Assignment 11- Solutions - Boston University...

Assignment 11- Solutions

1)

The picture above shows a source of light (the red dot), with the light shining on the origin,

which is marked by the position of the green dot. The horizontal axis is the interface separating

one medium from another. Some of the light reflects back into the first medium, and some

refracts into the second medium.

Choose the three correct statements about this situation from the list below.

The first medium (where the source is) has a higher index of refraction than the second

medium.

FALSE- since The angle inside the blue region is smaller than outside ��

The first medium has the same index of refraction as the second medium.

FALSE- same.

The first medium has a lower index of refraction than the second medium.

TRUE-same reason.

If the index of refraction of the second medium was increased, the angle of the reflected

beam would change.

FALSE- for the reflected wave the interface has acted like a plane mirror, it is reflected at the

same angle, regardless of the second medium’s refraction index.

Changing the index of refraction of the second medium would have no effect on the

angle of the reflected beam.

TRUE- as the previous one, interfaces acts like a mirror.

If the light source was moved to the position of the purple dot in the second medium, and

the light was incident on the origin again, the light would definitely experience total internal

reflection.

FALSE- Snell’s law doesn’t change if we reverse the direction of the light rays. As long as the

angle in the blue region is the same we have the same equation:

�� sin �� sin �� And so the outgoing ray will be at the same angle. The only difference is that we won’t have the

reflected ray outside (instead we get one inside.)


the light was incident on the origin again, some light would refract out into the upper medium,

traveling toward the position of the red dot.

TRUE- described above.


the light was incident on the origin again, some light would refract out into the upper medium,

but the light would not travel toward the position of the red dot.

FALSE

2)

The picture above shows a source of light in the lower medium (the purple dot), with the light

shining on the origin, which is marked by the position of the green dot. The horizontal axis is the

interface separating one medium from another. In this case, all of the light reflects back into the

lower medium.

Choose all the correct statements about this situation from the list below.

The lower medium (where the source is) has a larger index of refraction than the upper

medium.

TRUE- We are observing total internal reflection while going from the lower medium to the

upper one. This means that there exists a critical angle beyond which this happens and critical

angles are only for going from larger index of refraction to smaller. So lower medium has larger

index.

The lower medium has the same index of refraction as the upper medium.

FALSE- read above

The lower medium has a smaller index of refraction than the upper medium.

FALSE- read above

The angle of incidence in this case is greater than the critical angle for total internal

reflection to occur.

TRUE- read above.

The angle of incidence in this case is smaller than the critical angle for total internal

reflection to occur.

FALSE- read above.

If the light source was moved to the position of the red dot in the upper medium, and the

light was incident on the origin again, the light would definitely experience total internal

reflection.

FALSE- there won’t be total internal reflection in going from lower index to higher and, as we

found before, the lower medium has higher index of refraction.


light was incident on the origin again, some light would refract into the lower medium, traveling

toward the position of the purple dot.

FALSE- Some light would indeed refract into the lower medium, but it won’t be towards the

purple dot. This we can see from Snell’s law (blue for the lower medium):

�� sin �� sin ��

But we know that for the �� in the picture no �� can be found (�� is beyond the critical

angle), and that is why it gets reflected. So if we put some �� in the equation, the �� that we

get will never be the angle in the diagram.


light was incident on the origin again, some light would refract into the lower medium, but the

light would not travel toward the position of the purple dot.

TRUE- see above.

3)

This picture shows a light source at point A, at the top left of the picture. The light which refracts

into the lower medium eventually reaches point B, at the bottom right of the picture. Point A is

located at x = 0 m, y = +5 m, and point B is located at x = +10 m, y = -5 m. The upper medium is

air, with an index of refraction of n = 1.00. The interface separating the two media lies along the

line y = 0. Take the speed of light in vacuum to be 3.00 x 108 m/s.

An interesting fact about light is that the path it takes

from A to B always minimizes the total light travel time.

In other words, any other path from A to B, including

the shortest-distance path, would take the light more

time. One can show that this is equivalent to Snell's law.

In the case shown in the picture, the time it takes the

light to travel from point A to the green point, on the

interface, is 26.0 ns. By using Snell's law and

considering the geometry, answer the following.

(a) What is the x-coordinate of the green point?

We can find the distance between A and the green spot

using � � ��:

��

��

��

��

�

� �

�� 3 � 10� �� 26.0 ns � 7.8 m

Now this is the hypotenuse of the upper triangle. For �� we thus have:

cos �� 57.8 ' sin �� (1 ) cos� �� 0.768

The horizontal position of the green dot O is therefore:

' *+ � 7.8 m � sin �� 5.99 m

x = 5.99 m.

(b) What is the index of refraction of the second medium?

To figure out the index we need to use Snell’s law and for that we need the sine of the angles

which is the ratio of opposite over hypotenuse and so:

��

� sin ��sin ��

�*+��

10 ) *+��

We first need to find OB, which is the hypotenuse of the lower triangle:

�� (5� - .10 ) *+/� � 6.41 m

�� 1.23

(c) What is the total time, in nanoseconds, it takes the light to travel from point A to point B?

��

� ��1/��

� 6.41 m � 1.233 � 10� �

�� 26.0 ��

Which is the same as the time in the first region (this is not a coincidence, if the vertical

distance is the same the times are the same.) So the total time will be 2 � 26 �� 52.2 ��.

4)

A lamp emitting red light is placed at the bottom of a tank of liquid with depth H = 18.0 cm and

index of refraction n = 1.56 (for red light). Above the liquid is air, which has an index of

refraction of 1.00. On the surface of the liquid above the light, an observer sees a bright circle

with radius R.

(a) Why does the observer see a bright circle on the

surface?

same as the Since the lamp is the region of higher index of

refraction, not all of its light-rays will be able to escape

into air. The ones that hit the interface at an angle larger

than the critical angle will undergo total internal

reflection and beyond that point the observer outside

doesn’t see light rays coming out. The angle depends on

the horizontal distance on the interface from the point

right above the lamp.

Because the lamp is circular. False.

Because the pupils in the observer's eyes are circular. False.

Because light striking the surface outside of the circle experiences total internal

reflection. True.

(b) Calculate the radius of the bright circle seen by the observer.

The outermost ray that can come out is at the critical angle, since beyond that the rays will

reflect. First, the ray going at �3 makes the hypotenuse 4:

4 � (5� - 6�

The radius R of the bright circle on the interface is related to �3 by:

�3

�3

5

6

4

But also:

sin1� �

5� -6� �' 5 �

(c) From the list below, choose the two correct statements. The first three statements relate to

light that travels from the lamp toward point A, along the solid line in the figure. The other three

statements relate to what happens when the lamp emits gree

the index of refraction of this liquid decreases as the wavelength of light increases

for most materials.

If a beam of light travels directly from the lamp to the surface at point A, it refracts out into

the air and goes away infinitely far from the lamp.

FALSE- A is beyond �3, total internal reflection.

If a beam of light travels directly from the lamp to

infinitely far from the lamp along the surface of the liquid.

FALSE- A is not at �3, total internal reflection.

If a beam of light travels directly from the lamp to the surface at point A, it reflects back

into the liquid.

If the lamp is adjusted to emit green light instead, the radius of the bright circle

decreases.

If the lamp is adjusted to emit green light instead, the radius of the bright circle stays the

same.

If the lamp is adjusted to emit green lig

sin �3 � 54 � 5

√5� - 6�

sin �3 � 1� � 1

1.56 � 0.641 � 5

√5� - 6� ' (5� - 6� � �5

6� � ��5� � .�� ) 1/5�

� 6√�� ) 1 � 18.0 cm

√1.56� ) 1 � 15.0 cm



statements relate to what happens when the lamp emits green light instead of red light. Note that

the index of refraction of this liquid decreases as the wavelength of light increases


the air and goes away infinitely far from the lamp.

, total internal reflection.

If a beam of light travels directly from the lamp to the surface at point A, it goes away

infinitely far from the lamp along the surface of the liquid.

, total internal reflection.




If the lamp is adjusted to emit green light instead, the radius of the bright circle increases.



n light instead of red light. Note that

the index of refraction of this liquid decreases as the wavelength of light increases - this is true


the surface at point A, it goes away




ht instead, the radius of the bright circle increases.

5)

Binoculars generally use pairs of prisms in which the light experiences total internal reflection.

Each prism (in blue on the diagram) is right-angled, with the other two angles being 45°. A

diagram of the path followed by light as it travels through the prisms to your eyes is shown in the

figure above. If the prisms are surrounded by air, determine the minimum index of refraction of

the prism material.

In the figure we see that the vertical light ray incident

on the long edge is perpendicular to the edge and

therefore the angle of incidence with respect to the

normal is �� 0 (notice it’s not 908!). Therefore from

Snells law �� sin �� sin �� we get �� 0 as well.

So the light goes through the long edge without

refraction. Now, in order to make sure that the light

emerging from the long edge does not leave the prism

through the shorter edges we need it to undergo total

internal reflection from the shorter edges. If we look

at the figure, this means that 458 has to be larger than

the critical angle of going from the prism to air. You

can observe that if the light reflects off the short edge

to the left at 458 it hits the right one also at 458 and the same thing repeats and the light leaves

the prism through the long edge to the top. So:

sin 458 � sin �3 � ��9:

� 1�

' √22 � 1

� ' � � 2√2 � √2 ; 1.41

So the minimum index is � � 1.41.

458

�

458

458 458

458

6)

As shown in the figure above, a beam of red and violet light is incident along the normal to one

surface of a right-angled triangular glass prism. The glass has an index of refraction of 1.52 for

red light, and 1.54 for violet light. The prism is surrounded by air, which has an index of

refraction of 1.00. Note that you will probably find it helpful to draw a sketch showing how the

red and violet beams travel from the point at which they enter the prism to the side ab of the

prism.

(a) If the angle at vertex a of the prism is = 24.0 degrees, determine the angles of refraction for

the red and violet beams that emerge from the prism from the side ab. Show these refracted

beams on your sketch. The angles of refraction are measured, as usual, from the normal to the

surface.

angle of refraction for the red light is

Just as in previous problem, From Snell’s law �� sin �� sin ��since the light coming from the left impinges on

the vertical edge at right angle, the angle of incidence

relative to the normal to the surface is �� 0 and not 90.

Therefore �� 0 too and the light goes through without

refraction.

After this it hits the hypotenuse at angle �9:. To finc this

angle we use the fact that in the figure < - �9: � 90 (the

dashed line is the normal to the interface. Then we notice

that < is also in the small right-angled triangle to the top

and < - 248 � 90. From these two relations we find

�9: � 248. This is the angle for both red and violet light.

To find the angle of refraction of red light we use Snell’s

law with the index for red:

�9:

��

248

�� 1

�9:

<

�9: sin �9: � �� sin �� 9: � �=�> � 1.52, �� 1 sin �� 1.52 sin 248 � 0.6182 �� 38.178

angle of refraction for the violet light is

�9: � �@9�� 1.54, �� 1 sin �� 1.54 sin 248 �� 38.788

(b) What is the smallest index of refraction the prism could have for light to experience total

internal reflection when it is first incident on the side ab, for the geometry described above?

Assume the light enters the prism as in the diagram.

For this to happen we need �9: � 248 to be the critical angle. Thus:

sin �� sin 248 � ��9:

� 1�

� � 2.459

7)

Two speakers, which are separated by a distance d = 3.00 m, produce sound waves with the

same amplitude, phase and frequency. You stand a distance of 4.00 m directly in front of the left

speaker, on the dashed line shown in the diagram.

(a) Assume the speed of sound to be 340 m/s. What is the lowest frequency the sound can have

to produce completely constructive interference at your location?

we need Δ4 � �B for constructive interference. But what � should we chose? We want the

lowest frequency. According to CB � � � 340�/� lowest frequency corresponds to the largest

wavelength. Now since Δ4 is also fixed from the picture, �B is a constant and largest B requires

choosing the smallest nonzero � so � � 1. Therefore:

Δ4 � (D� - �� ) D � B B � (3� - 4� ) 4 � 5 ) 4 � 1m C � �

B � 340 6E (b) What is the lowest frequency the sound can have to produce completely destructive

interference at your location?

This time the smallest nonzero � is 1/2. So:

Δ4 � (D� - �� ) D � 12 B

B � 2m C � �

B � 170 6E 8)

Two identical speakers are pointed at one another, along the x-axis. The speakers, which are in

phase with one another, broadcast identical sound waves at a frequency of 170 Hz. Assume the

speed of sound to be 340 m/s. Speaker A is positioned at x = 0, and speaker B is located at x =

+6.20 m.

248

�9:

D

�

(D� - ��

(a) In between the speakers, there are a few locations at which the two waves produce

completely destructive interference. What is the x-coordinate of the point closest to speaker B at

which completely destructive interference occurs?

First let us find the wavelength:

B � �C � 340

170 � 2�

We need to find points of destructive interference which satisfy Δ4 � F� - ��G B

Δ4 � 4� ) 4� � * ) .� ) */ � 2* ) � � H� - 12I B

2* � � - H� - 12I B

We are looking for the point closest to B and in between the two sources. So we have to have:

* J � K 2* J 2� ' H� - 1

2I B J �

Closest to B means the largest x that satisfies the above condition and in turn the largest �.

Now all we need to do is to count from n�0 up and find the largest:

λ�2m

Hn- 12I λK λ

2 �1m , 3λ2 �3m , 5λ

2 �5m , 7λ2 �7m…

Since � � 6.20 m here the largest n that fits is the one that gives 5�. Using that, we find:

4� � *

�

4� � � ) *

� �

2* � � - H� - 12I B � 6.20 m - 5m � 11.2m

* � 5.6m

(b) In between the speakers, there are also a few locations at which the two waves produce

completely constructive interference. How many locations are there, along the line between the

speakers, at which completely constructive interference occurs?

We know that at the point half-way between the two sources the waves interfere

constructively. The next points of constructive interference after that, moving towards 2

happens where Δ4 � �B which are at:

* � 12 .� - �B/ � 1

2 .� - � � 2m/ � �2 - �.1�/ J �

�.1�/ J �2 � 3.10 m

� N 3

The largest � that makes * fit in the distance � this time is � � 3 since � � 6.20 m. From the

other side, approaching source 1, the last � we can have will be )3. So the possible �’s are:

)3, )2, )1,0,1,2,3 and their number is

2�OPQ - 1 � 2 � 6 - 1 � 7

For sources that are in phase, this is always the case. (If they become slightly, or completely

out of phase, we lose the one in the middle and the number will be 2�OPQ. )

andrew

Text Box

Another approach to this problem is to start with the midpoint between the speakers, which has a path-length difference of zero, and will thus experience constructive interference. Starting at the midpoint, walking half a wavelength (which is 1 m) toward one source will increase one path length by half a wavelength, and decrease the other by half a wavelength - this gives a path-length difference of a full wavelength. Each addition half wavelength you walk brings you to another spot where constructive interference happens. The way the problem was set up, the places where constructive interference happens are located at x = 0.1 m, 1.1 m, 2.1 m, 3.1 m, etc. The places where destructive interference happens are halfway between the places where constructive interference happens, so those are at x = 0.6 m, 1.6 m, 2.6 m, 3.6 m, etc.

9)

The pattern above is produced by interference between identical waves emitted by two sources.

Red denotes positive displacement of the medium, while blue denotes negative displacement.

Black indicates zero displacement. At the orange point near the top, we

interference is occurring. The distance the orange point is from the left source is the length of the

purple line; the distance the orange point is from the right source is the length of the green line.

(a) Select all the true statements about this situation.

For constructive interference to occur at the orange point, the net displacement must be non

zero at all times at that point.

FALSE- only the nodes have the same amplitude (zero) at all times and the anti

will be oscillating between a max and min.



Black indicates zero displacement. At the orange point near the top, we say that constructive



ments about this situation.

For constructive interference to occur at the orange point, the net displacement must be non

only the nodes have the same amplitude (zero) at all times and the anti-

be oscillating between a max and min.



say that constructive



For constructive interference to occur at the orange point, the net displacement must be non-

-nodes (peaks)

For constructive interference to occur at the orange point, the distance from each source to

the orange point must be an integer number of wavelengths.

FALSE- not the individual 4� and 4� but 4� ) 4� is the thing that has to be �B. Each of 4� and

4� are free to have values different from �B.

For constructive interference to occur at the orange point, the path-length difference for

the orange point must be an integer number of wavelengths. The path-length difference is the

distance the point is from the left source minus the distance the point is from the right source.

TRUE- see above. The waves must overlap in phase with each other.

In this particular case, the orange point is two wavelengths farther from the left source

than it is from the right source.

TRUE- from the figure, we know that the region around the perpendicular bisector to the line

joining the sources has constructive interference. The region where the orange dot sits is the

second bright region to the right of the bright region in the middle and:

Δ4 � 4� ) 4= � 0B K �R� STURV� Δ4 � 4� ) 4= � 1B K 1�� STURV� SRUD� Δ4 � 4� ) 4= � 2B K 2�� STURV� SRUD�

So the distance of the orange dot from the source to the left (4�) is greater that its distance

from the one to the right (4=) by -2B.

(b) In this case, what is the distance between the two sources, in terms of wavelengths?

The two sources are sitting on bright regions (although it’s somewhat hard to recognize). So

they are on peaks. Let’s say the red regions are maxima and blue ones are minima. From one

maximum to the next is one B. If we count the number of red regions on the line joining the

sources (disregarding the first one the left source is sitting on) we will find their distance in

terms of B: 5B

B B B B B


Red denotes positive displacement of the medium, while blue denotes negative dis

Black indicates zero displacement. At the orange point near the top, we say that destructive


purple line; the distance the orange point is from


Because the orange point is in a region of destructive interference, the net displacement

at the orange point is zero at all times.

TRUE- at points of destructive interference (nodes) the waves are 180 out of phase at all times.

For destructive interference to occur at the orange point, the distance from each source to the

orange point must be an integer number of wavelengths plus half a wav

FALSE- since sources are in phase,


Red denotes positive displacement of the medium, while blue denotes negative dis






at the orange point is zero at all times.

points of destructive interference (nodes) the waves are 180 out of phase at all times.


orange point must be an integer number of wavelengths plus half a wavelength.

since sources are in phase, Δ4 � F� - ��G B gives nodes, not integer multiples of





the right source is the length of the green line.


points of destructive interference (nodes) the waves are 180 out of phase at all times.


gives nodes, not integer multiples of B.

For destructive interference to occur at the orange point, the path-length difference for

the orange point must be an integer number of wavelengths plus half a wavelength. The path-

length difference is the distance the point is from one source minus the distance the point is from

the other source.

TRUE- see above.

In this particular case, the orange point is 1.5 wavelengths farther from the left source than it

is from the right source.

FALSE- like the previous problem, the central region has Δ4 � 0. The orange line is on the

second dark region to the left, so: Δ4 � 4� ) 4= � ) 1

2 B K 1�� WSX STURV� YTC� Δ4 � 4� ) 4= � ) 3

2 B K 2�� WSX STURV� YTC�

Since it’s closer to the source to the left 4� J 4=.

(b) In this case, the sources are separated by a distance of 15.0 cm, and the distance the orange

point is from the source on the right is 24.6 cm. How far is the orange point from the source on

the left?

We just found that Δ4 � 4� ) 4= � ) Z� B. We are given 4= � 24.6 cm. If we knew B we could

solve for 4�. But the figure is the same as the previous problem and there we found that the

distance between the sources was � � 5B and here we know that � � 15.0 cm. So we get:

B � 15.0 cm5 � 3.0 cm

4� � 4= ) 32 B � 24.6 cm ) 9

2 cm � 20.1 cm

Assignment 11- Solutions - Boston University...

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