| 1. |
List 5 examples of situations you have encountered outside
of the classroom that exemplify the law of reflection. You do not
have to restrict your answers to reflection of light but can include reflection
of any object, such as a pool ball banking off the side of a pool table.Student responses will
vary widely. Many may involve balls: a basketball shot made
off the backboard, a tennis ball bouncing on the court or off a racket,
a volleyball being passed from one player to another, pool balls bouncing
off each other and off the sides of the table, etc. Others may involve
reflection of light: adjusting rearview and side mirrors in a car
to see particular parts of the road, a dentist's mirror, etc. Some
students may also think of the reflection of sound (such as echoes) or
of water waves reflecting from the side of a pool. The point of the
question is to point out the myriad of situations in which the Law of Reflection
holds.
|
| 2. |
Two famous British scientists, Isaac Newton and Robert
Hooke, debated the nature of light in the 1600s. Newton claimed light
was composed of tiny "corpuscles", or particles. Hooke claimed light
was a continuous wave. Think about the behavior of light when it
is reflected by (a) by a smooth surface, and (b) by a rough surface.
In each case, is the behavior of light more like particles, more like a
wave, or explained equally well by either theory? Justify your answer.
Newton's theory of light
is nicely explained in the book Einstein and Bohr found in the References
page. Students are not expected to have familiarity with the arguments,
but hopefully they will recognize that both particles and waves obey the
law of reflection from both smooth and rough surfaces.
|
| 3. |
We see objects when light reflected by them reaches our
eyes. Do you think this reflection by most objects is total reflection
or diffuse reflection? Explain.
Since most objects are
visible from all directions, light is scattered by them in all directions.
Thus the reflection is diffuse.
|
| 4. |
Laser light is generally not visible as it travels through
air. (If you have access to a laser or to a laser pointer, verify
this for yourself.) Yet if you shine a laser through chalk dust,
the beam is visible. Explain why this occurs.
Chalk dust provides multiple
surfaces from which the laser beam can reflect. Since the laser beam
will strike dust particles at a wide range of incident angles, the light
will be diffusely reflected and thus visible from a variety of angles.
Without the dust, no reflection occurs, so no laser light reaches the observer.
|
| 5. |
If a laser beam is sent across a classroom, only students
in the direct line of the beam would be able to see that the laser is shining.
(Do NOT try to verify this - you should NEVER look into a laser!)
But if the beam strikes a wall, the entire class will be able to see the
spot made by the beam on the wall. Explain why this occurs.
Laser light is by definition
well-columnated, so it will not be emitted toward students other than those
in the direct path of the beam. The wall, however, scatters the light
in all directions due to diffuse reflection. (An apparently smooth
wall looks very bumpy to a laser beam.) The reflected light is thus
visible to the entire class as a spot on the wall.
|
| 6. |
 |
A scientist looking into a flat mirror hung perfectly
perpendicular to the floor cannot see her feet but can only see down to
the hem of her labcoat. Will she be able to see her feet if she backs
away from the mirror? What if she moves toward the mirror?
A drawing of light rays may help you explain your answer.
This one will be tricky
for students who have not studied ray diagrams in more detail than this
material provides. The scientist's view is limited by what light
can reflect off the bottom of the mirror and reach her eye as constrained
by the law of reflection. In this problem, the bottom of the mirror
is located halfway between the labcoat hem and the eye. In the ray
diagram to the left below, the (red) light ray from the hem of the coat
is reflected by the mirror into the scientist's eye. The (blue) light
ray from her shoe, however, is reflected above her head. As the scientist
moves back, the mirror will continue to be vertically located halfway between
the hem and the eye, and both the eye and the hem will continue to be equal
distances from the mirror. Thus the angle of light incident on the
mirror from the hem and the angle of reflection by the mirror to the eye
will change together. The scientist will not be able to see any more
of herself as she backs up. The ray diagram on the right below shows
that the ray from the hem still reaches the eye but the ray from the shoe
still reflects over her head. Due to the difficulties in drawing
accurate ray diagrams for two-dimensional images, students might claim
that the diagram on the right shows more of the scientist's image in the
mirror. But one must carefully draw the light rays from each point
on the image to get an accurate picture. Such treatment should show
that no more and no less of the scientist is imaged as she moves back from
the mirror.
 
|
|
| 7. |
A stream of tennis balls striking a metal plate will
exhibit total reflection, while the same stream of tennis balls reflecting
off of an old, cracked sidewalk will exhibit diffuse reflection.
What characteristic(s) of a surface distinguish(es) whether tennis balls
exhibit diffuse or total reflection when striking that surface?
If the irregularities in
a surface are close to the same size as the tennis ball (greater than a
few percent of the part of the ball in contact with the surface), they
can influence the average force on the ball and so visibly affect the direction
it travels. Smaller bumps will have a less significant effect overall.
The cracks in the sidewalk are big compared with tennis balls, so whether
or not a ball lands on the crack makes a big difference in its final trajectory.
The irregularities in a metal plate are generally much tinier than a tennis
ball, so the balls will all see the surface as smooth.
|
| 8. |
A stream of tennis balls striking a concrete wall will
exhibit total reflection while a laser beam of light striking the same
wall will be scattered in all directions. What characteristic(s)
of an object distinguish(es) whether that object exhibits diffuse or total
reflection when striking a given surface?
This is similar to the
previous question. Since the tennis balls are so much bigger than
the irregularities in a concrete wall, each tennis ball will have approximately
the same angle of incidence (when averaged over the portion of the ball
in contact with the wall). Light, on the other hand, interacts at
a much smaller scale. The wavelength of light provides a general
scale for interactions, and the irregularities in a concrete wall are generally
much larger than the wavelengths of visible light. Thus the individual
interactions of light with the wall will occur at different angles of incidence,
leading to diffuse reflection.
|
| 12. |
What are the physical limits on the index of refraction?
(i.e., what values of n are physically impossible to achieve?)
Explain your answer.
Since the speed of light
v
in any medium must always be less than or equal to the speed of light in
vacuum c, the index of refraction n = c/v will always
be greater than or equal to 1.0. Some students may include negative
indexes due to the mention of left-handed materials. The reading directs
students to consider only positive indices, but an intelligent inclusion of
negative indices could be acceptable here.
|
| 13. |
According to the theory of relativity, information cannot
move between two points any faster than c, the speed of light in
vacuum. Yet a shadow can move much faster than c. (a)
Explain how a shadow can move faster than c, the speed of the light
that causes the shadow. (b) How does the shadow moving faster
than c not violate the limit on information transfer, when a shadow
could conceivably carry information?
(a) Imagine a (very)
high-powered, wide-angle flashlight with a beam that originates 10.0 cm
in diameter but spreads to 20 meters in diameter at a distance of 10 meters
from the lamp. You shine this super-flashlight at a large white wall
10 m away, then pass your finger in front of the lamp. If your finger
moves at a speed of 1 m/s, it passes completely through the 10.0 cm beam
in 0.10 seconds. On the wall, the shadow of your finger also passes
through the beam in 0.10 seconds. But the shadow must travel through
a distance of 20 meters in that 0.10 seconds. The shadow therefore
travels at 200 m/s, much faster than your finger which produces the shadow.
If, instead of your finger, a high-speed bullet passes through the light
beam, the resulting shadow will move even faster. A bullet traveling
at 1000 m/s will require only 0.0001 seconds to cross the 10.0 cm beam,
and the shadow on the wall will therefore move 20 meters in 0.0001 seconds.
That's a speed of 200,000 m/s!
The speeds of the shadow in the
above example are still well below c, but you can extend this analogy
to a wall 20 m away (where the beam is 40 m in diameter) or to a wall 2,000,000
m away. Of course, a flashlight that would have measureable intensity
over a wide angle at that distance would blind individuals near it, but
one could still imagine the situation. It is, after all, analagous
to a star that radiates uniformly in all directions with an appreciable
intensity at exceptionally far distances. A planet (or a giant cloud
of interstellar dust) passing between that star and the observer would
cast a shadow, however small, that could move faster than c at the
observer's location.
(b) A shadow moving faster
than c does not violate the limit on information transfer, because
the shadow does not occur on the wall at the exact time that the finger
passes through the beam. When the finger first breaks the beam, light
(emitted before the finger entered the beam) is still traveling to the
wall. The "information" contained in the shadow, i.e., the
finger's presence in the beam or the absence of light from the beam, cannot
be "transferred" to the wall in less time that it takes light to travel
from the finger to the wall.
|
| 14. |
Two famous British scientists, Isaac Newton and Robert
Hooke, debated the nature of light in the 1600s. Newton claimed light
was composed of tiny "corpuscles", or particles. Hooke claimed light
was a continuous wave. Think about the behavior of light when it
travels from one medium to another. In particular, consider (a) the
slowing of light, and (b) the refraction, or bending, of light. In
each case, is the behavior of light more like particles, more like a wave,
or explained equally well by either theory? Justify your answer.
Newton's theory of light
is nicely explained in Sachs' book Einstein and Bohr found in the
References page. Students are not expected to have familiarity with
the arguments, so this question should be evaluated on the basis of critical
thinking skills demonstrated rather than an agreement with Newton's arguments.
Newton argued that refraction occurs because of interactions between the
corpuscles of light and particles of the medium through which it travels.
He expected that light would interact more in denser mediums, and predicted
that light would therefore speed up due to the stronger forces of
the denser medium. This produces a bending of light opposite that
which is observed, which Newton never did successfully explain, according
to Sachs.
|
| 15. |
The frequency f of a beam of light is related
to the speed v and the wavelength l of
the light as follows: v = lf.
How will this frequency change when the light moves from a denser medium
to a rarer medium?
One could approach this
question several ways. The fact that frequency does not change is
mentioned in the reading assignment (under the wavelength equation on the
page Refraction of Light),
so students could just cite that page. An optional link
from that statement explains that this lack of change in the frequency
is due to conservation of energy at the surface. Alternatively, the
student could use the relationships describing how speed changes and how
wavelength changes to show mathematically that frequency does not change.
|
| 16. |
Two famous British scientists, Isaac Newton and Robert
Hooke, debated the nature of light in the 1600s. Newton claimed light
was composed of tiny "corpuscles", or particles. Hooke claimed light
was a continuous wave. Think about the behavior of light when it
approaches total internal reflection. For angles of incidence slightly
less than the critical angle, light is partially reflected and partially
transmitted. Is the behavior of light in this situation more like
particles, more like a wave, or explained equally well by either theory?
Justify your answer.
Newton's theory of light
is nicely explained in Sachs' book Einstein and Bohr found in the
References page. Students are not expected to have familiarity with
the arguments, so this question should be evaluated on the basis of critical
thinking skills demonstrated rather than an agreement with Newton's argments.
The partial reflection could be explained by particle theories if one views
the interface between the media as porous, so that particles either make
it through the open portions of the interface or reflect from closed portions.
The wave explation is more subtle, requiring that a certain percentage
of any wave be transmitted and the rest reflected.
|
| 17. |
You have a glass beaker full of an unknown liquid.
How might you determine the liquid's composition using only a laser, a
protractor, a ruler, a pencil, and a reference guide containing optical
properties of various liquids?
The possibilities are endless,
but a critical angle experiment is likely to yield the best results.
But one needs to recognize the measured critical angle applies to a glass/liquid
interface, not to an air/liquid interface.
|
| 18. |
Why couldn't you use a square piece of glass to measure
the critical angle of the glass? A diagram will help explain your
answer.
 |
The opposing sides of a square
piece of glass are parallel, so the angle of refraction as light enters
the glass (q2
in diagram) equals the angle of incidence as light exits the glass (q1').
By symmetry then, the angle of incidence as light enters the glass (q1)
equals the angle of refraction as light exits the glass (q2').
If total internal reflection were achieved as the light exits the glass,
(q1') would
be undefined and therefore (q1)
would be undefined as well. So you couldn't get light into a square
piece of glass that would produce TIR on the opposite side. |
|
| 19. |
Light travels from a medium with n = 1.25 into
a medium of n = 1.34, at an angle of 27°
from the normal to the interface of the two media. (a) Will
the speed of the light increase, decrease, or remain the same? (b)
Will the wavelength of the light increase, decrease, or remain the same?
(c) Will the light bend toward the normal, away from the normal, or not
at all?
(a) Decrease.
Light travels more slowly in a denser medium (a medium with a higher index
of refraction).
(b) Decrease. As the
light slows down, the wavefronts "pile up" and the wavelength decreases.
(c) Toward the normal.
Snell's Law shows that light makes a smaller angle in the medium with the
higher index, unless the light enters along the normal or total internal
reflection occurs. We are told that the first does not occur, and
the second is not possible when traveling from a rarer to a denser medium.
|
| 20. |
Light travels from a medium with n = 1.63 into
a medium of n = 1.42, along the normal to the interface of the two
media. (a) Will the speed of the light increase, decrease,
or remain the same? (b) Will the wavelength of the light increase,
decrease, or remain the same? (c) Will the light bend toward the
normal, away from the normal, or not at all?
(a) Increase.
Light travels more quickly in a rarer medium (a medium with a lower index
of refraction).
(b) Increase. As the
light speeds up, the wavefronts spread out and the wavelength increases.
(c) Not at all. Snell's
Law shows that light traveling along the normal (with an angle of incidence
of 0°) is not refracted. Qualitatively, the entire wavefront
changes speed together, so no bending occurs.
|
| 21. |
 |
Light incident on a crown glass prism makes
an angle of 23.0° with the normal to the
surface, as shown to the left. The prism is surrounded by air.
(a) What is the angle of refraction inside the glass? (b) What is
the angle of reflection at the surface? |
(a) 14.9°.
For light traveling from air to crown glass, n1
= 1.00 and n2 = 1.52. The angle
of incidence q1
is 23.0°, so Snell's Law requires
(1.00) sin 23.0° = (1.52) sin
qq2,
so
q2
= sin-1[(0.658) sin 23°] = 14.9°
(b) 23.0°. The angle
of reflection always equals the angle of incidence, no matter what portion
of the beam is reflected and what portion is transmitted. Some students
may believe there can be no reflection when refraction occurs, but in general
part of the beam is reflected and obeys the Law of Reflection, while part
is refracted and obeys Snell's Law.
|
| 22. |
Laser light travels through air and enters a crown glass tube.
The path of the laser in air is not visible, but the glass scatters enough light
to determine the path of the light in the tube. The angle between the
light and the surface of the tube where the light enters is found to be
54.0° inside the glass. What was the angle
between the light and the surface
of the tube in the air outside the tube?
26.7°. Since angles
are measured to the normal, the angle of refraction qq2
inside the tube is 90.0° - 54.0°, or 36.0°. n1
of air is 1.00 for air, and n2 of
the tube is 1.52. Snell's Law gives
(1.00) sin q1
= (1.52) sin 36.0°, so
q1
= sin-1[(1.52) sin 36.0°] = 63.3°.
The question asks for the angle between
light and the surface, so we must find the complement of this incident
angle: 90.0° - 63.3° = 26.7°.
|
| 23. |
Light traveling through the crown glass tube
of the previous question eventually encounters the side edge of the tube,
which is perpendicular to the surface at which the light entered the tube.
The angle of incidence on this side edge is 54.0°.
Will the light escape the glass?
No. The critical
angle for the glass/air interface iss
qc
= sin-1 (n2/n1)
= sin-1 (0.658) = 41.1°.
Since the light strikes the side edge
at an angle greater than the critical angle, total internal reflection
occurs, and the light is trapped (and so does not escape).
|
| 24. |
The crown glass tube of the previous problems is submerged in
ethyl alcohol, and the angle at which
the light enters the glass is adjusted to maintain a 54.0°
angle of incidence on the side edge of the tube. (a) Will
light striking the side of the tube with an angle of incidence of 54.0°
escape? (b) What is the lowest value of n the medium outside the
tube could have that would allow light with that 54.0° incident angle to escape?
NOTE: This is a rather contrived
experiment, since changing the medium outside of the tube also changes
the relationship between the angle at which the light enters the tube and
the angle at which it strikes the side edge of the tube. This complexity
will be dealt with in the next section on Optical Fibers.
(a) Yes. The
critical angle for the glass/alcohol interface is ;
qc
= sin-1 (n2/n1)
= sin-1 (1.36/1.52) = 63.5°.
Since the light strikes the side edge
at an angle less than the critical angle, total internal reflection does
not occur, and much of the light escapes..
(b) 1.23. To make 54.0°
the critical angle requires
n2
= 1.52 sin 54.0° = 1.23,
which is less than water and close to
nothing in our Table.
The medium surrounding the tube would have to be pretty rare to allow that
light to escape.
|
| 25. |
What interface(s) between two materials from the
Table will result in a critical angle of 39.5°?
For each pair of materials, indicate which material the light should leave and which it should enter.
From diamond to salt, or from lead to heavy flint glass. The 39.5°
critical angle requires
n2/n1=sin 39.5° = 0.636 = 1/1.57.
We are looking for two materials with indices differing by almost 60%.
The only substances in the table that have indices greater than 1.57 are
asphalt, heavy flint glass, diamond and lead. Asphalt and heavy flint
glass have indices that are about 1.57*1.04, but there is no material in the
table with an index of 1.04. Thus the light must be leaving either diamond
or lead. Diamond's index is 2.42 = 1.57*1.54, while lead's is 2.6 =
1.57*1.66. Salt has an index of 1.54, so the light could be leaving
diamond and entering salt:
qc
= sin-1 (n2/n1)
= sin-1 (1.54/2.42) = 39.5° (diamond to salt).
Heavy flint glass has a listed index of 1.65, so the critical angle for lead to heavy flint glass
appears close to the given angle:
qc
= sin-1 (n2/n1)
= sin-1 (1.65/2.6) = 39.4° (lead to heavy flint glass).
This lead-glass critical angle equals the given angle within the two-digit uncertainty of lead's index.
|
| 26. |
Cubic zirconia has an index of refraction of 2.15.
(a) Without doing any calculations, explain whether the critical
angle for a cubic zerconia-air interface will be greater than or less than
the critical angle for a diamond-air interface. (Diamond has an index
of 2.42.) (b) Calculate the critical angles for the two interfaces
and check your answer to (a).
(a) Greater than.
sin qc = (n2/n1).
The smaller the index of refraction n1,
the larger the sine of the critical angle, and the larger the critical
angle itself.
(b) 27.7° (cubic zirconia)
and 24.4° (diamond). For both cases, the light enters air, so
n2
= 1.00. For cubic zirconia,
n1
= 2.15, so
qc
= sin-1 (1.0/2.15) = 27.7°.
For diamond, n1
= 2.42, so
qc
= sin-1 (1.0/2.42) = 24.4°.
As expected, the critical angle for
cubic zirconia to air is greater than the critical angle for diamond to
air.
|
| 27. |
Sometimes numerical aperture is expressed as just the
sine of the cut-off angle: NA = sin q0max.
How is this different from the definition given in your on-line reading?
When might the two expressions be equivalent? How typical do you
think such a condition is?
The definition in the reading
uses the more general definition: NA = n0
sin q0max.
This contains the index of the material outside the end of the fiber.
The two expressions are equivalent when n0
= 1, i.e., when this outer material is air or vacuum. This
condition is common enough for many texts to assume it is the case and
use the simpler expression given in this question. While the sides
of optical fibers are coated with several protective layers, light generally
enters into the end of the fiber directly from air or vacuum.
|
| 28. |
List three advantages of using a cladding
other than air when making optical fibers.
1. A solid cladding
provides protection for the core/cladding interface, minimizing the loss
from bumps at that interface.
2. When the ratio of the
indices is close to one, the fiber has a small cut-off angle and transmits
only that light that is traveling nearly parallel to the fiber sides.
This decreases the number of modes transmitted and minimizes modal dispersion.
3. A solid cladding is easier
to cover with insulation and water-protective layers. If a layer
of air were needed between the core and the protective layers, manufacturing
the fibers would be much more difficult.
|
| 29. |
You are talking about the exciting things you are learning
in this course to a friend with a poor science background. He asks,
"Why can't you just send light down an evacuated tube with a glass coating?
After all, light travels fastest in vacuum, so this setup would seem the
best." Write down an explanation of why a glass-lined evacuated tube
would not guide light, using arguments and language your friend would understand.
Light striking the glass
coating would travel right through it, just as light passes through a window
of your house. In order for light to be guided and trapped in a tube,
the inner material must be more dense than the outer material. When
this is the case, the first edge of a light wave to leave the dense material
will speed up, turning the light wave back toward the dense material.
If everything is set up correctly, it becomes impossible for the light
wave to leave the dense material, so all the light is reflected by the
boundary.
|
| 30. |
When having lunch with a friend, you mention how intriguing
optical fibers can be. In particular, you describe the photographs
on the on-line reading page "What
a Difference a Cladding Makes". Your friend wants to know why
submerging the plastic fiber core in water causes light to escape, when
the same beam at the same angle was trapped before the core was submerged.
How do you explain this phenomenon to your friend?
When light travels from plastic to water, its
path doesn't bend as much as it does when light travels from plastic to air.
Thus light traveling from the plastic into water will be trapped in the plastic
for a smaller range of incident angles than when the light travels into air. It
turns out that angles smaller than the critical angle for the plastic-air interface
cannot physically be achieved when the light enters the fiber at the end.
When the fiber is submerged in water, however, light can hit the wall of the
fiber at an angle less than the new (smaller) critical angle.
|
| 31. |
What is the numerical aperture of a fiber with a core
index of 1.62 and a cladding index of 1.55?
NA = (n12
- n22)1/2
= [(1.62)2 - (1.55)2]1/2
= 0.471
|
| 32. |
A fiber has a numerical aperture of 0.358. (a)
What is the cut-off angle when light enters the fiber from air? (b)
What is the cut-off angle for light entering the fiber from water?
(a) For air, n0
= 1.00, and
q0max
= sin-1(NA/n0)
= sin-1[(0.358)/(1.00)] = 21.0°.
(b) For water, n0
= 1.33, and
q0max
= sin-1(NA/n0)
= sin-1[(0.358)/(1.33)] = 15.6°.
|
| 33. |
A given fiber has a cut-off angle (when light enters
the fiber from air) of 32°. (a) What is the numerical aperture
of the fiber? (b) If the core of the fiber has an index of refraction
of 1.56, what is the index of the cladding?
(a) NA = n0
sin
q0max
= (1.00) sin(32°) = 0.53.
(b) Using the calculated value of
NA, we find
n22
= n12 - NA2.
= (1.56)2 - (0.530)2
= 2.152.
n2
= 1.47.
|
| 34. |
Measuring the cut-off angle for light entering a particular
fiber from air yields a value for numerical aperture of 0.567. What
is the numerical aperture of this fiber when light enters the fiber from
water?
0.567. While it is
tempting to look at the expression NA = n0
sin q0max and
conclude that the numerical aperture increases when the value of n0
changes from 1.00 for air to 1.33 for water, this is not the case.
The numerical aperture depends only on the core and cladding indices,
according to the relationship NA = (n12
- n22)1/2,
and is thus a property of the fiber, independent of how the fiber is used.
When he value of n0 increases, the
value of sin q0max
must decrease accordingly to keep the product constant.
|
| 35. |
A given fiber has a numerical aperture of 0.652.
Will light entering the fiber (from air) at an angle of entry of 63°
be trapped in the fiber?
No. The cut-off angle
for this fiber (for light entering from air) is found from
q0max
= sin-1(NA/n0)
= sin-1[(0.652)/(1.00)] = 40.7°.
The given angle of 63° is larger
than this cut-off angle, so the light is outside of the cone of acceptance
and will escape.
|
| 36. |
  The
on-line reading page "What
a Difference a Cladding Makes" shows photographs of light being trapped
inside a water-clad fiber (redisplayed on the left) and of light escaping
the same water-clad fiber (redisplayed on the right). These redisplayed
photos also indicate the angle of entry for each case: 37.9°
when light is trapped, and 52.3° when light escapes. (Test
them with a protractor to confirm!) (a) Based on these given
values and the images, what possible values could the cut-off angle for
this (water-clad) fiber take? (b) One value in that possible range
should be 45°. If the cut-off angle is
indeed 45.0°, what is the index of refraction of the plastic fiber
core? (The index of the water cladding is 1.33.) (c)
Repeat step (b) for the minimum and maximum possible values of the cut-off
angle you indicated in part (a). How different are these limits from
the value calculated in (b)?
(a) The only limit
we know for certain is that the cut-off angle is somewhere between 37.9°
and 52.3°. Since the light escaping the fiber in the second image
does so at a sizeable angle from the surface, we can surmise that 52.3°
is not right at the cut-off angle. I would be careful, however, of
concluding that the cut-off angle is significantly smaller than 52°.
The path of the escaping light is sensitive to small changes in the angle
of entry. (The 45° given in part (b) is just an estimate and
not intended to be taken as the "correct" value. As we find in part
(c), the index is rather insensitive to the cut-off angle.)
(b) Setting the two expressions
for NA equal to each other yields
n0
sin q0max =
(n12 - n22)1/2,
where we know n0
= n2 = 1.33 for the water cladding.
Using the 45° given, we find
n12
= n02 sin2q0max
+ n22 = (1.33)2*(sin2(45.0°)
+ 1) = 2.653
n1
= 1.63
(c) If the minimum limit of 37.9°
is the cut-off angle, our equation becomes
n12
= (1.33)2*(sin2(37.9°)
+ 1) = 2.436
n1
= 1.56
If instead the maximum limit of 52.3°
is the cut-off angle, we get
n12
= (1.33)2*(sin2(52.3°)
+ 1) = 2.876
n1
= 1.70
These limits based on the two measurements
are not that far from the value of the index calculated from the mid-range
angle of 45°. The index of the plastic is somewhere around 1.6.
|
| 37. |
 The
on-line reading page "What
a Difference a Cladding Makes" shows photographs of light being trapped
inside an air-clad fiber even for angles of entry approaching 90°.
One of these photos is redisplayed to the left. (a) Show that
no cut-off angle can be defined (and thus all light remains trapped) when
an air-clad (n2
= 1.00) fiber has a core index of 1.60, provided the light enter the plastic
from air as well (i.e., that n0
= 1.00). (b) What is the minimum value of the core index that will
trap all light when air is used as a cladding? If you have solved
part (c) of the previous problem, comment on the consistency between this
minimum core index and the lower bound on your calculated values of n
for the plastic.
(a) Using the given
values of n1, n2,
and n0, we find
sin2q0max
= [n12 - n22]/n02
= [(1.60)2 - (1.00)2]/(1.00)2
= 1.56.
This gives a value of 1.25 for sin
q0max
, which is impossible. Thus, no cut-off angle exists, all angles
fall within the cone of acceptance, and all light is trapped.
(b) This minimum value of
the core index can be found by setting sin q0max
equal to 1. We find then
n1min2
= n02 sin2q0max
+ n22 = (1)(1) + 1 =
2.
n1min
= 1.41.
Any (transparent) material with an
index of refraction greater than 1.41 will trap all light that enters it
by TIR when the material is surrounded by air. The range of possible
values for the index of the plastic, found in part (c) of the preceeding
problem, is indeed greater than 1.41.
|
| a. |
What is the speed of light inside the fiber's core?
Show your work.
The definition of the index
of refraction gives
v = c/n
= (3.00 x 108 m/s)/(1.40) = 2.14 x 108
m/s.
|
| b. |
What is the wavelength of light inside the fiber's core?
Show your work.
The wavelength is affected
by the index of refraction in a similar manner:
l1
= l0/n
= (633 nm)/(1.40) = 452 nm.
(Aside: The light in the fiber
still appears red to an observer. The observer's eye is not in the
fiber but rather is back in air, so light that reaches the eye has gone
back to its 633 nm wavelength.)
|
| c. |
 In
which direction does the light bend as it enters the fiber (toward the
top wall, toward the bottom wall, or neither)? Justify your answer.
Sketch the approximate path that the light takes before it hits the inner
wall (top or bottom) of the fiber, clearly showing the correct bending.
Toward the bottom wall.
The index of the core is greater than the index of air, so light traveling
from light into the core will bend toward the normal.
|
| d. |
Should the index of refraction of the cladding be greater
or less than the index of refraction of the core if the fiber is to guide
the light through total internal reflection? Explain.
Less than. Total
internal reflection occurs when light moves from a denser medium to a rarer
medium.
|
| e. |
The critical angle for the core-cladding interface is
found to be 59.0°. What is the index of refraction of the cladding?
Show your work.
n2
= n1 sin qc
= (1.40) sin (59.0°) = 1.20
|
| f. |
 In
your sketch of the light in the fiber, label the angle that should be compared
to the critical angle (at the n1/n2
interface) when determining whether the light will remain trapped in the
fiber. Call this angle q1.
What do we call q0 when q1 equals the critical angle?
The light will hit the
top edge at the critical angle if the light enters the fiber at the cut-off
angle.
|
| g. |
 When
the light strikes the inner wall of the fiber, some (or all) of it may
be reflected. Sketch the path of the reflected light in the fiber.
What condition must be met by the reflected light?
The angle of reflection
must equal the angle of incidence.
|
| h. |
What is the numerical aperture of this fiber? Show
your work.
NA = (n12
- n22)1/2
= [(1.40)2 - (1.20)2]1/2
= 0.721
|
| i. |
Will light entering the fiber at an angle
q0
= 30° be trapped in the fiber? How do you know this?
(Hint: Consider the conditions
on
q0.)Yes. 30° is less
than the cut-off angle of 46.1°, found from
q0max
= sin-1(NA/n0)
= sin-1[(0.721)/(1.00)] = 46.1°.
|
| j. |
 Use
Snell's Law to determine the angle of refraction at the entrance to the
fiber if the light enters the fiber at an angle of q0
= 30°.
The on-line reading labeled
this angle of refraction as q'1.
Using this notation, we see
n0
sin q0
= n1 sin q'1,
so
q'1
= sin-1[(n0/n1)
sin q0] = sin-1[(1.00/1.40)sin
30°] = 20.9°.
|
| k. |
 Use
geometry to determine the angle of incidence at which light strikes the
upper edge of the core.
The angle of incidence
at the upper edge (called q1
in the reading) is the complement of q'1
from the previous part, since q'1
and the angle at the upper edge are the two non-right angles of a right
triangle. Thus,
q1
= 90° - q'1
= 90° - 20.9° = 69.1°.
|
| l. |
Compare your answer to the previous question to the critical
angle for the core-cladding interface. Will light be trapped inside
the fiber? Is this consistent with the conclusion you drew earlier?
This angle of incidence
q1
is greater than the critical angle of 59° given earlier, so this light
will indeed be trapped in the fiber. This is consistent with the
conclusion based on the cut-off angle.
|
