chapter 23-light: geometric optics - university of...

Chapter 23- Light: Geometric Optics

Final Final ExamExam

Tuesday, Dec. 16Tuesday, Dec. 16thth

Time: 14:00Time: 14:00--17:00h17:00h

Room: CLRoom: CL--110110

Equation sheet available atEquation sheet available athttp://ilc2.phys.uregina.ca/~barbi/academic/phys109/phys109.html

There will be 8 problems divided into two sets:

Set 1: 5 problems

Set 2: You have to solve 3 out of 5 problems.

Remember, this is a comprehensive exam.

Tutorial next week will revise several subjects covered over

the semester.

Location: CL-110

Time: Wednesday, 4:30pm - 5:30pm

I will be in a research trip between Dec. 7-11.I will be in a research trip between Dec. 7-11.

However, feel free to come to my office on the following

days:

o 12th (afternoon);

o 15th (all day) and

o 16th (morning).

AssignmentAssignment 1111

Textbook (Giancoli, 6th edition), Chapter 23:

page 659: problems 9, 15 and 20

page 660: problem 47



page 663: problems 83 and 86

You do not need to hand in this assignment.

Solutions will be posted on Dec. 11th.

Comprehensive List of ProblemsComprehensive List of Problems

In addition to assignment 11, the following problems are

recommended in preparation for the final exam:

Chapter 3: 21, 31, 33

Chapter 4: 29, 34, 47, 64

Chapter 5: 18, 19, 23, 24Chapter 5: 18, 19, 23, 24

Chapter 6: 31, 41, 44, 85

Chapter 7: 24, 32, 43, 63

Chapter 8: 25, 49, 60, 83

Chapter 9: 4, 18, 21, 27

These problems will help but are not enough to prepare you for the final � Try

to solve as many other problems as you can.

Changes in OfficeChanges in Office--hours hours

The following changes will take place until the end of the semester

Office-hours:

- Monday, 12:00-13:00h

- Wednesday, 14:00-15:00h

- Friday, 13:00-14:00h

Old assignments and midterm exams

can be found in the Physics Office

(LB-226)(LB-226)

Solutions have been posted on

the web:

All marks, including assignments, have

been posted on the web.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdfhttp://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdf

Please, verify that all your marks have

been entered in the list.

Chapter 23

• The Ray Model of Light

• Reflection; Image Formed by a Plane Mirror

• Formation of Images by Spherical Mirrors

• Index of Refraction

• Refraction: Snell’s Law

• Total Internal Reflection; Fiber Optics

• Thin Lenses

• The Thin Lens Equation; Magnification

• Lensmaker’s Equation

Recalling Recalling LastLast LecturesLecturesRecalling Recalling LastLast LecturesLectures

Formation of Images by Spherical MirrorsFormation of Images by Spherical Mirrors

Eq. 23-2 applies to any ray that makes a small angle with the principal axis.

(23-2)

Eq. 23-2 applies to any ray that makes a small angle with the principal axis.

� These rays are called paraxial rays.


1) A ray parallel to the axis; after

reflection it passes through

the focal point;

2) A ray through the focal point;

after reflection it is parallel to

the axis

3) A ray perpendicular to the

mirror (radial direction); it

reflects back on itself

(the ray is in the direction of the

⇒ i=

r= 0

(the ray is in the direction of the

normal ⇒ θi= θ

r= 0 ).

The point where these three rays

cross is the image I’ . All other rays

from O’ will also cross I’.


(23-3)

Eq. 23-3 is called the mirror equation and relates the object and image

distances to the focal length ( f = r/2 )


We can also find the magnification, m, of a mirror as the ratio of image height

to the object height.

Note that we have introduced a sign in the above equation.

(23-4)

Note that we have introduced a sign in the above equation.

Do NOT confuse it with the equation obtained few slides ago.

The negative sign here is a convention to indicate that the image is inverted.


In our example, the object is between the center of curvature and the focal point

� its image is larger, inverted, and real (in front of the mirror).


If an object is outside the center of curvature of a concave mirror

� its image will be inverted, smaller, and real (in front of the mirror).


If an object is inside the focal point

� its image will be upright, larger, and virtual (behind the mirror).


Sign conventions;

Assuming the object height ho to be positive, the sign convention we use are:

• The image height hi is positive if the image is upright, and negative if inverted,

relative to the object;

• d ( d ) is positive if image ( object ) is in front of the mirror;• di ( do ) is positive if image ( object ) is in front of the mirror;

• di ( do ) is negative if image ( object ) is behind the mirror.

Note that the magnitude is positive for an upright image and

negative for an inverted image.


For a convex mirror, the image is always virtual,

upright, and smaller.

The equations 23-2 to 23-4 also applies. However,

both the focal length and radius of curvature

should be considered negatives for convex mirrors.

TodayTodayTodayToday

Index of RefractionIndex of Refraction

Light (and any other form of electromagnetic wave) travels at a speed of

c = 300,000 Km/s in vacuum.

However, in general it slows somewhat when traveling through a medium.

It is useful to define a quantity called index of

refraction n of the medium in which light

propagates as the ratio of the speed of light

in vacuum to the speed of light in this medium:in vacuum to the speed of light in this medium:

Since v is always smaller than c� n > 1

(23-5)


When light travelling in a transparent medium strikes a boundary with another

medium, part of the light is reflected and other part can be transmitted into the

other medium. The direction of the ray of light can change

direction if the new medium has different index of refraction.

This is called refraction.

The angle the outgoing ray makes with the normal to

the surface is called the angle of refraction the surface is called the angle of refraction

( θ2 in the figure) .


Refraction is responsible for some optical illusions. For example, the observer in

the figure thinks the foot of the person standing in the water is located at a

higher position than it really is.

This happens because the ray is refracted and change direction. However, the

observer still thinks that the ray is travelling in a straight path (the ray model)

from the foot of the person.

Refraction Refraction –– Snell’s LawSnell’s Law

The angle of refraction depends on the indices of refraction, and was

experimentally proven to be related to the angle θθθθ1 of incidence by the formula:

Where

θ = angle of incidence

(23-6)

θ1 = angle of incidence

θ2 = angle of refraction

n1 = index of refraction of medium 1

n2 = index of refraction of medium 2

Eq. 23-6 is known as Snell’s law or basic law of refraction.

Total Internal Reflection; Fiber OpticsTotal Internal Reflection; Fiber Optics

We can use eq. 23-6 and write:

This equation tells us that for the same angle of incidence θθθθ1 and index of

refraction n1 , the angle of refraction θθθθ2 is larger for smaller index of refraction n2 .

Also, for given indices of refraction n and n , there is a critical angle of Also, for given indices of refraction n1 and n2 , there is a critical angle of

incidence θθθθc for which the angle of refraction is 900.

(23-7)


If the angle of incidence is larger than the critical angle, NO transmission

occurs.

This is called total internal reflection.


Total internal reflection is the principle behind fiber optics.

Light will be transmitted along the fiber even if it is not straight.

An image can be formed using multiple small fibers.

Thin Lenses; Ray TracingThin Lenses; Ray Tracing

Lenses are very important in optics: they are used in eyeglasses, telescopes,

cameras, medical instruments, etc.

The most common lenses are circular with two faces, each being portion of a

sphere. Like in mirrors, the lenses surfaces can be convex, concave or plane.

Thin lenses are those whose thickness is small

compared to their radius of curvature. They may compared to their radius of curvature. They may

be either converging (a) or diverging (b).

Lenses work based on the Snell’s law.

A lens is usually made of material such as glass

or transparent plastic such that its index

of refraction is greater than of the air.


Some properties of lenses are similar to those defined for mirrors:

• The axis of a lens is the straight line passing through the center of the length

and perpendicular to its two surfaces.

• Focal point F is the point where parallel rays

converge after passing through a lens.

• Focal length is the distance from F to the lens.• Focal length is the distance from F to the lens.

The focal length is the same on both sides of

the lens, even if they have different radius of

curvature. If parallel rays falls on a lens at an

angle, they will converge at a point Fa as shown

in figure (b).

In addition, we can define:

• Focal plane: this is the plane that contains the

points F and Fa.


Converging lenses are those thicker in the center than at the edges.Snell’s law

apply to

these rays

n2

n1n1

Parallel rays are brought to a focus by

a converging lens.


Diverging lenses are those thinner in the center than at the edges.

Parallel light diverge; the focal point is

that point where the diverging rays that point where the diverging rays

would converge if projected back.


Optometrists and ophthalmologists specify the strength of an eyeglass by the

reciprocal of the focal length. This quantity is called power P of the lens:

Lens power is measured in diopters, D.

(23-8)

1 D = 1 m-1


Ray tracing for thin lenses is similar to that for mirrors. We have three key rays:

1- This ray comes in parallel

to the axis and exits

through the focal point.

2- This ray comes in through

the focal point and exits

parallel to the axis.

3- This ray goes through the

center of the lens and is

undeflected.


For a diverging lens, we can use the same three rays; the image is upright

and virtual (the rays do not actually pass through the image).

Converging lenses form real images

as the rays do really pass through the

image.


The thin lens equation is the same as the mirror equation:

The sign conventions are slightly different:

(23-9)

1. The focal length is positive for converging lenses and negative for

diverging.

2. The object distance is positive when the object is on the same side as

the light entering the lens otherwise it is negative.

3. The image distance is positive if the image is on the opposite side from

the light entering the lens (the image is said real); otherwise it is negative

(the image is said virtual).

4. The height of the image is positive if the image is upright and negative

otherwise.


The magnification formula is also the same as that for a mirror:

The magnification is positive if the image is upright and negative otherwise.

(23-10)

The Lensmaker’s equation relates the lens’s index of refraction and the radii of

curvatures of its two surfaces to its focal length:

Where:

Lensmaker’sLensmaker’s EquationEquation

(23-11)

Where:

R1 and R2 = radii of curvature of each of the lens’s surfaces.

n = index of refraction of the length.

Sign convention:

The radius of curvature is positive for convex surfaces (center of curvature

behind the lens) and negative for concave surfaces (center of curvature in

front of the lens).

OpticsOptics

Problem 23-24 (textbook): The speed of light in ice is 2.29 × 108 m/s. What is

the index of refraction of ice?

OpticsOptics

Problem 23-24 (textbook):

We find the index of refraction from

;c

vn

=

( )83.00 10 m s×( )8

83.00 10 m s

2.29 10 m s ,n

×× =

1.31.n =

OpticsOptics

Problem 23-27 (textbook): A diver shines a flashlight upward from beneath the

water at a 42.5° angle to the vertical. At what angle does the light leave the

water?

OpticsOptics


We find the angle of refraction in the water from

1 1 2 2sin sin ;n nθ θ=

( ) ( )1.33 sin 42.5 sin ,θ° = 1.00( ) ( ) 21.33 sin 42.5 sin ,θ° = 1.00

264.0 .θ = °

OpticsOptics

Problem 23-45 (textbook): A certain lens focuses light from an object 2.75 m

away as an image 48.3 cm on the other side of the lens. What type of lens is it

and what is its focal length? Is the image real or virtual?

OpticsOptics


The image is formed on the opposite side of the lens, therefore it is a

converging lens.

We find the focal length of the lens from

1 1 1;

+ =

According to the sign convention for lens, images formed on the opposite side of

a lens are such that di > 0 � the image is real.

o i

;d d f

+ =

1 1 1,

275cm 48.3cm f

+ =

41.1cm.f = +

OpticsOptics

Problem 23-11 (textbook): A dentist wants a small mirror that, when 2.20 cm

from a tooth, will produce a 4.5 х upright image. What kind of mirror must be

used and what must its radius of curvature be?

OpticsOptics


We find the image distance from the magnification:

i i

o o

;h d

mh d

−= =

( )i4.5 ,

2.20cm

d−+ = i

9.90cm.d = −

We find the focal length from

Because the focal length is positive, the mirror is concave with a radius of

o i

1 1 1;

d d f

+ = ( ) ( )

1 1 1,

2.20cm 9.90cm f

+ =

− 2.83cm.f =

( )2 2 2.83cm 5.7cm.r f= = =

OpticsOptics

Problem 23-30 (textbook): An aquarium filled with water has flat glass sides

whose index of refraction is 1.52. A beam of light from outside the aquarium

strikes the glass at a 43.5° angle to the perpendicular (Fig. 23–49). What is the

angle of this light ray when it enters

(a) the glass, and then

(b) the water? (b) the water?

(c) What would be the refracted angle if the ray entered the water directly?

OpticsOptics


(a) We find the angle in the glass from the refraction at the air–glass surface:

(b) Because the surfaces are parallel, the refraction angle from the first surface

is the incident angle at the second surface. We find the angle in the water from

the refraction at the glass–water surface:

1 1 2 2sin sin ;n nθ θ= ( ) ( ) 2

1.00 sin 43.5 sin ,θ° = 1.52 226.9 .θ = °

(c) If there were no glass, we would have

Note that, because the sides are parallel, is independent of the presence of the

glass.

2 2 3 3sin sin ;n nθ θ= ( ) ( ) 3

1.52 sin 26.9 sin ,θ° = 1.33 331.2 .θ = °

1 1 3 3sin sin ;n nθ θ ′= ( ) ( ) 3

1.00 sin 43.5 sin ,θ ′° = 1.33

331.2 .θ ′ = °

OpticsOptics

Problem 23-47 (textbook): A stamp collector uses a converging lens with focal

length 24 cm to view a stamp 18 cm in front of the lens.

(a) Where is the image located?

(b) What is the magnification?

OpticsOptics


(a) We locate the image from

o i

1 1 1;

d d f

+ = i

1 1 1,

18cm 24cmd

+ =

i

72cm.d = −

The negative sign means the image is 72 cm behind the lens (or on the same

side as the light entering the lens) � the image is virtual.

(b) We find the magnification from

( )( )

i

o

72cm4.0.

18cm

dm

d

−−= = − = +

chapter 23-light: geometric optics - university of...

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