussion in Lesson 4.




Rainbow Formation

One of nature's most splendid masterpieces is the rainbow. A rainbow is

an excellent demonstration of the dispersion of light and one more piece

of evidence that visible light is composed of a spectrum of wavelengths,

each associated with a distinct color. To view a rainbow, your back must be to the sun as you

look at an approximately 40 degree angle above the ground into a region of the atmosphere

with suspended droplets of water or even a light mist. Each individual droplet of water acts as

a tiny prism which both disperses the light and reflects it back to your eye. As you sight into

the sky, wavelengths of light associated with a specific color arrive at your eye from the

collection of droplets. The net effect of the vast array of droplets is that a circular arc

of ROYGBIV is seen across the sky. But just exactly how do the droplets of water disperse and

reflect the light? And why does the pattern always appear as ROYGBIV from top to bottom?

These are the questions which we will seek to understand on this page of The Physics

Classroom Tutorial. To understand these questions, we will need to draw upon our

understanding of refraction, internal reflection and dispersion.

The Path of Light Through a Droplet

A collection of suspended water droplets in the atmosphere serve as a refractor of light. The

water represents a medium with a different optical density than the surrounding air. Light

waves refract when they cross over the boundary from one medium to another. The decrease

in speed upon entry of light into a water droplet causes a bending of the path of light towards

the normal. And upon exiting the droplet, light speeds up and bends away from the normal.

The droplet causes a deviation in the path of light as it enters and exits the drop.

There are countless paths by which light rays from the sun can pass through a drop. Each path

is characterized by this bending towards and away from the normal. One path of great

significance in the discussion of rainbows is the path in which light refracts into the droplet,

internally reflects, and then refracts out of the droplet. The diagram at the right depicts such a

path. A light ray from the sun enters the droplet with a slight downward trajectory. Upon

refracting twice and reflecting once, the light ray is dispersed and bent downward towards an

observer on earth's surface. Other entry locations into the droplet may result in similar paths

or even in light continuing through the droplet and out the opposite side without significant

internal reflection. But for the entry location shown in the diagram at the right, there is an

optimal concentration of light exiting the airborne droplet at an angle towards the ground. As

in the case of the refraction of light through prisms with nonparallel sides, the refraction of

light at two boundaries of the droplet results in the dispersion of light into a spectrum of colors.

http://www.physicsclassroom.com/Class/refrn/u14l4a.cfm#p5





http://www.physicsclassroom.com/Class/light/u12l2a.cfm#visspec


The shorter wavelength blue and violet light refract a slightly greater

amount than the longer wavelength red light. Since the boundaries are

not parallel to each other, the double refraction results in a distinct

separation of the sunlight into its component colors.

The angle of deviation between the incoming light rays from the sun and the refracted rays

directed to the observer's eyes is approximately 42 degrees for the red light. Because of the

tendency of shorter wavelength blue light to refract more than red light, its angle of deviation

from the original sun rays is approximately 40 degrees. As shown in the diagram, the red light

refracts out of the droplet at a steeper angle toward an observer on the ground. There are a

multitude of paths by which the original ray can pass through a droplet and and subsequently

angle towards the ground. Some of the paths are dependent upon which part of the droplet the

incident rays contact. Other paths are dependent upon the location of the sun in the sky and

the subsequent trajectory of the incoming rays towards the droplet. Yet the greatest

concentration of outgoing rays is found at these 40-42 degree angles of deviation. At these

angles, the dispersed light is bright enough to result in a rainbow display in the sky. Now that

we understand the path of light through an individual droplet, we can approach the topic of

how the rainbow forms.

The Formation of the Rainbow

A rainbow is most often viewed as a circular arc in the sky. An observer on the ground

observes a half-circle of color with red being the color perceived on the outside or top of the

bow. Those who are fortunate enough to have seen a rainbow from an airplane in the sky may

know that a rainbow can actually be a complete circle. Observers on the ground only view the

top half of the circle since the bottom half of the circular arc is prevented by the presence of

the ground (and the rather obvious fact that suspended water droplets aren't present below

ground). Yet observers in an airborne plane can often look both upward and downward to view

the complete circular bow.

The circle (or half-circle) results because there are a collection of suspended droplets in the

atmosphere which are capable concentrating the dispersed light at angles of deviation of 40-

42 degrees relative to the original path of light from the sun. These droplets actually form a

circular arc, with each droplet within the arc dispersing light and reflecting it back towards the

observer. Every droplet within the arc is refracting and dispersing the entire visible light

spectrum (ROYGBIV). As described above, the red light is refracted out of a droplet at steeper

angles towards the ground than the blue light. Thus, when an observer sights at a steeper

angle with respect to the ground, droplets of water within this line of sight are refracting the

red light to the observer's eye. The blue light from these same droplets is directed at a less

http://www.physicsclassroom.com/Class/refrn/U14l4b.cfm#p4%23p4


steep angle and is directed along a trajectory which passes over the observer's head. Thus, it

is the red light which is seen when looking at the steeper angles relative to the ground.

Similarly, when sighting at less steep angles, droplets of water within this line of sight are

directing blue light to the observer's eye while the red light is directed downwards at a more

steep angle towards the observer's feet. This discussion explains why it is the red light which is

observed at the top and on the outer perimeter of a rainbow and the blue light which is

observed on the bottom and the inner perimeter of the rainbow.

Rainbows are not limited to the dispersion of light by raindrops. The splashing of water at the

base of a waterfall caused a mist of water in the air which often results in the formation of

rainbows. A backyard water sprinkler is another common source of a rainbow. Bright sunlight,

suspended droplets of water and the proper angle of sighting are the three necessary

components for viewing one of nature's most splendid masterpieces.


Mirages

Most of our discussion of refraction in this unit has pertained to the refraction of light at a

distinct boundary. As light is transmitted across the boundary from one material to another,

there is a change in speed which causes a change in direction of the light wave. The

boundaries which we have been focusing on have been distinct interfaces between two

recognizably different materials. The boundary between the glass of a fish tank and the

surrounding air or the boundary between the water in a pool and the surrounding air are

examples of distinct interfaces between two recognizably different materials.



It has been mentioned in our discussion that the refraction or bending of light occurs at the

boundary between two materials; and once a light wave has crossed the boundary it travels in

a straight line. The discussion has presumed that the medium is a uniformmedium. A uniform

medium is a medium whose optical density is everywhere the same within the medium. A

uniform medium is the same everywhere from its top boundary to its bottom boundary and

from its left boundary to its right boundary. But not every medium is a uniform medium, and

the fact that air can sometimes form a nonuniform medium leads to an interesting refraction

phenomenon - the formation of mirages.

A mirage is an optical phenomenon which creates the illusion of water and results from the

refraction of light through a nonuniform medium. Mirages are most commonly observed on

sunny days when driving down a roadway. As you drive down the roadway, there appears to

be a puddle of water on the road several yards (maybe one-hundred yards) in front of the car.

Of course, when you arrive at the perceived location of the puddle, you recognize that the

puddle is not there. Instead, the puddle of water appears to be another one-hundred yards in

front of you. You could carefully match the perceived location of the water to a roadside

object; but when you arrive at that object, the puddle of water is still not on the roadway. The

appearance of the water is simply an illusion.

Mirages occur on sunny days. The role of the sun is to heat the roadway to high temperatures.

This heated roadway in turn heats the surrounding air, keeping the air just above the roadway

at higher temperatures than that day's average air temperature. Hot air tends to be less

optically dense than cooler air. As such, a nonuniform medium has been created by the

heating of the roadway and the air just above it. While light will travel in a straight line through

a uniform medium, it will refract when traveling through a nonuniform medium. If a driver

looks down at the roadway at a very low angle (that is, at a position nearly one-hundred yards

away), light from objects above the roadway will follow a curved path to the driver's eye as

shown in the diagram below.

Light which is traveling downward into this less optically dense air begins to speed up. Though

there isn't a distinct boundary between two media, there is a change in speed of a light wave.

As expected, a change in speed is accompanied by a change in direction. If there were a

distinct boundary between two media, then there would be a bending of this light ray away

from the normal. For this light ray to bend away from the normal(towards the boundary), the

ray would begin to bend more parallel to the roadway and then bend upwards towards the



cooler air. As such, a person in a car sighting downward at the roadway will see an object

located above the roadway.

Of course, this is not a usual event. When was the last time that you looked downward at a

surface and saw an object above the surface? While not a usual event, it does happen. For

instance, suppose you place a mirror on the floor and look downward at the floor; you will see

objects located above the floor due to the reflection of light by the mirror. Even a glass window

placed on the floor will reflect light from objects above the floor. If you look downward at the

glass window at a low enough angle, then you will see objects located above the floor. Or

suppose that you are standing on the shore of a calm pond and look downward at the water;

you might see objects above the pond due to the reflection of light by the water.

So when you experience this sunny day phenomenon, your mind must quickly make sense of

how you can look downward at the roadway and see an object located above the road. In the

process of making sense of this event, your mind draws upon past experiences. Searching the

database of stored experiences, your mind is interested in an explanation of why the eye can

sight downward at a surface and see an object which is located above the surface. In the

process of searching, it comes up with three possible explanations based upon past

experiences. Your mind subtly ponders these three options.

There is a mirror on the road. Someone must have for some reason placed a mirror on the road. The mirror is reflecting light and that is why I see an image of the oncoming truck when I look downward at the road.

There is a glass window on the road. My gosh, do you believe it! Someone has left a glass window on the road. The glass window is reflecting light and that is why I see an image of the oncoming truck when I look downward at the road.

There is water on the road. It must have rained last night and there is a puddle of water left on the road. The water is reflecting light and that is why I see an image of the oncoming truck when I look downward at the road.

Of the three possible explanations of the image of the truck, only one makes a lot of sense to

the mind - there is water on the road. After all, while both glass windows and mirrors can

reflect light, nowhere in your mind's database of past experiences is there an account of a

mirror or glass window being seen on a roadway. Yet there are plenty of times that a water

puddle has been observed to be present on a roadway. Smart person that you are, you then

conclude that there is a puddle of water on the road which is causing you to see objects

located above the road when you sight downward at the road. The illusion is complete.

Jump To Lesson 5: Image Formation by Lenses

Image Formation by Lenses

The Anatomy of a Lens

If a piece of glass or other transparent material takes on the appropriate shape, it is possible

that parallel incident rays would either converge to a point or appear to be diverging from a

point. A piece of glass which has such a shape is referred to as a lens.

A lens is merely a carefully ground or molded piece of transparent material which refracts

light rays in such as way as to form an image. Lenses can be thought of as a series of tiny

refracting prisms, each of which refracts light to produce their own image. When these prisms

act together, they produce a bright image focused at a point.

There are a variety of types of lenses. Lenses differ from one another in terms of their shape

and the materials from which they are made. Our focus will be upon lenses which are

symmetrical across their horizontal axis - known as the principal axis. In this unit, we will

categorize lenses as converging lenses and diverging lenses. A converging lensis a lens

which converges rays of light which are traveling parallel to its principal axis. Converging

lenses can be identified by their shape; they are relatively thick across their middle and thin at

their upper and lower edges. A diverging lens is a lens which diverges rays of light which are

traveling parallel to its principal axis. Diverging lenses can also be identified by their shape;

they are relatively thin across their middle and thick at their upper and lower edges.


A double convex lens is symmetrical across both its horizontal and vertical axis. Each of the

lens' two faces can be thought of as originally being part of a sphere. The fact that a double

convex lens is thicker across its middle is an indicator that it will converge rays of light which

travel parallel to its principal axis. A double convex lens is a converging lens. A double

concave lens is also symmetrical across both its horizontal and vertical axis. The two faces of

a double concave lens can be thought of as originally being part of a sphere. The fact that a

double concave lens is thinner across its middle is an indicator that it will diverge rays of light

which travel parallel to its principal axis. A double concave lens is a diverging lens. These two

types of lenses - a double convex and a double concave lens will be the only types of lenses

which will be discussed in this unit of The Physics Classroom Tutorial.

As we begin to discuss the refraction of light rays and the formation of images by these two

types of lenses, we will need to use a variety of terms. Many of these terms should be familiar

to you because they have already been discussed during Unit 13. If you are uncertain of the meaning of the terms, spend some time reviewing them so that their meaning is firmly internalized in your mind. They will be essential as we proceed through Lesson 5. These terms describe the various parts of a lens and include such words as

Principal axis Vertical PlaneFocal Point Focal Length

If a symmetrical lens is thought of as being a slice of a sphere, then there would be a line

passing through the center of the sphere and attaching to the mirror in the exact center of the

lens. This imaginary line is known as the principal axis. A lens also has an imaginary vertical

axis which bisects the symmetrical lens into halves. As mentioned above, light rays incident

towards either face of the lens and traveling parallel to the principal axis will either converge

or diverge. If the light rays converge (as in a converging lens), then they will converge to a

point. This point is known as the focal point of the converging lens. If the light rays diverge

(as in a diverging lens), then the diverging rays can be traced backwards until they intersect at

http://www.physicsclassroom.com/Class/refln/u13l3a.cfm#vocab

http://www.physicsclassroom.com/Class/index.cfm

a point. This intersection point is known as the focal point of a diverging lens. The focal point

is denoted by the letter F on the diagrams below. Note that each lens has two focal points -

one on each side of the lens. Unlike mirrors, lenses can allow light to pass through either face,

depending on where the incident rays are coming from. Subsequently, every lens has two

possible focal points. The distance from the mirror to the focal point is known as thefocal

length (abbreviated by f). Technically, a lens does not have a center of curvature (at least not

one which has any importance to our discussion). However a lens does have an imaginary

point which we refer to as the 2F point. This is the point on the principal axis which is twice as

far from the vertical axis as the focal point is.

As we discuss the characteristics of images produced by converging and diverging lenses,

these vocabulary terms will become increasingly important. Remember that this page is here

and refer to it as often as needed.


Refraction by Lenses

We have already learned that a lens is a carefully ground or molded piece of transparent

material which refracts light rays in such as way as to form an image. Lenses serve to refract

light at each boundary. As a ray of light enters a lens, it is refracted; and as the same ray of

light exits the lens, it is refracted again. The net effect of the refraction of light at these two

boundaries is that the light ray has changed directions. Because of the special geometric

shape of a lens, the light rays are refracted such that they form images. Before we

approach the topic of image formation, we will investigate the refractive ability of converging

and diverging lenses.

First lets consider a double convex lens. Suppose that several rays of light approach the lens;

and suppose that these rays of light are traveling parallel to the principal axis. Upon reaching

the front face of the lens, each ray of light will refract towards the normal to the surface. At

http://www.physicsclassroom.com/Class/refrn/u14l5a.cfm#vocab

http://www.physicsclassroom.com/Class/refrn/u14l5a.cfm#types



this boundary, the light ray is passing from air into a more dense medium (usually plastic or

glass). Since the light ray is passing from a medium in which it travels fast (less optically

dense) into a medium in which it travels relatively slow (moreoptically dense), it will bend

towards the normal line. This is the FST principle of refraction. This is shown for two incident

rays on the diagram below. Once the light ray refracts across the boundary and enters the

lens, it travels in a straight line until it reaches the back face of the lens. At this boundary,

each ray of light will refract away from the normal to the surface. Since the light ray is passing

from a medium in which it travels slow (more optically dense) to a medium in which it travels

fast (less optically dense), it will bend away from the normal line; this is the SFA principle of

refraction.

The above diagram shows the behavior of two incident rays approaching parallel to the principal axis. Note that the two rays converge at a point; this point is known as the focal point of the lens. The first generalization which can be made for the refraction of light by a double convex lens is as follows:

Refraction Rule for a Converging Lens

Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel

through the focal point on the opposite side of the lens.

Now suppose that the rays of light are traveling through the focal point on the way to the lens.

These rays of light will refract when they enter the lens and refract when they leave the lens.

As the light rays enter into the more dense lens material, they refract towards the normal; and

as they exit into the less dense air, they refract away from the normal. These specific rays will

exit the lens traveling parallel to the principal axis.









The above diagram shows the behavior of two incident rays traveling through the focal point on the way to the lens. Note that the two rays refract parallel to the principal axis. A second generalization for the refraction of light by a double convex lens can be added to the first generalization.

Refraction Rules for a Converging Lens Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.

Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.

The Thin Lens Approximation

These two "rules" will greatly simplify the task of determining the image location for objects

placed in front of converging lenses. This topic will be discussed in the next part of Lesson 5.

For now, internalize the meaning of the rules and be prepared to use them. As the rules are

applied in the construction of ray diagrams, do not forget the fact that Snells' Law of refraction

of light holds for each of these rays. It just so happens that geometrically, when Snell's Law is

applied for rays which strike the lens in the manner described above, they will refract in close

approximation with these two rules. The tendency of incident light rays to follow these rules is

increased for lenses which are thin. For such thin lenses, the path of the light through the lens

itself contributes very little to the overall change in the direction of the light rays. We will use

this so-called thin-lens approximation in this unit. Furthermore, to simplify the construction of

ray diagrams, we will avoid refracting each light ray twice - upon entering and emerging from

the lens. Instead, we will continue the incident ray to the vertical axis of the lens and refract

the light at that point. For thin lenses, this simplification will produce the same result as if we

were refracting the light twice.



Rules of Refraction for Diverging Lenses

Now let's investigate the refraction of light by double concave lens. Suppose

that several rays of light approach the lens; and suppose that these rays of

light are traveling parallel to the principal axis. Upon reaching the front face of

the lens, each ray of light will refract towards the normal to the surface. At this boundary, the

light ray is passing from air into a more dense medium (usually plastic or glass). Since the light

ray is passing from a medium in which it travels relatively fast (less optically dense) into a

medium in which it travels relatively slow (more optically dense), it will bend towards the

normal line. This is the FST principle of refraction. This is shown for two incident rays on the

diagram below. Once the light ray refracts across the boundary and enters the lens, it travels

in a straight line until it reaches the back face of the lens. At this boundary, each ray of light

will refract away from the normal to the surface. Since the light ray is passing from a medium

in which it travels relatively slow (more optically dense) to a medium in which it travels fast

(less optically dense), it will bend away from the normal line. This is the SFA principle of

refraction. These principles of refraction are identical to what was observed for the double

convex lens above.

The above diagram shows the behavior of two incident rays approaching parallel to the

principal axis of the double concave lens. Just like the the double convex lens above, light

bends towards the normal when entering and away from the normal when exiting the lens. Yet,

because of the different shape of the double concave lens, these incident rays are not

converged to a point upon refraction through the lens. Rather, these incident rays diverge

upon refracting through the lens. For this reason, a double concave lens can never produce a

real image. Double concave lenses produce images which are virtual. This will be discussed in

http://www.physicsclassroom.com/Class/refrn/U14l5b.cfm#doubconv%23doubconv











more detail in the next part of Lesson 5. If the refracted rays are extended backwards behind

the lens, an important observation is made. The extension of the refracted rays will intersect at

a point. This point is known as the focal point. Notice that a diverging lens such as this double

concave lens does not really focus the incident light rays which are parallel to the principal

axis; rather, it diverges these light rays. For this reason, a diverging lens is said to have a

negative focal length.

The first generalization can now be made for the refraction of light by a double concave lens:

Refraction Rule for a Diverging Lens

Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in

line with the focal point (i.e., in a direction such that its extension will pass through the focal point).

Now suppose that the rays of light are traveling towards the focal point on the way to the lens.

Because of the negative focal length for double concave lenses, the light rays will head

towards the focal point on the opposite side of the lens. These rays will actually reach the lens

before they reach the focal point. These rays of light will refract when they enter the lens and

refract when they leave the lens. As the light rays enter into the more dense lens material,

they refract towards the normal; and as they exit into the less dense air, they refract away

from the normal. These specific rays will exit the lens traveling parallel to the principal axis.

The above diagram shows the behavior of two incident rays traveling towards the focal point on the way to the lens. Note that the two rays refract parallel to the principal axis. A second generalization for the refraction of light by a double concave lens can be added to the first generalization.


Refraction Rules for a Diverging Lens Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (i.e., in a direction such that its extension will pass through the focal point).

Any incident ray traveling towards the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.

A Third Rule of Refraction for Lenses

The above discussion focuses on the manner in which converging and diverging lenses refract

incident rays which are traveling parallel to the principal axis or are traveling through (or

towards) the focal point. But these are not the only two possible incident rays. There are a

multitude of incident rays which strike the lens and refract in a variety of ways. Yet, there are

three specific rays which behave in a very predictable manner. The third ray which we will

investigate is the ray which passes through the precise center of the lens - through the point

where the principal axis and the vertical axis intersect. This ray will refract as it enters and

refract as it exits the lens, but the net affect of this dual refraction is that the path of the light

ray is not changed. For a thin lens, the refracted ray is traveling in the same direction as the

incident ray and is approximately in line with it. The behavior of this third incident ray is

depicted in the diagram below.

Now we have three incident rays whose refractive behavior is easily predicted. These three

rays lead to our three rules of refraction for converging and diverging lenses. These three rules are summarized below.

Refraction Rules for a Converging Lens Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.


http://www.physicsclassroom.com/Class/refrn/U14l5b.cfm#thinlens%23thinlens

An incident ray which passes through the center of the lens will in affect continue in the same direction that it had when it entered the lens.

Refraction Rules for a Diverging Lens Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (i.e., in a direction such that its extension will pass through the focal point).



These three rules of refraction for converging and diverging lenses will be applied through the

remainder of this lesson. The rules merely describe the behavior of three specific incident rays.

While there are a multitude of light rays being captured and refracted by a lens, only two rays

are needed in order to determine the image location. So as we proceed with this lesson, pick

your favorite two rules (usually, the ones which are easiest to remember) and apply them to

the construction of ray diagrams and the determination of the image location and

characteristics.

Next Section: Image Formation RevisitedJump To Lesson 6: The Eye


Image Formation Revisited

One major principle discussed in both Unit 13 and Unit 14 of The Physics Classroom Tutorial is

the line of sight principle:

In order to view an object, you must sight along a line at that object; and when you do light will come from that object to your eye along the line of sight.

This very principle is combined with rules of reflection and refraction in order to explain how an

image is formed and what the characteristics of such images will be.

Plane Mirror Image Formation

In the plane mirror Lesson of Unit 13, it was mentioned that an image is formed by a plane

mirror as light emanates from an object in a variety of directions. Some of this light reaches

the mirror and reflects off the mirror according to the law of reflection. Each one of these rays

of light can be extended backwards behind the mirror where they will all intersect at a point

(the image point). Any person who is positioned along the line of one of these reflected rays

can sight along the line and view the image - a representation of the object. Thus,

an image location is a location in space where all the reflected light appears to come from.


http://www.physicsclassroom.com/Class/refln/u13l1b.cfm


http://www.physicsclassroom.com/Class/refrn/U14l5c.cfm

Since light from the object appears to diverge from this location, a person who sights along a

line at this location will perceive a replica or likeness of the actual object. In the case of plane

mirrors, the image is said to be avirtual image. Virtual images are images which are formed

in locations where light does not actually reach. Light does not actually pass through the

location on the other side of the plane mirror; it only appears to an observer as though the

light were coming from this position.

Curved Mirror Image Formation

We have also seen how images are created by the reflection of light off curved mirrors.

Suppose that a light bulb is placed in front of a concave mirror; the light bulb will emit light in a

variety of directions, some of which will strike the mirror. Each individual ray of light will reflect

according to the law of reflection. Upon reflecting, the light will converge at a point. At the

point where the light from the object converges, a replica or likeness of the actual object is

created; this replica is known as the image. Once the reflected light rays reached the image

location, they begin to diverge. The point where all the reflected light rays converge is known

as the image point. Not only is it the point where light rays converge, it is also the point where

reflected light rays appear to an observer to be coming from. Regardless of the observer's

location, the observer will see a ray of light passing through the real image location. To view

the image, the observer must line her sight up with the image location in order to see the

image via the reflected light ray. The diagram below depicts several rays from the object

reflecting from the mirror and converging at the image location. The reflected light rays then

begin to diverge, with each one being capable of assisting an individual in viewing the image

of the object.

For plane mirrors, virtual images are formed. Light does not actually pass through the virtual

image location; it only appears to an observer as though the light was emanating from the

virtual image location. The image formed by this concave mirror is a real image. When a real

http://www.physicsclassroom.com/Class/refln/u13l2b.cfm#virtual


image is formed, it still appears to an observer as though light is diverging from the real image

location. Only in the case of a real image, light is actually passing through the image location.

Converging Lens Image Formation

Converging lenses can produce both real and virtual images while diverging images can only

produce virtual images. The process by which images are formed for lenses is the same as the

process by which images are formed for plane and curved mirrors. Images are formed at

locations where any observer is sighting as they view the image of the object through the lens.

So if the path of several light rays through a lens is traced, each of these light rays will

intersect at a point upon refraction through the lens. Each observer must sight in the direction

of this point in order to view the image of the object. While different observers will sight along

different lines of sight, each line of sight intersects at the image location. The diagram below

shows several incident rays emanating from an object - a light bulb. Three of these incident

rays correspond to ourthree strategic and predictable light rays. Each incident ray will refract

through the lens and be detected by a different observer (represented by the eyes). The

location where the refracted rays are intersecting is the image location.

In this case, the image is a real image since the light rays are actually passing through the

image location. To each observer, it appears as though light is coming from this location.

Diverging Lens Image Formation

Diverging lens create virtual images since the refracted rays do not actually converge to a

point. In the case of a diverging lens, the image location is located on the object's side of the

http://www.physicsclassroom.com/Class/refrn/u14l5b.cfm#3rules

lens where the refracted rays would intersect if extended backwards. Every observer would be

sighting along a line in the direction of this image location in order to see the image of the

object. As the observer sights along this line of sight, a refracted ray would come to the

observer's eye. This refracted ray originates at the object, and refracts through the lens. The

diagram below shows several incident rays emanating from an object - a light bulb. Three of

these incident rays correspond to our three strategic and predictable light rays. Each incident

ray will refract through the lens and be detected by a different observer (represented by the

eyes). The location where the refracted rays are intersecting is the image location. Since

refracted light rays do not actually exist at the image location, the image is said to be a virtual

image. It would only appear to an observer as though light were coming from this location to

the observer's eye.

Images of Objects Which Do Not Occupy a Single Point

The above discussion relates to the formation of an image by a "point object" - in this case, a

small light bulb. The same principles apply to objects which occupy more than one point in

space. For example, a person occupies a multitude of points in space. As you sight at a person

through a lens, light emanates from each individual point on that person in all directions. Some

of this light reaches the lens and refracts. All the light which originates from one single point

on the object will refract and intersect at one single point on the image. This is true for all

points on the object; light from each point intersects to create an image of this point. The

result is that a replica or likeness of the object is created as we sight at the object through the

lens. This replica or likeness is the image of that object. This is depicted in the diagram below.



Now that we have discussed how an image is formed, we will turn our attention to the use of

ray diagrams to predict the location and characteristics of images formed

byconverging and diverging lenses.


Converging Lenses - Ray Diagrams

One theme of the Reflection and Refraction units of The Physics Classroom Tutorial has been

that we see an object because light from the object travels to our eyes as we sight along a line

at the object. Similarly, we see an image of an object because light from the object reflects off

a mirror or refracts through a transparent material and travel to our eyes as we sight at the

image location of the object. From these two basic premises, we have defined the image

location as the location in space where light appears to diverge from. Because light emanating

from the object converges or appears to diverge from this location, a replica or likeness of the

object is created at this location. For both reflection and refraction scenarios, ray diagrams

have been a valuable tool for determining the path of light from the object to our eyes.

In this section of Lesson 5, we will investigate the method for drawing ray diagrams for objects

placed at various locations in front of a double convex lens. To draw these ray diagrams, we

will have to recall the three rules of refraction for a double convex lens:

Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.


An incident ray which passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens.




http://www.physicsclassroom.com/Class/refrn/u14l5ea.cfm

http://www.physicsclassroom.com/Class/refrn/u14l5da.cfm

Earlier in this lesson, the following diagram illustrating the path of light from

an object through a lens to an eye placed at various locations was shown.

In this diagram, five incident rays are drawn along with their corresponding refracted rays.

Each ray intersects at the image location and then travels to the eye of an observer. Every

observer would observe the same image location and every light ray would follow the Snell's

Law of refraction. Yet only two of these rays would be needed to determine the image location

since it only requires two rays to find the intersection point. Of the five incident rays drawn,

three of them correspond to the incident rays described by our three rules of refraction for

converging lenses. We will use these three rays through the remainder of this lesson, merely

because they are the easiest rays to draw. Certainly two rays would be all that is necessary;

yet the third ray will provide a check of the accuracy of our process.

Step-by-Step Method for Drawing Ray Diagrams

The method of drawing ray diagrams for double convex lens is described below. The

description is applied to the task of drawing a ray diagram for an object located beyondthe 2F

point of a double convex lens.

1. Pick a point on the top of the object and draw three incident rays traveling towards the lens.

Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third incident ray such that it travels directly to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.



http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#rules%23rules

2. Once these incident rays strike the lens, refract them according to

the three rules of refraction for converging lenses.

The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray which traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. And the ray which traveled to the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.

3. Mark the image of the top of the object.

The image point of the top of the object is the point where the three refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point. (See notebelow.)

4. Repeat the process for the bottom of the object.

One goal of a ray diagram is to determine the location, size, orientation, and type of image which is formed by the double convex lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.

Some students have difficulty understanding how the entire image of an object can be

deduced once a single point on the image has been determined. If the object is merely a

vertical object (such as the arrow object used in the example below), then the process is easy.

The image is merely a vertical line. In theory, it would be necessary to pick each point on the

object and draw a separate ray diagram to determine the location of the image of that point.

That would require a lot of ray diagrams as illustrated in the diagram below.

http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#note%23note

http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#rules%23rules

Fortunately, a shortcut exists. If the object is a vertical line, then the image is also a vertical

line. For our purposes, we will only deal with the simpler situations in which the object is a

vertical line which has its bottom located upon the principal axis. For such simplified situations,

the image is a vertical line with the lower extremity located upon the principal axis.

The ray diagram above illustrates that when the object is located at a position beyondthe 2F

point, the image will be located at a position between the 2F point and the focal point on the

opposite side of the lens. Furthermore, the image will be inverted, reduced in size (smaller

than the object), and real. This is the type of information which we wish to obtain from a ray

diagram. These characteristics of the image will be discussed in more detail in the next section

of Lesson 5.

Once the method of drawing ray diagrams is practiced a couple of times, it becomes as natural

as breathing. Each diagram yields specific information about the image. The two diagrams

below show how to determine image location, size, orientation and type for situations in which

the object is located at the 2F point and when the object is located between the 2F point and

the focal point.

http://www.physicsclassroom.com/Class/refrn/u14l5db.cfm


It should be noted that the process of constructing a ray diagram is the

same regardless of where the object is located. While the result of the ray

diagram (image location, size, orientation, and type) is different, the same

three rays are always drawn. The three rules of refraction are applied in

order to determine the location where all refracted rays appear to diverge

from (which for real images, is also the location where the refracted rays

intersect).

Ray Diagram for Object Located in Front of the Focal Point

In the three cases described above - the case of the object being located beyond 2F, the case

of the object being located at 2F,and the case of the object being located between 2F and F -

light rays are converging to a point after refracting through the lens. In such cases, a real

image is formed. As discussed previously, a real image is formed whenever refracted light

passes through the image location. While diverging lenses always produce virtual images,

converging lenses are capable of producing both real and virtual images. As shown above, real

images are produced when the object is located a distance greater than one focal length from

the lens. A virtual image is formed if the object is located less than one focal length from the

converging lens. To see why this is so, a ray diagram can be used.

A ray diagram for the case in which the object is located in front of the focal point is shown in

the diagram at the right. Observe that in this case the light rays diverge after refracting

through the lens. When refracted rays diverge, a virtual image is formed. The image location

can be found by tracing all light rays backwards until they intersect. For every observer, the

refracted rays would seem to be diverging from this point; thus, the point of intersection of the

extended refracted rays is the image point. Since light does not actually pass through this

point, the image is referred to as a virtual image. Observe that when the object in located in

front of the focal point of the converging lens, its image is an upright and enlarged image

which is located on the object's side of the lens. In fact, one generalization which can be made

about all virtual images produced by lenses (both converging and diverging) is that they are

always upright and always located on the object's side of the lens.

Ray Diagram for Object Located at the Focal Point

Thus far we have seen via ray diagrams that a real image is produced when an object is

located more than one focal length from a converging lens; and a virtual image is formed when

an object is located less than one focal length from a converging lens (i.e., in front of F). But

http://www.physicsclassroom.com/Class/refrn/u14l5c.cfm#real

http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#atC%23atC

http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#atC%23atC

http://www.physicsclassroom.com/Class/refrn/U14l5da.cfm#method%23method

what happens when the object is located at F? That is, what type of image is formed when the

object is located exactly one focal length from a converging lens? Of course a ray diagram is

always one tool to help find the answer to such a question. However, when a ray diagram is

used for this case, an immediate difficulty is encountered. The diagram below shows two

incident rays and their corresponding refracted rays.

For the case of the object located at the focal point (F), the light rays neither converge nor

diverge after refracting through the lens. As shown in the diagram above, the refracted rays

are traveling parallel to each other. Subsequently, the light rays will not converge to form a

real image; nor can they be extended backwards on the opposite side of the lens to intersect

to form a virtual image. So how should the results of the ray diagram be interpreted? The

answer: there is no image!! Surprisingly, when the object is located at the focal point, there is

no location in space at which an observer can sight from which all the refracted rays appear to

be coming. An image cannot be found when the object is located at the focal point of a

converging lens.

Next Section: Converging Lenses - Object-Image RelationsJump To Lesson 6: The Eye


Converging Lenses - Object-Image Relations

Previously in Lesson 5, ray diagrams were constructed in order to determine the

general location, size, orientation, and type of image formed by double convex

lenses. Perhaps you noticed that there is a definite relationship between the

image characteristics and the location where an object placed in front of a double

convex lens. The purpose of this portion of the lesson is to summarize these object-image

relationships. The best means of summarizing this relationship is to divide the possible object

locations into five general areas or points:

Case 1: the object is located beyond the 2F point

Case 2: the object is located at the 2F point

Case 3: the object is located between the 2F point and the focal point (F)



http://www.physicsclassroom.com/Class/refrn/U14l5db.cfm

Case 4: the object is located at the focal point (F)

Case 5: the object is located in front of the focal point (F)

Case 1: The object is located beyond 2F

When the object is located at a location beyond the 2F point, the image will always be located

somewhere in between the 2F point and the focal point (F) on the other side of the lens.

Regardless of exactly where the object is located, the image will be located in this specified

region. In this case, the image will be an inverted image. That is to say, if the object is right-

side up, then the image is upside down. In this case, the image is reduced in size; in other

words, the image dimensions are smaller than the object dimensions. If the object is a six-foot

tall person, then the image is less than six feet tall. Earlier in Unit 13, the

term magnification was introduced; the magnification is the ratio of the height of the object

to the height of the image. In this case, the magnification is a number with an absolute value

less than 1. Finally, the image is a real image. Light rays actually converge at the image

location. If a sheet of paper was placed at the image location, the actual replica or likeness of

the object would appear projected upon the sheet of paper.

Case 2: The object is located at 2F

When the object is located at the 2F point, the image will also be located at the 2F point on the

other side of the lens. In this case, the image will be inverted (i.e., a right-side-up object results

in an upside-down image). The image dimensions are equal to the object dimensions. A six-

foot tall person would have an image which is six feet tall; the absolute value of the

magnification is exactly 1. Finally, the image is a real image. Light rays actually converge at

the image location. As such, the image of the object could be projected upon a sheet of paper.

Case 3: The object is located between 2F and F

When the object is located in front of the 2F point, the image will be located beyond the 2F

point on the other side of the lens. Regardless of exactly where the object is located between C

and F, the image will be located in the specified region. In this case, the image will be inverted

(i.e., a right-side-up object results in an upside-down image). The image dimensions are larger

than the object dimensions. A six-foot tall person would have an image which is larger than six

feet tall. The absolute value of the magnification is greater than 1. Finally, the image is a real

image. Light rays actually converge at the image location. As such, the image of the object

could be projected upon a sheet of paper.

http://www.physicsclassroom.com/Class/refln/u13l2b.cfm#mag

Case 4: The object is located at F

When the object is located at the focal point, no image is formed. As discussed earlier in

Lesson 5, the refracted rays neither converge or diverge. After refracting, the light rays are

traveling parallel to each other and cannot produce an image.

Case 5: The object is located in front of F

When the object is located at a location in front of the focal point, the image will always be

located somewhere on the same side of the lens as the object. Regardless of exactly where in

front of F the object is located, the image will always be located on the object's side of the lens

and somewhere further from the lens. The image is locatedbehind the object. In this case, the

image will be an upright image. That is to say, if the object is right-side up, then the image

will also be right-side up. In this case, the image is enlarged; in other words, the image

dimensions are greater than the object dimensions. A six-foot tall person would have an image

which is larger than six feet tall. The magnification is greater than 1.Finally, the image is a

virtual image. Light rays diverge upon refraction; for this reason, the image location can only

be found by extending the refracted rays backwards on the object's side the lens. The point of

their intersection is the virtual image location. It would appear to any observer as though light

from the object were diverging from this location. Any attempt to project such an image upon

a sheet of paper would fail since light does not actually pass through the image location.

It might be noted from the above descriptions that there is a relationship between the object

distance and object size and the image distance and image size. Starting from a large value,

as the object distance decreases (i.e., the object is moved closer to the lens), the image

distance increases; meanwhile, the image height increases. At the 2F point, the object

distance equals the image distance and the object height equals the image height. As the

object distance approaches one focal length, the image distance and image height approaches

infinity. Finally, when the object distance is equal to exactly one focal length, there is no

image. Then altering the object distance to values less than one focal length produces images

which are upright, virtual and located on the same side of the lens as the object. Finally, if the

object distance approaches 0, the image distance approaches 0 and the image height

ultimately becomes equal to the object height. These patterns are depicted in the diagram

below. Eight different object locations are drawn in red and labeled with a number; the

corresponding image locations are drawn in blue and labeled with the identical number.

http://www.physicsclassroom.com/Class/refrn/u14l5da.cfm#atF

http://www.physicsclassroom.com/Class/refrn/u14l5da.cfm#atF



1. Identify the means by which you can use a converging lens to form a real image.

2. Identify the means by which you can use a converging lens to form a virtual image.

3. A converging lens is sometimes used as a magnifying glass. Explain how this works;

specifically, identify the general region where the object must be placed in order to produce

the magnified effect.


Diverging Lenses - Ray Diagrams

Earlier in Lesson 5, we learned how light is refracted by double concave lens in a manner that

a virtual image is formed. We also learned about three simple rules of refraction fordouble

concave lenses:

Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (i.e., in a direction such that its extension will pass through the focal point).







These three rules will be used to construct ray diagrams. A ray diagram is a tool used to

determine the location, size, orientation, and type of image formed by a lens. Ray diagrams for

double convex lenses were drawn in a previous part of Lesson 5. In this lesson, we will see a

similar method for constructing ray diagrams for double concave lenses.

Step-by-Step Method for Drawing Ray Diagrams

The method of drawing ray diagrams for a double concave lens is described below.

1. Pick a point on the top of the object and draw three incident rays traveling towards the lens.

Using a straight edge, accurately draw one ray so that it travels towards the focal point on the opposite side of the lens; this ray will strike the lens before reaching the focal point; stop the ray at the point of incidence with the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third ray to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.

2. Once these incident rays strike the lens, refract them according to the three rules of

refraction for double concave lenses.

The ray that travels towards the focal point will refract through the lens and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray which traveled parallel to the principal axis on the way to the lens will refract and travel in a direction such that its extension passes through the focal point on the object's side of the lens. Align a straight edge with the point of incidence and the focal point, and draw the second refracted ray. The ray which traveled to the exact center of the lens will continue to travel in the same direction. Place arrowheads upon the rays to indicate their direction of travel. The three rays should be diverging upon refraction.

3. Locate and mark the image of the top of the object.

The image point of the top of the object is the point where the three refracted rays intersect. Since the three refracted rays are diverging, they must be extended behind the lens in order to intersect. Using a straight edge, extend each of the rays using dashed lines. Draw

http://www.physicsclassroom.com/Class/refrn/U14l5ea.cfm#rules%23rules

http://www.physicsclassroom.com/Class/refrn/U14l5ea.cfm#rules%23rules



the extensions until they intersect. All three extensions should intersect at the same location. The point of intersection is the image point of the top of the object. The three refracted rays would appear to diverge from this point. This is merely the point where all light from the top of the object would appear to diverge from after refracting through the double concave lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point. See note below.

4. Repeat the process for the bottom of the object.

The goal of a ray diagram is to determine the location, size, orientation, and type of image which is formed by the double concave lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the lens as the image of the top of the object. At this point the complete image can be filled in.

Some students have difficulty understanding how the entire image of an object can be

deduced once a single point on the image has been determined. If the object is merely a

vertical object (such as the arrow object used in the example below), then the process is easy.

The image is merely a vertical line. This is illustrated in the diagram below. In theory, it would

be necessary to pick each point on the object and draw a separate ray diagram to determine

the location of the image of that point. That would require a lot of ray diagrams as illustrated in

the diagram below.

Fortunately, a shortcut exists. If the object is a vertical line, then the image is also a vertical

line. For our purposes, we will only deal with the simpler situations in which the object is a

vertical line which has its bottom located upon the principal axis. For such simplified situations,

the image is a vertical line with the lower extremity located upon the principal axis.

http://www.physicsclassroom.com/Class/refrn/U14l5ea.cfm#note%23note

The ray diagram above illustrates that the image of an object in front of a double concave

lens will be located at a position behind the double concave lens. Furthermore, the image will

be upright, reduced in size (smaller than the object), and virtual. This is the type of information

which we wish to obtain from a ray diagram. The characteristics of this image will be discussed

in more detail in the next section of Lesson 5.

Once the method of drawing ray diagrams is practiced a couple of times, it becomes as natural

as breathing. Each diagram yields specific information about the image. It is suggested that

you take a few moments to practice a few ray diagrams on your own and to describe the

characteristics of the resulting image. The diagrams below provide the setup; you must merely

draw the rays and identify the image. If necessary, refer to the method described above.

Next Section: Diverging Lenses - Object-Image RelationsJump To Lesson 6: The Eye


Diverging Lenses - Object-Image Relations

Previously in Lesson 5, ray diagrams were constructed in order to determine the location, size,

orientation, and type of image formed by double concave lenses (i.e., diverging lenses). The

ray diagram constructed earlier for a diverging lens revealed that the image of the object was

virtual, upright, reduced in size and located on the same side of the lens as the object. But will

these always be the characteristics of an image produced by a double concave lens? Can

convex lenses ever produce real images? Inverted images? Magnified Images? To answer

these questions, we will look at three different ray diagrams for objects positioned at different

locations along the principal axis. The diagrams are shown below. (Note that only two sets of

incident and refracted rays were used in the diagram in order to avoid overcrowding the

diagram with rays.)


http://www.physicsclassroom.com/Class/refrn/u14l5ea.cfm


http://www.physicsclassroom.com/Class/refrn/U14l5eb.cfm

http://www.physicsclassroom.com/Class/refrn/U14l5ea.cfm#method%23method

http://www.physicsclassroom.com/Class/refrn/u14l5eb.cfm

The diagrams above shows that in each case, the image is

located on the object' side of the lens

a virtual image

an upright image

reduced in size (i.e., smaller than the object)

Unlike converging lenses, diverging lenses always produce images which share these

characteristics. The location of the object does not affect the characteristics of the image. As

such, the characteristics of the images formed by diverging lenses are easily predictable.

Another characteristic of the images of objects formed by diverging lenses pertains to how a

variation in object distance effects the image distance and size. The diagram below shows five

different object locations (drawn and labeled in red) and their corresponding image locations

(drawn and labeled in blue).

The diagram shows that as the object distance is decreased, the image distance is decreased

and the image size is increased. So as an object approaches the lens, its virtual image on the

same side of the lens approaches the lens as well; and at the same time, the image becomes

larger.


The following questions pertain to the image characteristics of all types of optical devices

discussed in the last two units - plane mirrors, concave mirrors, convex mirrors,converging


http://www.physicsclassroom.com/Class/refln/u13l4c.cfm

http://www.physicsclassroom.com/Class/refln/u13l3e.cfm


lenses, and diverging lenses. Use your understanding of the object-image relationships for

these three types of mirrors and two types of lenses to answer these questions.

1. How can a plane mirror, concave mirror, convex mirror, converging lens and/or diverging

lens be used to produce an image which has the same size as the object?


lens be used to produce a magnified image?


lens be used to produce an upright image?

4. How can a a plane mirror, concave mirror, convex mirror, converging lens and/or diverging

lens be used to produce a real image?

5. The image of an object is found to be upright and reduced in size. What type of mirror

and/or lens is used to produce such an image?

The Eye


The Anatomy of the Eye

The human eye is a complex anatomical device that remarkably

demonstrates the architectural wonders of the human body. Like a

camera, the eye is able to refract light and produce a focused

image that can stimulate neural responses and enable the ability

to see. In Lesson 6, we will focus on the physics of sight. We will

use our understanding of refraction and image formation to

understand the means by which the human eye produces images of distant and nearby

objects. Additionally, we will investigate some of the common vision problems which plague

humans and the customary solutions to those problems. As we proceed through Lesson 6, we

will apply our understanding of refraction and lenses to the physics of sight.

The eye is essentially an opaque eyeball filled with a water-like fluid. In the front of the eyeball

is a transparent opening known as the cornea. The cornea is a thin membrane which has an

index of refraction of approximately 1.38. The cornea has the dual purpose of protecting the

eye and refracting light as it enters the eye. After light passes through the cornea, a portion

of it passes through an opening known as the pupil. Rather than being an actual part of the

eye's anatomy, the pupil is merely an opening. The pupil is the black portion in the middle of

the eyeball. It's black appearance is attributed to the fact that the light which the pupil allows

to enter the eye is absorbed on the retina (and elsewhere) and does not exit the eye. Thus, as

you sight at another person's pupil opening, no light is exiting their pupil and coming to your

eye; subsequently, the pupil appears black.

Like the aperture of a camera, the size of the pupil opening can be adjusted by the dilation of

the iris. The iris is the colored part of the eye - being blue for some people and brown for

others (and so forth); it is a diaphragm which is capable of stretching and reducing the size of

the opening. In bright-light situations, the iris adjusts its size to reduce the pupil opening and

limit the amount of light which enters the eye. And in dim-light situations, the iris adjusts so as

to maximize the size of the pupil opening and increase the amount of light which enters the

eye.

Light which passes through the pupil opening, will enter the crystalline lens. The crystalline

lens is made of layers of a fibrous material which has an index of refraction of roughly 1.40.

Unlike the lens on a camera, the lens of the eye is able to change its shape and thus serves to

fine-tune the vision process. The lens is attached to theciliary muscles. These muscles relax

and contract in order to change the shape of the lens. By carefully adjusting the lenses shape,

the ciliary muscles assist the eye in the critical task of producing an image on the back of the

eyeball.

The inner surface of the eye is known as the retina. The retina contains the rods and cones

which serve the task of detecting the intensity and the frequency of the incoming light. An

adult eye is typically equipped with up to 120 million rods which detect the

intensity of light and about 6 million cones which detect the frequency of light.

These rods and cones send nerve impulses to the brain. The nerve impulses travel through a

network of nerve cells. There are as many as one-million neural pathways from the rods and

cones to the brain. This network of nerve cells is bundled together to form the optic nerve on

the very back of the eyeball.

Each part of the eye plays a distinct part in enabling humans to see. The ultimate goal of such

an anatomy is to allow humans to focus images on the back of the retina. This task is

discussed in the next part of Lesson 6.

The Eye

Image Formation and Detection

Earlier in Lesson 6, we learned that the eye consists of a cornea (thin outer membrane), a lens

attached to ciliary muscles, and a retina (inner surface equipped with nerve cells). These four

parts of the eye are the most instrumental in the task of producing images which are

discernible by the brain. In order to facilitate the ability to see, each part must enable the eye

to refract light so that is produces a focused image on the retina.

It is a surprise to most people to find out that the lens of the eye is not where all the refraction

of incoming light rays takes place. Most of the refraction occurs at the cornea. The cornea is

the outer membrane of the eyeball which has an index of refraction of 1.38. The index of

refraction of the cornea is significantly greater than the index of refraction of the surrounding

air. This difference in optical density between the air the corneal material combined with the

fact that the cornea has the shape of a converging lens is what explains the ability of the

cornea to do most of the refracting of incoming light rays. The crystalline lens is able to alter

its shape due to the action of the ciliary muscles. This serves to induce small alterations in the

amount of corneal bulge as well as to fine-tune some of the additional refraction which occurs

as light passes through the lens material.

The bulging shape of the cornea causes it to refract light in a manner to similar to adouble

convex lens. The focal length of the cornea-lens system varies with the amount of contraction

(or relaxation) of the ciliary muscles and the resulting shape of the lens. In general, the focal

length is approximately 1.8 cm, give or take a millimeter. As learned in our discussion

of convex lenses in Lesson 5, the image location, size, orientation, and type is dependent upon

the location of the object relative to the focal point and the 2F point of a lens system. Since the

object is typically located at a point in space more than 2-focal lengths from the "lens," the







image will be located somewhere between the focal point of the "lens" and the 2F point. The

image will be inverted, reduced in size, and real. Quite conveniently, the cornea-lens system

produces an image of an object on the retinal surface. The process by which this occurs is

known as accommodation and will be discussed in more detail in the next part of Lesson 6.

Fortunately, the image is a real image - formed by the actual convergence of light rays at a

point in space. Vision is dependent upon the stimulation of nerve impulses by an incoming

light rays. Only real images would be capable of producing such a stimulation. Finally, the

reduction in the size of the image allows the entire image to "fit" on the retina. The fact that

the image is inverted poses no problem. Our brain has become quite accustomed to this and

properly interprets the signal as originating from a right-side-up object.

The use of the lens equation and magnification equation can provide an idea of the

quantitative relationship between the object distance, image distance and focal length. For

now we will assume that the cornea-lens system has a focal length of 1.80 cm (0.0180 m). We

will attempt to determine the image size and image location of a 6-foot tall man (ho=1.83 m)

who is standing a distance of approximately 10 feet away (do= 3.05 meters). (The lens

equation is derived geometrically upon the assumption that the lens is a thin lens. The lens of

the eye is anything but thin and as such the lens equation does not provide a truly accurate

model of the eye lens. Despite this fact, we will use the equation as a simplified approximation

of the mathematics of the eye.)

Like all problems in physics, begin by the identification of the known information.

do = 3.05 m ho = 1.83 m f = 0.0180 m

Next identify the unknown quantities which you wish to solve for.

di = ??? hi = ???

To determine the image distance, the lens equation can be used. The following lines represent

the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di



1/(0.0180 m) = 1/(3.05 m) + 1/di

55.6 m-1 = 0.328 m-1 + 1/di

55.2 m-1 = 1/di

di = 0.0181 m = 1.81 cm

To determine the image height, the magnification equation is needed. Since three of the four

quantities in the equation (disregarding the M) are known, the fourth quantity can be

calculated. The solution is shown below.

hi/ho = - di/do

hi /(1.83 m) = - ( 0.0181 m)/(3.05 m)

hi = - (1.83 m) • ( 0.0181m)/(3.05 m)

hi = -0.0109 m = -1.09 cm

From the calculations in this problem it can be concluded that if a 1.83-m tall person is

standing 3.05 m from your cornea-lens system having a focal length of 1.8 cm, then the image

will be inverted, 1.09-cm tall (the negative values for image height indicate that the image is

an inverted image) and located 1.81 cm from the "lens".

Now of course, if the person is standing further away (or closer to) the eye, the image size and image distance will be adjusted accordingly. This is illustrated in the following table for the same 6-foot tall (1.83 m) person.

Dependence of himage and dimage on dobject

(focal length is fixed at 1.8 cm)

Object Distance Image Distance Image Height

1.00 m 1.83 cm 3.35 cm

3.05 m 1.81 cm 1.09 cm

100 m 1.80 cm 0.329 cm

The results of these calculations (as illustrated in the table above) illustrate two important

principles concerning the ability of the eye to discern objects which are both close up and far

away. First, the distance between the observer and the object will greatly influence the image

size (height and width of the image formed on the retina) and quality. Objects which are

viewed at close proximity produce images which are larger than distant objects. Such an

image is typically spread over the entirety of the retina and even at times would extend

beyond the extremities of the retina. Thus, the full dimensions of a 6-foot tall person cannot be

seen if he/she stands 1-meter away (unless the eyeballs are rolled in their socket). On the

other hand, the entire image of the same person can easily be seen when standing 100-meters

away. In this instance, the image takes up a small amount of space on the back of the retina.

The drawback however is that the finer details of the image are lost due to the reduction in the

size of the image. Since the image is stimulating a smaller region of nerve cells, some details

are lost since they fail to provide sufficient stimulation to allow the brain to discern them. For

such an object distance, the small image which is created of a man might make it impossible

to discern that the man's fly is open or that his socks and belt don't match.

Second, the varying distance between the observer and the object poses some potential

problems for the human eye. Objects located varying distances from a lens system with a fixed

focal length produce images which are varying distances from the lens. Yet, the eye must

always produce an image on the retina - a location which is always the same distance away

from the cornea. The eye cannot afford to allow changes in the image distance. So how does

an eye always focus images with the same dimage regardless of the fact that the dobject is

different? How can an object 100 meters away be focused the same distance from the cornea-

lens system as an object which is 1 meter away? The answer: the cornea-lens system is able to

change its focal length. The ciliary muscles of the eye serve to contract and relax, thus

changing the shape of the lens. This serves to to allow the eye to change its focal length and

thus appropriately focus images of objects which are both close up and far away. This process

is known as accommodation and is the focus of the next part of Lesson 6.

The Eye

The Wonder of Accommodation

While the entire surface of the retina contains nerve cells, there is a small portion with a

diameter of approximately 0.25 mm where the concentration of rods and cones is greatest.

This region, known as the fovea centralis, is the optimal location for the formation of the

image. The eye typically rotates in its socket in order to focus images of objects at this

location. The distance from the outer surface of the cornea (where the light undergoes most of

its refraction) to the central portion of the fovea on the retina is approximately 2.4 cm. Light

entering the cornea must produce an image with a distance of 2.4 cm from its outer edge.

Unlike a camera, which has the ability to change the distance between the film (the detector)

and the lens, the distance between the retina (the detector) and the cornea (the refractor) is

fixed. The image distance is unchangeable. Subsequently, the eye must be able to alter the

focal length in order to focus images of both nearby and far away objects upon the retinal

surface. As the object distance changes, the focal length must be changed in order to keep the

image distance constant.


The ability of the eye to adjust its focal length is known

as accommodation. Since a nearby object (small dobject) is typically focused

at a further distance (large dimage), the eye accommodates by assuming a

lens shape that has a shorter focal length. This reduction in focal length will cause more

refraction of light and serve to bring the image back closer to the cornea/lens system and upon

the retinal surface. So for nearby objects, the ciliary muscles contract and squeeze the lens

into a more convex shape. This increase in the curvature of the lens corresponds to a shorter

focal length. On the other hand, a distant object (large dobject) is typically focused at a closer

distance (small dimage). The eye accommodates by assuming a lens shape that has a longer

focal length. So for distant objects the ciliary muscles relax and the lens returns to a flatter

shape. This decrease in the curvature of the lens corresponds to a longer focal length. The data table below demonstrates how a changing focal length would be required to maintain a constant image distance of 1.80 cm.

Dependence of f upon dobject

(dimage is fixed at 1.80 cm)

Object Distance Focal Length

0.25 m 1.68 cm

1 m 1.77 cm

3 m 1.79 cm

100 m 1.80 cm

Infinity 1.80 cm

(The values above have been calculated using the lens equation. The lens equation represents a simplified mathematical model of the eye.)

The ability of the eye to accommodate is automatic. Furthermore, it occurs instantaneously.

Focus on a far away object and quickly turn your attention to a nearby object; observe that

there is no noticeable delay in the ability of the eye to bring the nearby object into focus.

Accommodation is a remarkable feat!

The power of a lens is measured by opticians in a unit known as a diopter. A diopter is the

reciprocal of the focal length.

diopters = 1/(focal length)

A lens system with a focal length of 1.8 cm (0.018 m) is a 56-diopter lens. A lens system with a

focal length of 1.68 cm is a 60-diopter lens. A healthy eye is able to bring both distant objects

and nearby objects into focus without the need for corrective lenses. That is, the healthy eye is

able to assume both a small and a large focal length; it would have the ability to view objects

with a large variation in distance. The maximum variation in the power of the eye is called

the Power of Accommodation. If an eye has the ability to assume a focal length of 1.80 cm

(56 diopters) to view objects many miles away as well as the ability to assume a 1.68 cm focal

length to view an object 0.25 meters away (60 diopters), then its Power of Accommodation

would be measured as 4 diopters (60 diopters - 56 diopters).

http://www.physicsclassroom.com/Class/refrn/u14l6a.cfm#ciliary

http://www.physicsclassroom.com/Class/refrn/u14l6a.cfm#ciliary

The healthy eye of a young adult has a Power of Accommodation of

approximately 4 diopters. As a person grows older, the Power of

Accommodation typically decreases as a person becomes less able to view

nearby objects. This failure to view nearby objects leads to the need for corrective lenses. In

the next two sections of Lesson 6, we will discuss the two most common defects of the eye

- nearsightedness and farsightedness.

The Eye

Nearsightedness and its Correction

The human eye's ability to accommodate allows it to view focused images of both nearby and

distant objects. As mentioned earlier in Lesson 6, the lens of the eye assumes a large

curvature (short focal length) to bring nearby objects into focus and a flatter shape (long focal

length) to bring a distant object into focus. Unfortunately, the eye's inability to a provide a

wide variance in focal length leads to a variety of vision defects. Most often, the defect occurs

at one end of the spectrum - either the inability to assume a short focal length and focus on

nearby objects or the inability to assume a long focal length and thus focus on distant objects.

Nearsightedness or myopia is the inability of the eye to focus on distant objects. The

nearsighted eye has no difficulty viewing nearby objects. But the ability to view distant objects

requires that the light be refracted less. Nearsightedness will result if the light from distant

objects is refracted more than is necessary. The problem is most common as a youth, and is

usually the result of a bulging cornea or an elongated eyeball. If the cornea bulges more than

its customary curvature, then it tends to refract light more than usual. This tends to cause the

images of distant objects to form at locations in front of the retina. If the eyeball is elongated

in the horizontal direction, then the retina is placed at a further distance from the cornea-lens

system. Subsequently the images of distant objects form in front of the retina. On the retinal

surface, where the light-detecting nerve cells are located, the image is not focused. These

nerve cells thus detect a blurry image of distant objects.

http://www.physicsclassroom.com/Class/refrn/u14l6c.cfm#accomdn

http://www.physicsclassroom.com/Class/refrn/u14l6c.cfm#accomdn



The cure for the nearsighted eye is to equip it with a diverging lens. Since the nature of the

problem of nearsightedness is that the light is focused in front of the retina, a diverging lens

will serve to diverge light before it reaches the eye. This light will then be converged by the

cornea and lens to produce an image on the retina.

(Note: In the diagram above that the light approaching the eye from a distant object is

traveling as a bundle of rays which are roughly parallel to each other. Suppose for a moment

that the distant object is the lettering on the chalk board in the front of the room as you sight

at it from the back of the room. Geometrically, whatever light rays from a particular letter or

word which reach your eye will be traveling roughly parallel to each other. This is not the case

when viewing nearby objects as demonstrated on the previous page.)


refraction of light,from lens and slabs

Documents