Optics

Chapter 9

Optics

Light waves  Light generally refers to the narrow band of

EM waves that can be seen by human beings. These are transverse waves.  The frequencies range from 4×1014 to 7.5×1014 hertz. 

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Light waves, cont’d  We typically express the wavelengths in

nanometers.

1 nanometer = 10−9 meter = 0.000 000 001 meter = 1 nm 

The wavelengths of visible light range from 750 nm (red light) to 400 nm (blue light).

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Light waves, cont’d  Keep in mind that:

We perceive different frequencies of light as different colors, and  White light is typically a combination of all frequencies of the visible spectrum. 

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Light waves, cont’d  To describe various properties of light, we

use two different methods: wavefronts, or  light rays. 

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Reflection  Specular reflection is reflection off of a very

smooth surface. 

Specular reflection occurs when the direction the light wave is traveling changes.

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Reflection, cont’d  To study reflection, we need some

terminology. 

The normal is an imaginary line drawn perpendicular to the mirror and touching it at the point where the incident ray strikes the mirror. 7

Reflection, cont’d The angle of incident is the angle between the incident ray and the normal.  The angle of reflection is the angle between the reflected ray and the normal. 

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Reflection, cont’d  The law of reflection states that the angle of

incidence equals the angle of reflection.

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Reflection, cont’d  Diffuse reflection occurs when light strikes a

surface that is not smooth and polished but rough. 

The light rays reflect off the random bumps and nicks in the surface.

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Reflection, cont’d  Objects have color because the light actually

penetrates into the material.  Some of the light is reflected and some is absorbed.

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Reflection, cont’d  A white surface takes the color of whatever

light strikes the surface. 

Shining a red light onto a white piece of paper makes the paper look red.

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Reflection, cont’d  A colored surface only reflects light of the

same “color” as the surface. 

A red surface looks red when you shine white light on it. 

It appears somewhat black if you shine blue on it.

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Diffraction  Diffraction occurs when a wave passes

through a hole or a slit.  It is only observable when the opening is not too much larger than the wavelength of the wave.

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Diffraction, cont’d Since the wavelength of visible light is so small, diffraction is not noticeable through a window.  You need a very narrow slight to see the diffraction of light. 





This figure shows the result of laser light after passing through a slit 0.008 cm wide. The screen was 10 meters from the slit.

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Interference  Interference occurs when two waves overlap.  Two waves are said to be “in phase” when

their peaks overlap.  When two waves are in phase, their amplitudes add.  This is called constructive interference.

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Interference, cont’d  Two waves are said to be “out of phase”

when the peaks of one wave overlap the valleys of the other..  When two waves are out of phase, their amplitudes cancel.  This is called destructive interference.

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Interference, cont’d  Here is a diagram illustrating a two-slit

interference experiment. Light is passed through two slits.  The light strikes a screen behind the slits.  An interference pattern appears on the slit. 

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Interference, cont’d 

Bright regions on the screen indicate constructive interference. 



The light from each slit constructively interfere.

Dark regions on the screen indicate destructive interference. 

The light from each slit destructively interferes.

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Interference, cont’d 

Here is an actual photograph of two-slit interference using a laser.

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Interference, cont’d  Thin-film interference occurs when light

passes through a thin layer of one substance next to another substance. 

Here, light passes from air, through a thin layer of oil floating on a surface of water.

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Interference, cont’d  Some of the light is reflected off the top

surface of the oil.  The rest of the light passes through the oil.  Some of the transmitted light is eventually reflected off the water’s surface and passes back into the air. 22

Interference, cont’d  The ray that passes through the oil travels a

greater distance.  If the two waves emerge in step, you have constructive interference.  If the two waves emerge out of step, you have destructive interference. 23

Interference, cont’d  Important factors that determine whether the

interference is constructive or destructive include: The wavelength of the light,  The thickness of the film, and  The angle at which the light strikes the film. 

 If a single wavelength of light is used, you

obtain bright and dark areas.  If several wavelengths of light are used (white light), you obtain different colors 24

Interference, cont’d  Here are two examples of thin-film

interference using multi-wavelength light.

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Polarization  Imagine a rope attached at one end to a wall.  The other end is free for you to move.  If you move your hand from side-to-side, we

would say that the wave is horizontally polarized.

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Polarization, cont’d  If you move your hand from up and down, we

would say that the wave is vertically polarized.  Polarization is only possible with transverse waves.

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Polarization, cont’d  Since light is a transverse wave, it can be

polarized.  A Polaroid filter, absorbs light passing through it unless the light is polarized in a particular direction.

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Polarization, cont’d  If vertically polarized light is passed through a

vertical polarizer, all the light passes.  If light polarized to 45º from the vertical is passed through a vertical polarized, only some of the light passes.

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Polarization, cont’d  If horizontally polarized light is passed

through a vertical polarizer, none of the light passes through the polarizer.

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Polarization, cont’d  Most light is unpolarized. 

It contains components with random polarizations.

 If unpolarized light is passed through a

vertical polarizer, then only those components that were originally vertically polarized pass through the filter. 31

Polarization, cont’d  LCDs use

perpendicular polarizers with a liquid crystal to either allow or block light. 

The liquid crystal changes the polarization of the light. 32

Plane mirrors  Most mirrors are plane mirrors.

They are flat and almost perfect reflectors of light.  The reflection appears to originate from behind the mirror. 

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One-way mirrors  A “one-way mirror” is made by partially

coating glass so that it reflects some of the light and allows the rest to pass through.  This is called a half-silvered mirror.

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One-way mirrors, cont’d  The one-way mirror acts as a mirror for the

side that is brightly lit.  It acts as a window on the dimly lit side.  If a bright light is turned on in the dimmer room, the “one-way” effect is lost.

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Curved mirrors  A concave mirror is a mirror that curves

inward on the reflecting side.  Parallel rays that reflect off the mirror are focused at a point called the focal point.

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Curved mirrors, cont’d  A concave mirror can be used to create an

image larger than that created by a plane mirror.

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Curved mirrors, cont’d  A convex mirror is curved outward on the

reflecting side.  It creates an image smaller than a plane mirror.

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Curved mirrors, cont’d  An advantage of a convex mirror is that it has

a wide field of view. 

Images of things spread over a wide area can be viewed in it.

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Curved mirrors, cont’d  Here is an example of

the image from a convex mirror.

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Astronomical mirrors  Many large telescopes use curved mirrors.  Here is a common design. 

The large, concave mirror focus the object on the smaller, convex mirror, and then to the eye.

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Astronomical mirrors, cont’d  A problem with large mirrors is spherical

aberration. 

This occurs when not all of the rays focus at the same point.

 This can be overcome

by using a parabolic mirror instead of a spherical mirror. 42

Astronomical mirrors, cont’d  The Hubble telescope originally had an

aberration problem, as seen in these two images.

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Refraction  Imagine a light ray passing through air as it

arrives at the surface of another transparent substance, e.g., glass. 

The boundary between the air and the glass is called an interface.

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Refraction, cont’d  Part of the ray passes into the glass while the

rest reflects.  The reflected law obeys the law of reflection.  The ray that passes into the glass obeys the law of refraction.

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Refraction, cont’d  The ray that passes into the glass is called

the refracted ray.  The angle between the refracted ray and the normal is the angle of refraction.

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Refraction, cont’d  The law of refraction states that:

a light ray is bent toward the normal when it enters a transparent medium in which light travels slower.  a light ray is bent away from the normal when it enters a transparent medium in which light travels faster. 

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Refraction, cont’d  This figure illustrates the effect

as the light passes from the glass into the air. 

Since light travels faster in air than glass, the ray is bent away from the normal.

 The principle of reversibility

states that the path of a light ray through a refracting surface is reversible. 48

Refraction, cont’d

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Refraction, cont’d  Why does light slow down when it passes into

a transparent medium? 

Recall that light is an electromagnetic wave. 

It is composed of oscillating electric and magnetic fields.

As this wave passes through the medium, it interacts with the electrons.  The electrons then begin to oscillate and radiate EM waves. 

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Refraction, cont’d These emitted waves tend to lag behind the incident waves.  Because of the lag, the speed at which the light propagates through the medium is reduced. 

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Refraction, cont’d  The reason a change in speed causes the

light to bend can be seen by examining the wavefronts as they cross the interface.

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Refraction, cont’d  The wavefronts have to stay parallel after

they enter the glass.

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Refraction, cont’d  Since the light travels slower in the glass, the

wavefronts must point in a different direction to remain aligned.

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Refraction, cont’d  This has the effect of reducing the

wavelength. The speed of light is determined by the type of medium.  The frequency must remain constant. 



If you shine red light into glass, it still looks like red light.

 So our formula still holds.

v= fλ

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Refraction, cont’d  Here is a graph of the angle of incidence

passing from air into glass.

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Example Example 9.1 The figure depicts a light ray going from air into glass with an angle of incidence of 60º. Find the angle of refraction.

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Example Example 9.1 ANSWER: Looking at Figure 9.34, we can find the refracted angle since we know the incident angle. So the refracted angle is 36º.

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Total internal reflection  Consider passing light from glass to air.  The light passes from the glass into the air.  As the angle of incidence increases, there

reaches a certain angle at which light no longer passes out of the glass. 

This means all the light is reflected back into the glass.

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Total internal reflection, cont’d  This phenomenon is called total internal

reflection.  The angle at which total internal reflection begins is called the critical angle.

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Total internal reflection, cont’d  At an angle of

43º, the angle of refraction equals 90º.  So, the critical angle is 43º. 

This could be seen from Table 9.34. 61

Example Example 9.2 A homeowner wishes to mount a floodlight on a wall of a swimming pool under water so as to provide the maximum illumination of the surface of the pool for use at night. At what angle with respect to the wall would the light be pointed?

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Example Example 9.2 ANSWER: To illuminate the surface, we want the light to be refracted so that the angle of refraction is along the surface. This corresponds to total internal reflection. So we need the critical angle for water into air.

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Example Example 9.2 ANSWER: Table 9.1 gives the critical angle as 48.6º.

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Fiber optics  Optical fibers use total internal reflection. 

They are flexible, coated strands of glass through which you can pass light.

 With total internal reflection, the light passes

down the strand without passing out the sides.  So most of the light exits the strand.

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Lenses and images  A lens is a piece of glass (or other material)

specially ground to alter the paths of light rays.  Imagine a block of glass round so that one end takes the shape of a segment of a sphere.

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Lenses and images, cont’d  The optical axis is the symmetry line.  Assume that parallel rays are striking the

convex side of the glass.  The incident rays are refracted. 

Above the optical axis are refracted down, those below are refract up.

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Lenses and images, cont’d  The overall effect is that the light is focused at

a point, called the focal point. 

One can work this out from the law of refraction by considering each ray.

 So a convex surface serves to converge

incoming light.

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Lenses and images, cont’d  Next, shine light on a concave surface of

glass.  The rays diverge as the light passes into the light.

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Lenses and images, cont’d  But they diverge as if they originated at a

point outside of the glass.  So a concave surface serves to diverge

incoming light.

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Lenses and images, cont’d  A converging lens causes parallel light rays to

converge to a point, called the focal point.  The distance from the center of the lens to the focal point is called the focal length. 

If a source is placed at the focal length, the rays will emerge parallel to each other. 71

Lenses and images, cont’d  A diverging lens causes parallel light rays to

diverge, appearing to originate from a point called the focal point.

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Lenses and images, cont’d  For both types of lenses, there are two focal

points. 

One on each side.

 As a rule, converging lenses are thicker at the

center than at the ends.

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Image formation  The main use of lenses is to form images.  We can draw a model of image formation

using just three rays.  The particular rays we use are chosen because they satisfy the law of refraction in a straightforward manner.

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Image formation, cont’d  The rays are:

A ray initially parallel to the optical axis passes through the focal point on the other side of the lens.  A ray that passes through the focal point (on the same side as the object) emerges parallel to the optical axis on the other side.  A ray that passes exactly through the optical axis emerges along the optical axis on the other side. 

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Image formation, cont’d  Here is a diagram of these three rays.  Where the rays intersect on the opposite side

of the lens defines the image.

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Image formation, cont’d  The distance from the lens to the object is

called the object distance. 

Denoted by the symbol s.

 The distance from the lens to the image is

called the image distance. 

Denoted by the symbol p.

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Image formation, cont’d  If we know the object focal length and object

distance, the image distance is

sf p= s− f

( lens formula )

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Example Example 9.3 In a slide projector, a slide is positioned 0.102 meters from a converging lens that has a focal length of 0.1 meter. At what distance from the lens must the screen be placed so that the image of the slide will be in focus?

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Example Example 9.3 ANSWER: The problem gives us:

s = 0.102 m f = 0.1 m

From the lens formula:

0.102 m ) ( 0.1 m ) ( sf p= = s− f 0.102 m − 0.1 m = 5.1 m.

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Image formation, cont’d  A real image is an image that can be

projected onto a screen. 

You can see the image on the opposite side of the lens. 

The image from a magnifying glass.

 A virtual image is an image that cannot be

projected onto a screen. 

You see the image by looking into the lens. 

Similar to the “mirror image” seen looking into a plane mirror. 81

Image formation, cont’d  These figures

demonstrate the difference between real & virtual images.

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Example Example 9.4 A converging lens with focal length 10 centimeters is used as a magnifying glass. When the object is a page of fine print 8 centimeters from the lens, where is the image?

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Example Example 9.4 ANSWER: The problem gives us:

s = 8 cm f = 10 cm

From the lens formula:

8 cm ) ( 10 cm ) ( sf p= = s− f 8 cm − 10 cm = −40 cm.

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Example Example 9.4 DISCUSSION: The negative sign indicates that the image is on the same side of the lens as the object. Therefore, it is a virtual image and must be viewed through the lens.

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Image magnification  Typically, a lens produces an image that is

not the same height as the object.  The magnification, M, of a lens is the ratio of the image height to the object height:

image height M= object height 86

Image magnification, cont’d  The magnification of an object also depends

on the object distance.  An alternative expression for magnification is:

−p M= s

If p is positive (image is to the right of the lens and real), M is negative — image is inverted.  If p is negative (image is to the left of the lens and virtual), M is positive — image is upright. 

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Example Example 9.5 Compute the magnification for the slide projector in Example 9.3 and for the magnifying glass in Example 9.4. Example 9.3

s = 0.102 m f = 0.1 m p = 5.1 m

Example 9.4

s = 8 cm f = 10 cm p = -40 cm 88

Example Example 9.5 ANSWER: For the projector:

For the glass:

− p −5.1 m M= = s 0.102 m = −50 − p − ( −40 cm ) M= = s 8 cm = +5 89

Example Example 9.5 DISCUSSION: The projector image is 50 times as large as the slide and is inverted (since M is negative). The magnifying glass image is 5 times larger and upright (since M is positive).

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Aberration  Recall that spherical aberration is the effect

that rays that pass through different part of the lens are focused at different locations.

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Aberration, cont’d  Chromatic aberration occurs when a lens

illuminated with white light focuses the various colors at different locations. 

This effect is due to dispersion (more later).

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The human eye  Light passes through the opening in the iris

and forms and image on the retina. 

The cornea and lens acts as a single converging lens.

 The eye muscles

change the thickness of the lens to change the focal length. 93

The human eye, cont’d  The eye muscles change the thickness of the

lens to change the focal length. A thinner lens is used for near objects.  A thicker lens is used for farther objects. 

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The human eye, cont’d  Myopia (nearsightedness) causes the lens to

focus distant objects in front of the retina.  A diverging lens corrects the problem.

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The human eye, cont’d  Hyperopia (farsightedness) causes the lens

to focus distant objects behind of the retina.  A converging lens corrects the problem.

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Dispersion and color  Dispersion is a phenomenon in which the

speed of light through a medium depends on the frequency of the light. 

Dispersion is why passing a white light through a prism causes the individual colors to be separated. 

The different frequencies are refracted at differing angles. 97

Dispersion and color, cont’d  The different frequencies are refracted at

different angles. 

Shorter wavelengths are refracted more than longer wavelengths.

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Dispersion and color, cont’d  This is because (in most materials), violet

light travels slower through the material than red light. 

In common glass, the speed of violet light is 1.95×108 m/s while for red light it is 1.97×108 m/s.  This difference in speed is only about 1%.

 In diamond, the difference in speeds is closer to

2%. 

This explains the “brilliance” of diamond.

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Atmospheric Optics  A rainbow occurs because of the dispersion

of sunlight through water droplets in the atmosphere.  Some general observations are: Rainbows are arcs of colored light, with red on the outside of the box and violet on the inside.  Rainbows are always seen against a background of water droplets, typically with the Sun toward your back. 

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Atmospheric Optics, cont’d  A secondary rainbow results from a similar

process.

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Atmospheric Optics, cont’d  A halo is a circular arc of

light surrounding the Sun or full Moon in winter.  When the temperature in the upper atmosphere drops below freezing, ice crystals form.

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Atmospheric Optics, cont’d  The blueness of the sky is caused by air

molecules scattering sunlight in all directions.  The oscillating electric field of an EM wave passing through the air causes the electrons of air molecules to oscillate.  As discussed, these electrons emit EM radiation.  This emitted light travels outward in all directions — thus the term scattered. 103

Atmospheric Optics, cont’d  The reason the sky is blue is because the air

molecules are more efficient at absorbing and radiating higher frequencies of light.  The amount of radiant energy scattered per second can be expressed as

Escattered 1 ∝ 4 t λ 

λ is the wavelength of the scattered light. 104

Atmospheric Optics, cont’d  The wavelength of blue light is about 0.7

times that of red light.  So there is about 4.2 (= 1/0.7^4) times as much blue in scattered sunlight as there is red.

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