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Unit 3 - Foundations of WavesChapter 6 - Light, Mirrors, and Lenses

Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 1 / 57

Section 6.1 - The Behaviour of Light

History of Light

Plato (428 BCE - 348 BCE) thought that light consisted of streamsof particles emitted from the eyes.

Aristotle (384 BCE - 322 BCE) thought the streams were emitted inthe other direction.

Newton (1642 - 1727) postulated that light consists of streams ofparticles.

Huygens (1629 - 1695) and others supported the idea that light doesnot consist of particles but rather of waves.

Einstein, Planck, Broglie, and Bohr (20th century) proposed andconfirmed that all particles also have a wave nature (and vice versa).This is called wave-particle duality.

Watch: “The History of Light”Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 2 / 57

https://www.youtube.com/watch?v=OLCqaWaV6jA

Sources of Light

Objects can be seen only when a source of light is present. There is adifference between luminous bodies, such as stars, incandescent andfluorescent lamps, which emit light, and illuminated bodies such as thetrees, grass and moon, which reflect light.

Most visible objects are illuminated by a luminous source and are seen bylight, which they reflect.


Man Made Light

Solids give off light when heated. When heated, a solid’s electrons do notmove in exactly the same way. The resulting light has a whole range offrequencies, which extend from ultraviolet to infrared, which our eyescannot see. These heated materials are known as incandescentsubstances.

Gasses such as neon and sodium give off light when electric currents arepassed through them or when they are heated, but unlike a solid, a gasgives off light at a unique frequency. Neon is red and sodium is orange.


Fluorescent lamps produce light through mercury vapours, whichproduces ultraviolet rays. These rays strike the interior of the tube or bulb,which is coated with chemicals called phosphors. These chemicals give offvisible light when bombarded by ultraviolet rays.


Natural Light

Stars produce light by the internal nuclear reactions occurring in thecentre of each star. The temperature of the star’s surface determines thecolour of the light. Very hot stars are white and cool stars are red.

Other sources of natural light are lightning, produced from electriccharges rushing through the atmosphere between clouds and the ground,Aurora Borealis, produced when air molecules interact with chargedparticles emitted from the Sun that are caught in Earth’s magnetic field,and bioluminescent animals such as fireflies and electric eels whose lightis produced by chemical reactions.


Transmission and Absorption of Light

There are three different classes of absorption:

Transparent - Materials such as glass, quartz, and air allow thepassage of light in straight lines.

Translucent - Materials such as paper and frosted glass allow light topass through but in different directions so that one cannot see objectsthrough them.

Opaque - Materials such as wood, brick and metals allow no light topass through.


Dispersion

When light is passed through transparent crystals brilliant colours of lightare produced. At first this phenomenon was attributed to the crystal butNewton was able to show that these colours where already present inlight. If a beam of white light is allowed to pass through a prism thecomponents of light are refracted at various amounts to produce a seriesof colours called a spectrum. This is called dispersion.


The Light Spectrum


Why is the Sky Blue?

The scattering of light is why the sky is blue during the day and redduring the sunset. Scattering (α) is inversely proportional to the 4thpower of the wavelength (α = 1

4λ).

1 The short waves of violet light are scattered easily compared to thelonger red waves. Therefore when light enters the earth’s atmospherethe frequency that is scattered most is that of blue.

2 At night since the light must travel a greater distance through theatmosphere because of the angle light enters the atmosphere the redwaves pass through the atmosphere while the other colours arecompletely scattered and the sun becomes progressively redder as itsets. (Video)


http://www.youtube.com/watch?v=qP2Vp1zj8H0&t=0m19s

How does light travel?

Light travels in straight lines. This is evident by observing the shadowsproduced by objects.

If the light source is a point then the shadow produced by an image will bedark and sharp since no light is reflected. It is important to note thatdirectly around the shadow is a region where the shadow is slightly fuzzy.This is caused by the diffraction of light. When light travels close to theedge of any object, or opening, it is bent in its path and travels in a newdirection. Recall: This is diffraction.


If the light source is not a point then two types of shadows will be created.A region where no light is reflected called the umbra and a region wherepartial reflection occurs which is called the penumbra.

Solar and Lunar eclipses follow the pattern above. The following diagramshows how the lunar eclipse will appear from different parts of the earth.


Application: Images in a Pinhole Camera

A pinhole camera is effectively a light-proof box with a small hole in oneside. Light from a scene passes through this single point and projects aninverted image on the opposite side of the box.

Key fact: If a source of light passes through a tiny hole than the imageproduced is inverted.


The relationship between the object and the image is given by the formula:

height of image

height of object=

distance of image from pinhole

distance of object from pinhole

Symbolically, we represent this as:

h1h2

=d1d2


Example: Determine the height of an image produced by an object12.2 m in height that is 19.7 m from a camera with a lens to film lengthof 38 mm.

h1h2

=d1d2

h1 =?; h2 = 12.2 m

d1 = 38 mm = 0.038 m; d2 = 19.7 m

h112.2 m

=0.038 m

19.7 m

h1 =(12.2 m)(0.038 m)

19.7 m

h1 = 0.024 m =⇒ h1 = 24 mm


Speed of Light

The speed of light can change depending on the medium that it istravelling through. Therefore the speed will differ in our atmosphere andspace. The modern value for the speed of light in a vacuum is

299 792 400 m/s. For calculation purposes, c = 3.00× 108 m/s . A light

year is the distance light will travel in 1 year.

=⇒ 1 light year = 9.46× 1015 m

Time of flight techniques Cavity resonancetechniques

Watch: “The Original Method”Watch: “Using a Microwave to Calculate the Speed of Light”


https://www.youtube.com/watch?v=DMKE5YGLnmc

https://www.youtube.com/watch?v=GH5W6xEeY5U

Example: Light takes 1.28 s to travel from the moon to the Earth. Whatis the distance between them?

Recall: v =∆d

∆t

Since we’re interested in the movement of light in space:v = c = 3.00× 108 m/s

∆t = 1.28 s

∆d = (v)(∆t) = (3.00× 108 m/s)(1.28 s) = 3.84× 108 m

Since I didn’t ask, you can leave the distance in this unit. If you wish, youcan convert it to a more appropriate unit: 384 000 km


Practice: How much time would be required for a spaceship travelling at60.0× 104 km/h to reach the closet known star, Proxima Centauri, 4.3light-years away? [See Section 6.1 Video]


Section 6.2 - Reflection of Light

When light is reflected off of a surface, the type of reflection depends onthe smoothness of the surface.

Regular reflection occurs when reflecting bodies are smooth such asa mirror.

Diffuse reflection occurs when the reflecting surface is light incolour but irregularly shaped.

Reflex reflection occurs when light is reflected from a manufacturedsurface such as traffic signs.


Laws of Reflection:

1 The angle of incidence is equal to the angle of reflection.

2 The incident ray, the reflected ray, and the normal to the reflectingsurface lie in the same plane.


Images in a Plane Mirror

When you look into a mirror, you see an image of your face apparentlylocated behind the mirror. If you move towards the mirror, your image willmove closer to the mirror so that your image is always the same distancefrom the mirror as the object. If you raise your right hand the image onthe mirror appears to be a left hand coming up to meet your right hand.

Letters viewed in a plane mirror appear reversed horizontally butnot vertically. This is called lateral inversion.


Example: A student stands 3.0 m in front of a plane mirror.

a) How far behind the mirror is his virtual image?

Draw a diagram of the situation:

Light that bounces off of the man needs to travel 3.0 m to hit themirror, then is reflected 3.0 m to reach the man’s eyes. Therefore, itappears to the man that his virtual image is 3.0 m behind the mirror.

b) If he steps forward 1.0 m, what distance will separate him from hisvirtual image?

Now he is 2.0 m from the mirror (and his virtual image appears 2.0 mbehind the mirror). Therefore, the total distance is 4.0 m.


Applications of Plane Mirrors

Plane mirrors have many applications one of which is a looking glass i.e.mirror. Other applications are:

Cameras

Periscope

Optical Lever

Kaleidoscope

See-through Mirrors

Theatrical Effects

Right Angle Images

Watch: “Magic with Mirrors”Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 23 / 57

http://www.youtube.com/watch?v=CoV6wrEkQZ4&t=6m37s

Reflection from Curved Surfaces

Two types of curved mirrors used are convex or concave.1 Concave refers to a reflecting surface that curves inwards. This type

of mirror will make parallel light rays converge.

2 Convex refers to a reflecting surface that curves outwards. This typeof mirror will make parallel light rays diverge.

The centre of curvature (C) is the centre of a curved mirror. The radius(R) of curvature is any straight line drawn from C to the curved surface.The geometric centre of a curved mirror is called the vertex (V) and theline drawn through C and V is called the principal axis.


Concave Mirrors

Rays of light that are parallel to the principal axis that strike a concavemirror are reflected through a common point. This common point is calledthe focus. The focus is a point that lies on the principal axis. Thedistance from the vertex to the focus is called focal length (f).

f =radius of curvature

2


Rules for Rays in Curved Mirrors

A ray parallel to the principal axis is reflected through the focus.

A ray that passes through the focus is reflected parallel to the focallength.

A ray that passes through point C is reflected back along the samepath.

You’ll use at least two of these rules to determine the location and characteristics

of an image given the starting position of some object in front of a curved mirror.


The characteristics of an image formed in a concave mirror depend on theposition of the object. There are essentially five cases:

Object Position Image Position Size of Image Orientation of Image Type of ImageObject distance isgreater than C

Image is betweenC and F

Smaller Inverted Real

Object is on C Image is on C Same size Inverted Real

Object is betweenC and F

Image is greaterthan C

Larger Inverted Real

Object is on F No image

Object is in frontof F

Image is behindthe mirror

Larger Upright Virtual


Case 1: Object > C

You’ll find a copy of each of these diagrams (Cases 1 - 5) in the Chapter 6 Practice

Problems package. You’ll be expected to predict the location of the image. To

draw these, follow the 3 laws outlined on slide 26. You need only use 2 of the

laws to determine the position and characteristics of the image.


Case 2: Object = C [See Section 6.2 Video]

Try and predict the location and characteristics of the image that would be reflected.

The solutions for the table above are found on slide 27. See the video for a step-

by-step walkthrough.


Case 3: F < Object < C [See Section 6.2 Video]


Case 4: Object = F [See Section 6.2 Video]


Case 5: Object < F [See Section 6.2 Video]


Convex Mirrors

Regardless of the position the object, the image is always upright, smalland virtual.

Note: “Virtual” means that the reflected light rays never converge. Thismeans you cannot capture the image on a screen; i.e. it doesn’t reallyexist. So why do we “see” it? Our eyes will extrapolate where the lightrays “would have” converged (see the dotted lines in the image below).Therefore, the image only exists in our brains.


Section 6.3 - The Refraction of Light

Light travels in lines only when the medium through which it travelsremains the same. If it passes through a different medium, such as fromair to water, the light rays will bend. This is called refraction. The indexof refraction is the term used to compare the amount of refraction oflight in mediums and is found using the following formula.

Index of refraction of a medium =speed of light in a vacuum

speed of light in the medium

n =c

v


Example: The index of refraction of water is 1.33. Calculate the speed oflight in water.

n =c

v

n = 1.33c = 3.00× 108 m/s

Note: c is a constant value. You will never need to solve for c .

v =?

v =c

n=

3.00× 108 m/s

1.33= 2.26× 108 m/s


Practice: What is the index of refraction of a solid in which the speed oflight is 1.24× 108 m/s? Identify the solid. Reference the table above.

[See Section 6.3 Video]


Snell’s Law

This is used to determine the direction that light will travel into a specificmedium.

n =sin θisin θR

θi - angle of incidence

θR - angle of refraction

n - index of refraction


Examples:

1) A ray enters a medium of unknown index of refraction at 30◦. Theangle of refraction is 24◦. Calculate the index of refraction to 2 s.f.

n =sin θisin θR

θi = 30◦

θR = 24◦

n =sin 30◦

sin 24◦= 1.2


2) A ray of light enters water (nw = 1.33). If the angle of incidence is30◦, what is the angle of refraction? Go to 2 s.f.

n =sin θisin θR

θi = 30◦

θR =?n = 1.33

Solving for θR will require us to use the sin−1 function (usually 2ndfunction “sin” on most calculators).

sin θR =sin θin

=⇒ sin θR =sin 30◦

1.33

θR = sin−1

(sin 30◦

1.33

)= θR = 22◦


Snell’s law can also be used to determine the direction light will travelfrom one medium to a different medium. In this case the formula used is

n1 and n2 correspond to the two different mediums of interest.


Example: Light travels from glass into air. The angle of refraction in airis 60.0◦. What is the angle of incidence in glass if the index of refractionfor glass is 1.52 and the index of refraction for air is 1.00?

n1 sin θi = n2 sin θR

1 = glass; 2 = airn1 = 1.52n2 = 1.00θi =?θR = 60.0◦

(1.52)(sin θi ) = (1.00)(sin 60.0◦)

sin θi =(1.00)(sin 60.0◦)

1.52

θi = sin−1

((1.00)(sin 60.0◦)

1.52

)= 34.7◦


Practice: [See Section 6.3 Video]

1) Light travels from glass into water. The angle of incidence in glass is40.0. What is the angle of refraction in water (nw = 1.33; ng = 1.52).

2) If the index of refraction for diamond is 2.42, what will be the angle ofrefraction in diamond for an angle of incidence in water of 60.0◦?(nw = 1.33)


Law of Refraction

The ratio of the sine of the angle of incidence to the sine of the angleof refraction is a constant (Snell’s Law)

The incidence ray and refracted ray are on opposite sides of thenormal at the point of incidence and all three are in the same plane.

When light travels from onetransparent medium into another,some reflection always occurs.This is partial reflection and partialrefraction. The degrees of reflectionand refraction that occur dependon the angle of incidence and thedensities of the media.


Total Internal Reflection and Critical Angle

As the angle of incidence increases, the intensity of a reflected raybecomes progressively stronger and the intensity of a refracted raybecomes progressively weaker. As the angle of incidence increases, theangle of refraction increases, eventually reaching a maximum of 90◦.

The point in which refraction ceases and the entire incident light isreflected internally is called total internal reflection.

When the angle of refraction is 90◦, the incident ray forms an angle ofincidence that has a unique value for each medium. This unique angle ofincidence is called the critical angle. For water, with an index ofrefraction of 1.33, the critical angle is approximately 49◦.

Watch: “Total Internal Reflection” What is the critical angle for glass?


https://www.youtube.com/watch?v=NAaHPRsveJk

In general the critical angle can be found using the formula:

sin ic =1

n

ic - critical angle

n - index of refraction

Example: If the index of refraction for zircon is 1.92, what is the criticalangle?

This is as simple as it gets - a two variable equation! You will have tomake use of the inverse sine function once again.

sin ic =1

n=⇒ sin ic =

1

1.92

ic = sin−1

(1

1.92

)= 31.4◦


Practice: Reference your table of indices of refraction where necessary.[See Section 6.3 Video]

1) What is the critical angle in flint glass when light passes from flintglass into air?

2) If the index of refraction for water is 1.33, what is the critical angle?

3) The critical angle for a medium is 40.5◦. What is the index ofrefraction of the medium? What is the medium?


Section 6.4 - Lenses and Optical Instruments

Types of Lenses


Refraction of Lenses

A lens is a circular piece of glass with a uniformly curved surface thatchanges the direction of light passing through. When a series of rayspasses through a lens, each ray is refracted by a different amount. Raysstriking the lens near the edge are bent the most because the curvature isgreatest at this point and least at the centre where the surface is flat.

In a converging (convex) lens, we have the following parts:

Optical Centre (O) - the centre of the lens.

Principal Axis (P) - a line drawn through the optical centreperpendicular to the surface.

Focus (F) - Parallel rays are refracted and converge at this point.

Focal length (f) - The distance from F to O.


Images Formed by Converging (Convex) Lenses1) When the object is beyond C (2 focal lengths away), the image

produced is between C and F and is real, inverted and smaller thanthe object.

2) When the object is at C, the image produced at C and is real, invertedand the same size as the object.


3) When the object is between 1 and 2 focal lengths away, the image isbeyond 2 focal lengths and is real, inverted and larger than the object.

4) When the object is at 1 focal length away no image is formed.

5) When the object is between the lens and 1 focal length away, the imageis virtual, upright and larger than the object.

These 5 cases are identical to the 5 cases for converging mirrors. You will be tested

on your knowledge of both, but I will have you only draw situations for mirrors.


Application: The Human Eye

The eye is almost a sphere about 3 cm in diameter. The cornea and lenscombine to focus an inverted image on the retina. The cells on the retinarespond to various intensities and colours of light and transmit thesesignals through the optic nerve to the portion of your brain located at therear of your head.


Ciliary muscles and suspensory ligaments adjust the shape of the lens tofocus images of objects that are far or near on the retina. This ability ofthe eye to change the focal length of the lens to adjust for far and nearobjects is called the power of accommodation.

The eye can only focus on one image at a time. If it is focused on a nearimage then images far away will be out of focus. The iris opens and closesthe eye opening called the pupil. This change in the pupils opening allowsvarious amounts of light to enter the eye similar to the aperture setting ona camera. The front part of the eye is the cornea. The fluid calledvitreous gel maintains the shape of the eye.


Defects in Vision

Farsightedness (hyperopia): is the inability to see nearby objects clearly.It usually occurs because the distance between the lens and the retina istoo short or if the cornea-lens is too weak and cannot focus the image onthe retina.


Nearsightedness (myopia): is the inability to see far images clearly. Itusually occurs because the distance between the lens and the retina is toolong or if the cornea-lens is too strong.


Correcting Defects in Vision:

Watch: Lasik Eye SurgeryMr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 56 / 57

https://www.youtube.com/watch?v=qoH0VHrOM9A&ab_channel=LondonVisionClinic

Astigmatism is the inability of the eye to focus light in different planes atthe same time. It is caused when the cornea or lens is not perfectlyspherical.

Colour Blindness is the inability of the eye to distinguish certain colours.

Cataracts are formed due to a gradual loss of transparency of the eye’slens.


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