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Eðlisfræði 2, vor 2007
33. The Nature and Propagation of Light
Assignment is due at 2:00am on Wednesday, January 17, 2007
Credit for problems submitted late will decrease to 0% after the deadline has passed.
The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.
The unopened hint bonus is 2% per part.
You are allowed 4 attempts per answer.
The basics of light propagation and waves
Understanding the Propagation of Light
Learning Goal: To understand ray diagrams, as well as basic reflection and refraction problems.
There are two ways of indicating, in a diagram, the path that light follows. One way is by using wavefronts
(shown in blue); the other is by using rays (shown in red) . Wavefronts represent, in a schematic way, the
successive peaks of the electromagnetic wave at a specific time. Light is a
transverse wave; it moves perpendicular to the wavefronts. Rays are used to
indicate the direction of motion of the light. Ray diagrams are typically used in
problems where the wave nature of light is not important, as will be the case in
geometric optics.
Notice in the diagram that the wavefronts get closer together inside of the glass.
This is because the speed of light in glass is less than that in air. The frequency
of a wave does not change when it propagates through different media, even
though its speed may change. (Waves can be neither created nor destroyed at the
boundary between different media; hence, the number of waves that strike the boundary per unit time must equal the
number of waves that leave the boundary per unit time.)
Let be the wave's speed, its wavelength, and its frequency. These quantities are related via the equation
Note that, if the wave speed decreases, the wavelength must also decrease for the frequency to remain constant.
.
Part A
What is the wavelength
glass is ?
of light in glass, if its wavelength in air is , its speed in air is , and its speed in the
Hint A.1 How to approach the problem
Hint not displayed
, , and .
Part B
Part not displayed
Part C
Part not displayed
Two important things happen to light when it strikes a transparent boundary: It gets reflected and it gets refracted.
When you see your reflection in glass, you are seeing the result of reflection from a transparent boundary. In the
figure , the ray moving toward the air/glass interface is called the incident ray. The ray leaving the boundary in air
is called the reflected ray. The ray leaving the boundary inside the glass is
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called the refracted ray.
Reflection from a mirror and reflection from a transparent boundary both obey
the law of reflection:
, where is the angle of incidence (the angle
between the incoming ray and the normal to the surface), and is the angle of
reflection (the angle between the normal line and the reflected ray) .
Part D
If light strikes the air/glass interface at an angle 32.0
to the normal, what is the angle of reflection, ?
The second important effect of light striking a transparent boundary is refraction. Refraction is the bending of light
caused by the difference in the speed of light between materials. When light moves into a medium with a higher
index of refraction (i.e., lower speed of light), the refracted ray has a smaller angle to the normal than the incident
ray. Snell's law gives this angle of refraction, :
. Since we are assuming that the speed of light in
air is very close to the speed of light in vacuum, you will use
in this
problem.
Part E
If light strikes the air/glass interface at an incidence angle of 32.0
1.50 for the index of refraction of glass.
, what is the angle of refraction, ? Use
Wavelength, Frequency, and Speed of Light in Different Media
A beam of light from a monochromatic laser shines into a piece of glass. The glass has a thickness and an index
of refraction . The wavelength of the laser light in vacuum is
and its frequency is . In this problem, the
Part A
How long does it take for a short pulse of the light to travel from one end of the glass to the other?
Hint A.1 How to approach the problem
Hint not displayed
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Hint A.2 The speed of light
Hint not displayed
Part A.3 Find the speed of light in the glass
Part not displayed
Express the time
in terms of
and .
An exercise reviewing geometry and reflections, followed by problems in reflection and refraction
Geometry and Reflections
Learning Goal: To learn and practice the geometry skills necessary for complex reflection setups.
The law of reflection has the very simple form
,
where
is the angle between the normal and the incident ray and
is the angle between the normal and the
reflected ray. Although the law itself is easy to use, many realistic situations involve successive reflections from
multiple surfaces. The law of reflection does not become any more complicated in such cases, but the geometry of
the rays does become complicated. Consider the case of light shining onto a mirror, which is attached to another
mirror at some angle , as shown in the figure . In this problem, we will find the angle at which light leaves the
arrangement of two mirrors.
Part A
If the light strikes the first mirror at an angle , what is the reflected angle ?
.
=
Part B
Now, find the angle (shown in the new figure ) in terms of . You can easily find
substitute your expression from Part A.
Hint B.1 Relationship between
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in terms of , then just
and
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listed for you, so just use the number "23" to indicate 23 degrees.
=
Part C
Now, find the angle
shown in the figure in terms of
and .
Hint C.1 Angles in a triangle
Hint not displayed
and .
=
Part D
Find the angle
picture.
shown in the figure in terms of
Hint D.1 Relationship between
and . You will need to assume that
, as it appears in the
and
Hint not displayed
Hint D.2 How to find
in terms of
Hint not displayed
and .
=
Virtually any reflection problem, no matter how intimidating it may seem, can be broken down into simple
parts by considering each individual reflection carefully.
Diffuse Reflection
The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called
specular reflection). However, you've likely been told that when you look at something, you are seeing light
reflected from the object that you are looking at. This is reflection of a different sort: diffuse reflection. In this
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problem, you will see how diffuse reflection actually arises from the same law of reflection that you are accustomed
to for reflections from mirrors.
Part A
Consider a spotlight shining onto a horizontal mirror . If the light from the spotlight strikes the mirror at an
angle to the normal, what angle to the normal would you expect for the
reflected rays?
.
This simple rule of reflection no longer seems to hold for diffuse reflection. Consider the same spotlight but now
reflecting from the surface of a table . Unlike the light reflected from the mirror, the light reflected from the table
seems to go in all directions. If it didn't, then you'd only be able to see tables
when you were at a specific angle to the lights above you! To understand why
the light reflects in all directions, you must first look at a slightly simpler
problem.
Consider a flat surface, inclined downward from the horizontal by an angle
The red line represents the surface and the
red dotted line indicates the normal to this
surface (the normal line). The two blue
dashed lines represent horizontal and
vertical. The angle between the incoming
ray and the vertical is . Throughout this
problem, assume that is larger than
but smaller than . (If you wish, you can
determine the correct sign rules to
.
Part B
Find the angle
between the reflected ray and the vertical.
Hint B.1 How to approach the problem
Hint not displayed
Part B.2 Find the angle between the normal line and vertical
Part not displayed
Part B.3 Find the angle between the incoming ray and the normal line
Part not displayed
Part B.4 Find the angle between the normal line and the reflected ray
Part not displayed
Express the angle between the reflected ray and the vertical in terms of
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and .
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Part C
Part not displayed
Underwater Optics
Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently,
scuba divers wear masks to allow them to form images properly underwater. However, this does affect the
perception of distance, as you will calculate.
Consider a flat piece of plastic (index of refraction ) with water (index of refraction ) on one side and air (index
of refraction ) on the other. If light is to move from the water into the air, it will be refracted twice: once at the
water/plastic interface and once at the plastic/air interface.
Part A
If the light strikes the plastic (from the water) at an angle , at what angle
the air)?
does it emerge from the plastic (into
Hint A.1 Angles inside the plastic
Hint not displayed
Hint A.2 Important theorem from geometry
Hint not displayed
Part A.3 Find the angle in the plastic
Part not displayed
Express your answer in terms of , , , and . Remember that the inverse sine of a
number should be entered as a s i n ( x ) in the answer box.
Humans estimate distance based on several different factors, such as shadows and relative positions. The most
important method for estimating distance, triangulation, is performed unconsciously. Triangulation is based on the
fact that light from distant objects strikes each eye at a slightly different angle. Your brain can then use that
information to determine the angle as shown in the figure . In the figure, points L and R represent your left and
right eyes, respectively. The distance between your eyes is , and the distance
to the object, point O, is .
Part B
What is the distance to the object in terms of
and ?
and .
Part C
Part not displayed
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Your eyes determine by assuming that and (in the figure) are equal. This is true, unless the light rays are
bent before they reach your eyes, as they are if you're wearing a scuba mask underwater.
Underwater, the situation changes, as shown in the figure . Your eyes will calculate an apparent distance
the angle that reaches your eyes, instead of the correct geometric angle .
This is the same that you calculated in Part A. Note that there are no
important geometric considerations arising from the refraction except the
substitution of for , because the refraction takes place so close to your
eyes. If the problem discussed someone looking out of the porthole of a
submarine, the geometry would become more complicated.
using
Part D
Find the ratio
. Since we are dealing with small angles, you may use the approximations
and
.
Part D.1 Use the small-angle approximations
Part not displayed
Part D.2 Find
Part not displayed
and
.
Part E
Part not displayed
A Sparkling Diamond
A beam of white light is incident on the surface of a diamond at an angle . Since the index of refraction depends
on the light's wavelength, the different colors that comprise white light will
spread out as they pass through the diamond. The indices of refraction in
diamond are
for red light, and
for blue light. The surrounding
air has
. Note that the angles in the figure are not to scale.
Part A
Calculate
, the speed of red light in the diamond. To four significant figures,
.
Part B
Calculate
, the speed of blue light in the diamond. To four significant figures,
.
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Part C
Derive a formula for , the angle between the red and blue refracted rays in the diamond.
Part C.1 Apply Snell's law for blue light
Part not displayed
Part C.2 Apply Snell's law for red light
Part not displayed
Part C.3 Compare the angles
Part not displayed
Express the angle in terms of
,
, and . Use
. Remember that the proper way to
enter the inverse sine of
in this case is a s i n ( x ) .
Part D
Part not displayed
Part E
Now consider , the angle at which the blue refracted ray hits the bottom surface of the diamond. If is larger
than the critical angle , the light will not be refracted out into the air, but instead it will be totally internally
reflected back into the diamond. Find .
Part E.1 Find the refracted angle when
Part not displayed
Part E.2 Apply Snell's law
Part not displayed
Part F
Part not displayed
Is Light Reflected or Refracted?
When light propagates through two adjacent materials that have different optical properties, some interesting
phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels
from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly
reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a
lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium. The
index of refraction of a material, denoted by , gives an indication of the speed of light in the material. It is defined
as the ratio of the speed of light in vacuum to the speed in the material, or
.
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Part A
When light propagates from a material with a given index of refraction into a material with a smaller index of
refraction, the speed of light
Hint A.1 Index of refraction
The index of refraction of a material is defined as the ratio of the speed of light in vacuum to the speed in
that particular material, or
.
Since it is the ratio of two positive quantities that have the same units, the index of refraction is a pure (positive)
number. Note that the speed of light in a certain material is inversely proportional to the index of refraction of
that material.
increases.
Part B
What is the minimum value that the index of refraction can have?
Hint B.1 How to approach the problem
Hint not displayed
between 0 and 1
The index of refraction of a material is always a positive number greater than 1 that tells us how fast the light
travels in the material. The greater the index of refraction of a material, the more slowly light travels in the
material.
An example of reflection and refraction of light is shown in the figure. An incident ray of light traveling in the
upper material strikes the interface with the lower material. The reflected ray
travels back in the upper material, while the refracted ray passes into the lower
material. Experimental studies have shown that the incident, reflected, and
refracted rays and the normal to the interface all lie in the same plane.
Moreover, the angle that the reflected ray makes with the normal to the
interface, called the angle of reflection, is always equal to the angle of
incidence. (Both of these angles are measured between the light ray and the
normal to the interface separating the two materials.) This is known as the law
of reflection.
The direction of propagation of the refracted ray, instead, is given by the angle that the refracted ray makes with
the normal to the interface, which is called the angle of refraction. The angle of refraction depends on the angle of
incidence and the indices of refraction of the two materials. In particular, if we let be the index of refraction of
the upper material and the index of refraction of the lower material, then the angle of incidence, , and the angle
of refraction, , satisfy the relation
.
This is the law of refraction, also known as Snell's law.
Part C
Now consider a ray of light that propagates from water (
) to air (
). If the incident ray strikes the water-air
interface at an angle
, which of the following relations regarding the angle of refraction, , is correct?
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Part C.1 Find an expression for the ratio of the sines of
and
Part not displayed
When light propagates from a certain material to another one that has a smaller index of refraction, that is,
, the speed of propagation of the light rays increases and the angle of refraction is always greater than the
angle of incidence. This means that the rays are always bent away from the normal to the interface separating
the two media.
Part D
Consider a ray of light that propagates from water (
) to glass (
). If the incident ray strikes the
water-glass interface at an angle
, which of the following relations regarding the angle of refraction is
correct?
Part D.1 Find an expression for the ratio of the sines of
Let the index of refraction of water be
of the sine of to the sine of .
and
and that of glass be . Use Snell's law to find an expression for the ratio
Express your answer in terms of some or all of the variables
,
, and
.
=
Now, note that for the water-glass interface
. Therefore,
.
When light propagates from a certain material to another one that has a greater index of refraction, that is,
, the speed of propagation of the light rays decreases and the angle of refraction is always smaller than the
angle of incidence. This means that the rays are always bent toward the normal to the interface separating the
two media.
Part E
Consider a ray of light that propagates from air (
) to any one of the materials listed below. Assuming that the
ray strikes the interface with any of the listed materials always at the same angle , in which material will the
direction of propagation of the ray change the most due to refraction?
Hint E.1 How to approach the problem
Hint not displayed
Part E.2 Find an expression for the sine of the angle of refraction
Part not displayed
ice (
) water (
) turpentine (
) glass (
) diamond (
)
The greater the change in index of refraction, the greater the change in the direction of propagation of light. To
avoid or minimize undesired bending of the light rays, light should travel through materials with matching
indices of refraction.
Is light always both reflected and refracted at the interface separating two different materials? To answer this
question, let's consider the case of light propagating from a certain material to another material with a smaller
index of refraction (i.e.,
).
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Part F
In the case of
, if the incidence angle is increased, the angle of refraction
Hint F.1 How to approach the question
Recall that, according to Snell's law, the sine of the angle of refraction is directly proportional to the sine of the
angle of incidence. Thus, as the angle of incidence is increased, the angle of refraction changes accordingly.
Moreover, since the angle of refraction is greater than the angle of incidence, as you found in Part C, the angle of
refraction can reach its maximum value sooner than the angle of incidence.
increases up to a maximum value of 90 degrees.
Since the light is propagating into a material with a smaller index of refraction, the angle of refraction, , is
always greater than the angle of incidence, . Therefore, as is increased, at some point will reach its
maximum value of 90 and the refracted ray will travel along the interface. The angle of incidence for which
is called the critical angle . For any angle of incidence greater than , no refraction occurs. The ray
no longer passes into the second material. Instead, it is completely reflected back into the original material.
This phenomenon is called total internal reflection and occurs only when light encounters an interface with a
second material with a smaller index of refraction than the original material.
Part G
What is the critical angle
for light propagating from a material with index of refraction of 1.50 to a material
with index of refraction of 1.00 ?
Part G.1 Find an expression for the sine of the angle of incidence
Part not displayed
= 0.730
In conclusion, light is always both reflected and refracted, except in the special situation when the conditions
for total internal reflection occur. In that case, there is no refracted ray and the incident ray is completely
reflected.
Scattering and polarized light
The process of scattering of light by a molecule (Rayleigh scattering) is an important physical phenomenon.
Instead of thinking of scattering as light simply bouncing off the molecule, one should think of it as an absorption
The probability for light to be scattered is proportional to the inverse of the wavelength to the fourth power,
This means that the shorter wavelengths (toward blue) get scattered more strongly than the longer wavelengths
(toward red).
.
Rayleigh scattering can explain why the daytime sky looks blue, the sunset looks red, and clouds are white. In the
afternoon you observe mostly scattered light (blue); in the evening you see mostly transmitted light (red). The
clouds have higher concentration of water and ice droplets. This means that light gets rescattered many times and all
wavelengths get a chance to scatter out of the clouds, adding up to white light.
Another effect that can be explained by light scattering is polarization. When you look at the sky with Polaroid
sunglasses it appears darker or brighter from different angles. This is because the scattered light is partially
polarized. The white light scattered from the clouds is unpolarized, because the light scatters randomly, multiple
times. The direction of its polarization becomes random and thus the light is unpolarized. This effect can be useful
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for making dramatic photographs of the sky.
Consider a photographer who wants to take a picture of an interesting cloud formation. To increase the ratio of the
clouds' intensity to that of the blue sky the photographer uses a polarizing filter.
Part A
How would the photographer use the polarizing filter to find out the direction of polarization of the light coming
from the blue sky? Her only reference is the polarization axis of the filter.
Rotate the filter until the light's intensity is minimum; light's polarization is along filter's
axis.
Rotate the filter until the light's intensity is maximum; light's polarization is along filter's
axis.
Part B
Find the angle between the filter's polarizing axis and the direction of polarization of light necessary to increase
the ratio of the clouds' intensity to that of the blue sky so that it is three times the normal value.
Part B.1 Find the intensity of light from the sky through the polarizing filter
What is , the intensity of light from the blue sky after it passes through the photographer's polarizing filter
with the polarizing axis at an angle to the direction of the light's polarization? The intensity of light from the
sky before it passes through the filter is .
Hint B.1.a Polarizers and electric fields
Hint not displayed
Hint B.1.b Definition of intensity
Hint not displayed
and .
=
Part B.2 Find the intensity of light from the the clouds through the polarizing filter
What is , the intensity of light from the clouds after it passes through the photographer's polarizing filter? The
intensity of light from the clouds before it passes through the filter is . Assume that light from the clouds in
unpolarized.
Hint B.2.a Components of randomly polarized light
When light is described as unpolarized this means that it has a random polarization, that is, no preferred axis.
Randomly polarized light has an equal chance of having components parallel to or perpendicular to the
polarizing axis of the filter.
.
=
Hint B.3 Ratio
You need to find an angle
at which
. Substitute expressions for
and that you have found.
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Birefringence in Calcite
Calcite (
) is a crystal with abnormally large birefringence. The index of refraction for light with electric field
parallel to the optical axis (called extraordinary waves or e-waves) is 1.4864. The index of refraction for light with
electric field perpendicular to the optical axis (called ordinary waves or o-waves) is 1.6584.
Part A
Find the critical angle
for e-waves in calcite.
Hint A.1 Snell's law
Hint not displayed
Hint A.2 Definition of critical angle
Hint not displayed
Part B
Part not displayed
Part C
Part not displayed
Part D
Part not displayed
Part E
Part not displayed
Electromagnetic radiation is emitted when a charged particle moves through a medium faster then the local speed of
light (which is always lower then the speed of light in vacuum). This radiation is known as Cerenkov radiation.
Cerenkov radiation is found in many interesting places such as particle detectors and nuclear reactors and can even
be seen by astronauts when cosmic rays traverse their eyes. It should be stressed that the particle is never going
faster then the speed of light in vacuum (or ), just faster then the speed of light in the material (which is always
less then ). The creation of Cerenkov radiation occurs in much the same way that a sonic boom is created when a
plane is moving faster then the speed of sound in the air. The various wavefronts that propagate in the material add
coherently to create an effective shock wave. In this problem you will become familiar with this type of radiation
and learn how to use its properties to get information about the particles that created it.
Part A
What is the threshold velocity
(which has an index of refraction
for creating Cerenkov light of a charged particle as it travels through water
)?
Part B
What is the threshold velocity
(with index of refraction
)?
for creating Cerenkov light of a charged particle as it travels through ethanol
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to three significant figures.
to three significant figures.
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Part C
Which of the following best explains why neutrally charged particles can't give off Cerenkov radiation?
When a charged particle passes straight through a medium faster than the local speed of light, it will emit the
Cerenkov radiation in a cone. Next we will calculate how the cone angle is correlated to the speed of the particle.
Part D
If a particle is traveling straight through a material with index of refraction at a speed , what is the angle that
the cone of light makes with the particle's trajectory? In other words what is the angle between the vector of the
propagating Cerenkov radiation and the vector in the direction of the propagating particle?
Hint D.1 Geometry of the problem
Hint not displayed
Hint D.2 Using the geometry
Hint not displayed
Ring-imaging Cerenkov detectors are devices that can accurately measure the velocity of charged particles as they
pass through them. They are very useful as subdetectors in large particle detector systems. Let us look at what
decisions need to go into designing a Cerenkov detector.
Part E
Suppose you wish to accurately measure the speed of high energy particles with velocities greater then 98% the
speed of light in vacuum. You can use a ring-imaging Cerenkov detector consisting of a thin slab of material
separated from an array of photomultiplier tubes (devices used to detect weak light signals) by an arbitrary open
space. The detector works on the principle that the Cerenkov radiation emitted in the thin slab will be a cone of
light that can be measured with the array of photomultiplier tubes. Your photomultiplier tubes, having a finite
width, can only resolve a finite change in the angle of the ring created by the Cerenkov radiation. Use these
constraints and the equation for from Part D to determine which of the
following substrate materials is best suited to giving you the greatest
precision in determining particle velocity.
Hint E.1 Using example velocities
Hint not displayed
Part F
Part not displayed
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Part G
Part not displayed
Summary
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3 of 10 problems complete (24.89% avg. score)
9.38 of 12 points
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