Mastering Physics HW 8 Ch 22 - Wave Optics

HW 8 Ch 22 Wave OpticsDue: 11:59pm on Tuesday, November 3, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

A message from your instructor...

The answer to Part C is "roughly constant." I am telling you because the graph is hard to see on the simulation.

PhET Tutorial: Wave Interference

Learning Goal:

To understand the cause of constructive and destructive interference for the doubleslit experiment, and to explain howthe interference pattern depends on the parameters of the emitted waves.

For this tutorial, use the PhET simulation Wave Interference. This simulation allows you to send waves through avariety of barriers and look at the resulting interference patterns.

Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file.Choose to run or open it.

You can choose between water waves, sound waves, or light. You can adjust the slit width and slit separation usingslider bars, and you can put a barrier containing one or two slits in front of the source of the waves. Clicking on AddDetector produces a plot showing the wave amplitude vs. time for the location of the detector, which can be dragged toany location.

Feel free to play around with the simulation. When you are done, click Reset All before beginning Part A.

Part A

Select the Light tab, with no barrier. Select a wavelength so that the light is red, and adjust the amplitude of thelight to the highest setting.

Light is a form of electromagnetic wave, containing oscillating electric and magnetic fields. Select Add Detector,which shows how the electric field oscillates in time at the location of the probe. The amplitude of the wave at thelocation of the probe is equal to the maximum electric field measured.

How does the amplitude of the wave depend on the distance from the source?

ANSWER:

CorrectThe amplitude of the electric field decreases with distance from the source. You can also see this byselecting Show Graph. As a result, the intensity of the light, which is proportional to the amplitude squared,also decreases with distance from the source.

Part B

Select Show Screen to place a screen on the right edge of the panel. You might also want to select the PlotIntensity graph to see the details more clearly.

Which statement best describes how the intensity of the wave depends on position along the screen?

ANSWER:

CorrectThe light wave spreads out circularly, so the amplitude of the wave simply depends on the distance from thescreen to the light source, as we saw in Part A. All locations on the screen are nearly the same distance tothe source, so the amplitude is roughly constant. The intensity of the light is proportional to the amplitudesquared. Therefore, since the amplitude is nearly constant, so is the intensity.

Part C

Now, select a barrier with one slit, and use the Barrier Location slider bar to place it roughly away fromthe light source (second tick mark on the slider bar). Adjust the slit width (using the slider bar) to roughly (first tick mark). Keep the wavelength of the light set to red.

Which statement best describes how the intensity of the wave depends on position along the screen?

ANSWER:

The amplitude decreases with distance.

The amplitude increases with distance.

The amplitude is constant.

The intensity is large near the middle of the screen, then decreases to nearly zero, and then increasesagain as the distance from thr middle of the screen increases.

The intensity is roughly constant.

The intensity is a maximum near the middle of the screen (directly to the right of the source) andsignificantly decreases above and below the middle of the screen.

1295 nm262 nm

Correct

Since the slit width is small compared to the wavelength, the light passing through the slit spreads out nearlycircularly, so the intensity behaves similarly to when there is no barrier (although the intensity is lower).

Part D

Now, select Two Slits and a slit separation of roughly . (Keep the slit widths and barrier location thesame as in Part C, and be sure the amplitude is still set to the highest setting).

Which statement best describes how the intensity of light on the screen behaves?

ANSWER:

Correct

Circular waves come out of each of the two slits. However, in contrast to one slit, an interference patternoccurs with two slits, due to the two spherical waves overlapping. This interference pattern has severallocations with a high intensity, alternating with locations where the intensity is nearly zero. If you look carefullyat the screen, you will see faint red lines, called fringes, where the intensity is relatively large.

Part E

In the previous part, you learned about the interference pattern that produces several fringes on the screen. Tomake the fringes more visible, adjust the wavelength of the light to make the light green. You should see a fringe inthe middle of the screen, and several others above and below the middle.Click on Show Graph (near the bottom of the window), and then press the blue pause button (at middlebottom ofthe screen). Use the measuring tape to measure the wavelength of the green light (you can measure from crest tocrest in the Electric field vs. Position plot). Make a note of this wavelength measurement to use as a referencewhen we compare two other distances next.

Now, measure the distance from the first bright fringe above the middle of the screen to the upper slit. Call thisdistance . Next, measure the corresponding distance to the lower slit, . The distances and are shownin the figure for clarity.

The intensity is large near the middle of the screen, then decreases to nearly zero, and then increasesagain as the distance from the middle of the screen increases.



1750 nm


The intensity is large near the middle of the screen, then decreases to nearly zero, and then increasesagain as the distance from the middle of the screen increases.


r1 r2 r1 r2

How do the distances and compare?

ANSWER:

CorrectSince the extra distance the wave travels from one of the slits is exactly equal to the wavelength, the crestsof one of the waves still meet up with the crests of the other wave, causing constructive interference. As youcan confirm, for every location where the amplitude is relatively very large, the difference in the distances isequal to some integer times the wavelength, so that the two waves meet up exactly in phase.

Part F

Compare the distances from the first location nearest the middle of the screen where the intensity is nearly zero(dark fringe) to each of the two slits. How do the distances compare?

r1 r2

The difference in the distances is equal to half the wavelength of the wave.

The distances are the same.

The difference in the distances is equal to a quarter of the wavelength of the wave.

The difference in the distances is equal to the wavelength of the wave.

ANSWER:

CorrectWith the difference in the distances equal to half the wavelength, the two waves are exactly out of phase, so acrest of one wave meets up with the trough of the other wave, causing the two waves to add up to nearly zero.This is destructive interference.

Part G

How does the distance between consecutive bright fringes depend on the wavelength of the light?

ANSWER:

CorrectAs you found in Part C, a bright fringe occurs when the difference in the distances to the slits is equal to aninteger times the wavelength. If the wavelength increases, this distance must increase, which causes thefringes to move further apart.

Part H

The distances are the same.

The difference in the distances is equal to a quarter of the wavelength of the wave.

The difference in the distances is equal to half the wavelength of the wave.

The difference in the distances is equal to the wavelength of the wave.

The fringes get closer together as the wavelength increases.

The spacing of the fringes does not change when the wavelength changes.

The fringes get farther apart as wavelength increases.

How does the distance between consecutive bright fringes depend on the slit separation?

ANSWER:

CorrectFor any location except the midpoint on the screen, the difference in the distances to the slits increases asthe slit separation increases. Therefore, as the slit separation increases, all fringes move toward the center ofthe screen in order for the difference in the distances to the two slits to remain constant.

Part I

How does the distance between the bright fringes depend on the slit width (for slit widths less than the wavelengthof the light)?

ANSWER:

CorrectFor slit widths that are small compared to the wavelength, the light wave diffracts through the slit and comesout as a circular wave. Decreasing the slit width simply blocks more light from going through the slit, butdoesn’t change the interference pattern.

Part J

How does the distance between the bright fringes depend on the amplitude of the wave?

Hint 1. How to approach the problem

Does the amplitude of the wave affect the path lengths from the fringes to the slits? Does the amplitudeaffect the phase of the light or its wavelength?

ANSWER:

The fringes get closer together as the slit separation increases.

The fringes get farther apart as the slit separation increases.

The spacing of the fringes does not change when the slit separation changes (just the brightnesschanges).

The spacing of the fringes does not change when the slit width changes.

The fringes get closer together as the slit width increases.

The fringes get farther apart as the slit width increases.

Correct

The interference pattern of fringes depends only on the wavelength, the slit separation, and the distance fromthe slits to the screen. Increasing the amplitude simply makes the fringes brighter. An equation thatsummarizes the results of the last several questions is , where is the distance betweenconsecutive bright fringes, is the wavelength, is the distance between the slits, and is the distancebetween the screen and the slits.

Part K

Does interference occur when water or sound waves encounter a barrier with two slits?

ANSWER:

Correct

Interference is a very common phenomenon that can occur with any type of wave.

PhET Interactive Simulations University of Colorado http://phet.colorado.edu

± Fringes from Different Interfering Wavelengths

Coherent light with wavelength 608 passes through two very narrow slits, and the interference pattern is observedon a screen a distance of 3.00 from the slits. The firstorder bright fringe is a distance of 4.84 from the center ofthe central bright fringe.

Part A

For what wavelength of light will the firstorder dark fringe (the first dark fringe next to a central maximum) beobserved at this same point on the screen?

Express your answer in micrometers (not in nanometers).


The fringes get farther apart as the amplitude increases.

The spacing of the fringes does not change when the amplitude changes (just the brightness changes).

The fringes get closer together as the amplitude increases.

Δy = λL/d Δyλ L d

Yes, interference also occurs for both of these types of waves.

No, it only occurs for light.

Interference also occurs for water waves, but not for sound waves.

Interference also occurs for sound waves, but not for water waves.

nmm mm

http://phet.colorado.edu/

For this problem we can use the wavelength of the first beam of light, as well as the dimensions of theinterference pattern that it creates, to determine the separation of the two slits. Using this information andthe dimensions of the interference pattern of the second beam of light, we can then determine the secondbeam's wavelength.

Hint 2. Interference pattern equation

The equation for the constructive interferencefringes from two slits projected on a screen is

,

where is the distance between the two slits, is the wavelength of light, and is the anglebetween the constructive peak and thecenterline. Note that .

For the destructive interference pattern, one canuse the equation

,

where all the variables are the same as for thecase of constructive interference.

Using the approximation , which is valid for small , will be helpful.

Hint 3. Correct order to use for destructive interference

You might have some confusion about whether to use or for the "firstorder" destructiveinterference fringe. The correct way to look at the situation is to use the main equation for destructiveinterference,

,

and note that if we use , we get the same answer as if we use (just with a minus sign).

This is due to the fact that we have arbitrarily defined the equation with instead of ,

which would also have worked just fine. Since we are looking for the first dark fringe, we can use and . Both give the same answer for the magnitude of the angle off the centerline.

ANSWER:

Correct

Notice that the answer is twice the first wavelength. This makes sense, because we are dealing with the samepoint on the screen, so the path difference, given by , is the same for each wavelength. Since thefirst wavelength experiences constructive interference, the path difference must equal . Therefore, for lightof wavelength , this same path difference is exactly half of its wavelength, giving destructive interference.

d

d sin( ) = mλθm

d λθm

mthm = 0, ±1, ±2, . . .

dsin( ) = (m+ )λθm12

sin(θ) ≈ tan(θ) θ

m = 0 m = ±1

dsin(θ) = (m+ )λ12

m = −1 m = 0

(m+ )12 (m− )1

2m = 0

m = −1

1.22 μm

d sin(θ)λ λ

2λ

Double Slit 1

Two lasers are shining on a double slit, with slit separation . Laser 1 has a wavelength of , whereas laser 2 has awavelength of . The lasers produce separate interference patterns on a screen a distance 5.10 away from theslits.

Part A

Which laser has its first maximum closer to the central maximum?

Hint 1. Path difference

The first maximum comes when the path difference between the two slits is equal to one full wavelength.Think about which laser has a smaller wavelength, and recall that the distance from the central maximum isproportional to the path difference.

ANSWER:

Correct

Part B

What is the distance between the first maxima (on the same side of the central maximum) of the twopatterns?

Express your answer in meters.

Hint 1. Find the location of the first maximum for laser 1

The first maximum corresponds to constructive interference, with (since the central maximumcorresponds to ). Using the smallangle approximation, what is the distance of this maximumfrom the central maximum for laser 1?


Hint 1. Angle to maxima

The angle to the th maximum is given by , where is the angle, is theseparation between the slits, and is the wavelength of the light.

Hint 2. Distance on screen

For a screen that is far from the slits, as in this problem, the distance on the screen from thecentral maximum is , where is the angle from the slits to the point on the screen and is the distance from the slits to the screen.

d d/20d/15 m

laser 1

laser 2

Δymax−max

m = 1m = 0 y1

m d sin(θ) = mλ θ dλ

yy = R sin(θ) θ

R

ANSWER:

Hint 2. Find the location of the first maximum for laser 2

Now that you have found the first maximum for laser 1, what is the location of the first maximum forlaser 2?






ANSWER:

ANSWER:

Correct

Part C

What is the distance between the second maximum of laser 1 and the third minimum of laser 2, onthe same side of the central maximum?


Hint 1. Find the location of the second maximum

If the central maximum corresponds to , then you should be able to figure out what the secondmaximum corresponds to. Using that, what is the distance to the second maximum of laser 1 from thecentral maximum?


= y1 m

y2


yy = R sin(θ) θ

R

= y2 m

= 8.54×10−2 Δymax−max m

Δymax−min

m = 0y1





ANSWER:

Hint 2. Find the value of the third minimum

The first minimum corresponds to (since there is no central minimum). What, then, is the value of for the third minimum? Recall that is always an integer.

Express your answer as a whole number.

ANSWER:

Hint 3. Find the location of the third minimum

Given that , what is the location of the third minimum?


Hint 1. Difference between the angle to a maximum and a minimum

Once you have the value of , the equation for the angle to a minimum is almost identical to theequation for the angle to a maximum. The only difference is that you use in place of ifyou want to find the location of a minimum instead of a maximum.

ANSWER:

ANSWER:


yy = R sin(θ) θ

R

= y1 m

m

m = 0m m

= mthird minimum

m = 2

mm + 1/2 m

= ythird minimum m

= 0.349 Δymax−min m

Correct

Double Slit with Reflections

A radar tower sends out a signal of wavelength . It is meters tall, and it stands on the edge of the ocean. A weatherballoon is released from a boat that is a distance out tosea. The balloon floats up to an altitude . In this problem,assume that the boat and balloon are so far away from theradar tower that the small angle approximation holds.

Part A

Due to interference with reflections off the water, certain wavelengths will be weak when they reach the balloon.What is the maximum wavelength that will interfere destructively?

Express your answer in terms of , , and .

Hint 1. Smallangle approximation

For small angles, , with measured in radians. Thus, you can find a simplerelationship between and and the doubleslit equations.

Hint 2. Interference with reflections

When light reflects off a surface with a higher index of refraction, the reflected beam gets an extra halfwavelength phase shift (that is, a phase shift of ). This causes expressions that used to yield interferencemaxima to yield interference minima!

Hint 3. Double slit approximation

Notice that due to the reflection, this problem looks almost like a double slit problem with slit width .Thus, the double slit equations should apply. Note, however, that the red (reflected) beam is in fact reflectedoff the water (and only appears to come from a second "slit").

λ x

dh

λ

h d x

sin(θ) ≈ tan(θ) ≈ θ θh d

π

2x

ANSWER:

Correct

Part B

What is the maximum wavelength that will interfere constructively?


ANSWER:

Correct

TwoSlit Interference

As Richard Feynman stated in his book on quantum mechanics,

Interference contains the heart and soul of quantum mechanics.

In fact, interference is a phenomenon of classical waves, easily perceived with sound or light waves. (It contains thesoul of quantum mechanics only after you swallow the preposterous notion that particles in motion are described by awave equation rather than the laws of Newtonian mechanics.)

In this problem, you will look at a classic wave interference problem involving electromagnetic waves. Young's doubleslit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one of the most elegantexperiments in physics (because it demonstrates the important concept of interference so simply). For the purposes ofthis problem, we assume that two long parallel slits extending along the z axis (out of the plane shown in the figure) areseparted by a distance . They are illuminated coherently, that is, in phase, by light with a wavelength , for exampleby a laser beam polarized in the z direction. (Lacking a laser, Young used an intense source diffracted by a slit toproduce coherent illumination of his double slits.)

The key point is that the electric field far downstream from the slit (e.g., at a large positive x value) is the sum of theelectric fields emanating from each of the two slits. Hence, the relative phase of these electric fields at someobservation point determines whether they add in phase (constructively) or out or phase (destructively).To refresh your memory about traveling waves, the electric field that is incident on the double slits from the leftis a function of and . Let us assume that it has amplitude . We will also assume a cosine trigonometric functionwith the arbitrary phase set equal to zero (i.e., at the point you find that ). Then

.

The argument of the cosine function is the phase . It can be written as , where is theangular frequency ( ) and is the wave number defined by . The phase increases by each timethe distance increases by . Moreover, the phase is constant for an observer moving in the positive x direction at the

= λ 2hxd

λ

h d x

= λ 4hxd

d λ

OE(x, t)

x t Eleftx = 0, t = 0 E = Eleft

E(x,t) = cos[ (x−ct)]Eleft2πλ

Φ(x, t) Φ(x, t) = kx − ωt ωω = 2πf k k = 2π/λ 2π

λ

speed of light, i.e., for whom .

Part A

Now consider the electric field observed at a point thatis far from the two slits, say at a distance from themidpoint of the segment connecting the slits, at an angle from the x axis. Here, far means that , a regime

sometimes called Fraunhofer diffraction.

The critical point is that the distances from the slits topoint are not equal; hence the waves will be out ofphase due to the longer distance traveled by the wavefrom one slit relative to the other. Calculate the phase

of the wave from the lower slit that arrivesat point .

Express your answer in terms of , , , , , , andconstants like .

Hint 1. Definition of phase

Recall that the phase of an electromagnetic wave (in air) is given by

,where is the wavelength and is the speed of light in air. Note that is the total distance between thereference point and the point of interest, not just the x component of this distance.

Hint 2. Calculate the extra distance

The distance from the lower slit to point is slightly longer than . That is, . Find thisextra distance . Be sure to exploit the fact that , which means that the path from the lower slit topoint is essentially parallel to the path from the middle of the slits to point . Hence all the pathdifference occurs in the vicinity of the slits.

Express your answer in terms of and .

x = ct

Or

θ r ≫ d

O

(O, t)ΦlowerO

d θ λ c r tπ

Φ(x, t)

Φ(x,t) = (x−ct)2πλ

λ c x

O r = r + δrxlowerδr r ≫ d

O O

d θ

ANSWER:

Hint 3. Find the phase of the lower wave

Find the phase of the oscillating electric field that arrives at point from the lower slit.

Express your answer in terms of , , , , , and constants such as .

ANSWER:

ANSWER:

Correct

Part B

Now calculate the phase of the wave from the upper slit that arrives at point .

Express your answer in terms of , , , , , , and constants such as .

Hint 1. Find the difference in distances

The distance from the upper slit to point is slightly shorter than . Find the distance by which the pathis shorter. Be sure to exploit the fact that . This means that the path from the upper slit to point isessentially parallel to the path from the middle of the slits to point . Hence all the path difference occurs in

= δr

(O, t)Φlower O

r δr λ c t π

= (O, t)Φlower

= (O, t)Φlower (r+ sin(θ)−ct)2πλ

d2

(O, t)Φupper O

d θ λ c r t π

O r δrr ≫ d O

O

the vicinity of the slits.

Express your answer in terms of and .

ANSWER:

ANSWER:

Correct

In order to make the math as simple as possible, we will define two phases:

and

.Then

and

.

Part C

Assuming that the maximum amplitude of the field at point for a wave originating from a single slit positionedmidway between the slits is , now find the magnitude of the combined field at due to the two slits shownin . You may ignore variations in the maximum amplitudeand consider only variations in phase of the wavesemerging from the slits.


d θ

= δr

= (O, t)Φupper (r− sin(θ)−ct)2πλ

d2

ϕ = (r−ct)2πλ

δϕ = dsin(θ)πλ

= ϕ + δϕΦlower

= ϕ − δϕΦupper

OE(r) E O

E(r) ϕ δϕ

Hint 1. Find the magnitude of the electric field due to the lower slit

What is , the magnitude of the electric field due to the lower slit? Recall that the general form of themagnitude of the electric field is the product of the amplitude and the cosine of the phase.


ANSWER:

ANSWER:

Correct

Two equivalent ways of expressing this answer are

and

.The reason for this is the identity

.This formula can be obtained by twice applying the standard formula

for the cosine of a sum.

Part D

The key aspect of twoslit interference is the dependence of the total intensity at point on the angle . Find thisintensity .

The formula for intensity is

.

Express your answer in terms of , , , and , where . Note: should becoded as cos(x)^2.

Hint 1. Amplitude of the electric field

Recall that can be written as the product of the amplitude and the cosine of the phase, where the phasedepends on time. In the above expression for , the term is a function of time, whereas the rest ofthe variables/functions are not. Therefore, the amplitude of is . Substitute for to findthe dependence of on .

ANSWER:

Elower

E(r) ϕ δϕ

= Elower

= E E(r)(cos(ϕ + δϕ) + cos(ϕ − δϕ))

E = E(r)[cos (ϕ + δϕ) + cos (ϕ − δϕ)]

E = 2E(r) cos ϕ cos δϕ

cos(A + B) + cos(A − B) = 2 cos(A) cos(B)

cos(X + Y ) = cos(X) cos(Y ) − sin(X) sin(Y )

O θI(θ)

I = cϵ0 (amplitude of E)2

2

Imax θ d λ = 2 cImax ϵ0 E(r)2 xcos2

EE cos(ϕ)

E 2E(r) cos(δϕ) δϕI θ

Correct

Part E

Twoslit interference is usually observed at small angles, and thus can be replaced by just . In this limit,the important observable is the spacing between successive minima (or maxima) of the interference pattern. Findthe angular spacing of the interference pattern.

Express your answer in terms of , , and any needed constants.


The intensity has the form

.

Since this depends on a cosine squared, the phase difference between two maxima or minima is , not ,as it would be for a simple cosine function. Thus, to find the angular separation, you must set the differencein phase between two different directions equal to :

.

Note that once you make the substitution for both phases, you will have a relatively simpleexpression involving , where .

ANSWER:

Correct

This is a famous formula:

.

The angular spacing of the interference pattern (in radians) is simply the ratio of the wavelength to the slitseparation.

A message from your instructor...

The answer to Part D is "1&3 only." It doesn't make a lot of sense. Incandescent bulbs are also going to be mutuallyincoherent, meaning the phase of each will not be coordinated with the other. But maybe I could see it this way

= I(θ) Imax (cos(( )dsin(θ)))πλ

2

sin(θ) θ

Δθ

d λ

Imax(cos( ))π sin(θ)d

λ

2

π 2π

π

( )− ( ) = ππ sin( )dθ2

λ

π sin( )dθ1

λ

sin(θ) = θΔθ Δθ = −θ2 θ1

= Δθ λd

Δθ = λd

"incoherent" doesn't apply because they are sources that are not making light of a single wavelength, and so comparingthe phases doesn't even make sense.

± Understanding TwoSource Interference

Learning Goal:

To understand the assumptions made by the standard twosource interference equations and to be able to use them ina standard problem.

For solving twosource interference problems, there exists a standard set of equations that give the conditions forconstructive and destructive interference. These equations are usually derived in the context of Young's double slitexperiment, though they may actually be applied to a large number of other situations. The underlying assumptionsupon which these equations are based are that two sources of coherent, nearly monochromatic light are available, andthat their interference pattern is observed at a distance very large in comparison to the separation of the sources.Monochromatic means that the wavelengths of the waves, which determine color for visible light, are nearly identical.Coherent means that the waves are in phase when they leave the two sources.

In Young's experiment, these two sources corresponded to the two slits (hence such phenomena are often called twoslit interference). Under these assumptions, the conditions for constructive and destructive interference are as follows:

for constructive interference

,

and for destructive interference

,

where is the separation between the two sources, is the wavelength of the light, is an arbitrary integer, and isthe angle between a line perpendicular to the line segment connecting the sources and the line from the midpoint of thatsegment to the point where the interference is being observed. These equations are often spoken of in terms of visiblelight, but they are, in fact, valid for any sort of waves, as longas the two sources fit the other criteria given.

Part A

Which of the following scenerios fits all of the criteria for the twosource interference equations to be valid?

ANSWER:

d sin θ = mλ (m = 0, ±1, ±2, …)

dsinθ = (m+ )λ (m = 0, ±1, ±2,…)12

d λ m θ

Correct

Part B

Which of the following statements explain why the twosource interference equations are not valid for an observerfar away from two red LED signal lights?

1. not monochromatic sources2. incoherent sources3. observed from a distance similar to or smaller than the separation between the sources

ANSWER:

Correct

Part C

Why are the twosource interference equations not valid for an observer on a road far away from two neighboringradio towers for different radio stations?

1. sources emit at different frequencies (i.e., not monochromatic sources)2. incoherent sources3. observed from a distance similar to or smaller than the separation between the sources

ANSWER:

An observer is standing far away from two red LED signal lights.

Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto ascreen with two slits in it and the light from the two slits finally shines onto a faraway screen.

An observer stands on a road far away from two neighboring radio towers for different radio stations.

Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto ascreen with two slits in it and the light from the two slits finally shines onto a nearby screen.

An observer stands on a road that runs five kilometers away from the two synchronized transmittingtowers for a radio station.

1 only

2 only

3 only

1 and 2 only

1 and 3 only

2 and 3 only

all three

Correct

Part D

Why are the twosource interference equations not valid for light from an incandescent bulb that shines onto ascreen with a single slit, and then the light shines onto a screen with two slits in it and the light from the two slitsfinally shines onto a nearby screen?

1. not monochromatic sources2. incoherent sources3. observed from a distance similar to or smaller than the separation between the sources

ANSWER:

Correct

Part E

Consider a road that runs parallel to the line connecting a pair of radio towers that transmit the same station(assume that their transmissions are synchronized), which has an AM frequency of kilohertz. If the road is kilometers from the towers and the towers are separated by meters, find the angle to the first point ofminimum signal ( ). Hint: A frequency of kilohertz corresponds to a wavelength of meters for radiowaves.

Express your answer in radians, to two significant figures.

ANSWER:

1 only

2 only

3 only

1 and 2 only

1 and 3 only

2 and 3 only

all three

1 only

2 only

3 only

1 and 2 only

1 and 3 only

2 and 3 only

all three

1000 5400 θ

m = 0 1000 300

Correct

Recall that the smallangle approximations only hold for very small angles. This angle is almost too big. Ifgreater than two significant figures of accuracy were needed in this problem, then it would no longer be valid touse the smallangle approximations.

Part F

If the angle in the twosource interference equations is small, then using smallangle approximations yields the

equation , where is the distance from the sources to the points where they are beingdetected (in Young's experiment the screen, in this example the road), and is the distance from the centralmaximum to the minimum of order . Use this equation to find the distance from the central maximum to theminimum in the previous part.

Express your answer in meters to two significant figures.

ANSWER:

Correct

Problem 22.32

Light from a heliumneon laser ( = 633 ) is used to illuminate two narrow slits. The interference pattern is observedon a screen 2.8 behind the slits. Twelve bright fringes are seen, spanning a distance of 52 .

Part A

What is the spacing (in ) between the slits?

Express your answer using two significant figures.

ANSWER:

All attempts used; correct answer withheld by instructor

Problem 22.38

A diffraction grating having 520 diffracts visible light at 34.0 .

= 0.38 radians θ

θ

= Rym(m+1/2)λ

dR

ymm

= 1900 y0 m

λ nmm mm

mm

= d .409 mm

lines/mm ∘

Part A

What is the light's wavelength?

Express your answer with the appropriate units.

ANSWER:

Correct

Problem 22.40

A tripleslit experiment consists of three narrow slits, equally spaced by distance and illuminated by light ofwavelength . Each slit alone produces intensity on the viewing screen at distance . Consider a point on the distant viewing screen such that the pathlength difference between any two adjacent slits is .

Part A

What is the intensity at this point?

Express your answer in terms of .

ANSWER:

Correct

Part B

What is the intensity at a point where the pathlength difference between any two adjacent slits is ?

Express your answer in terms of .

ANSWER:

Correct

Problem 22.70

The intensity at the central maximum of a doubleslit interference pattern is . The intensity at the first minimum iszero.

= 538 λ nm

dλ I1 L

λ

I1

= I 9I1

λ/2

I1

= I I1

4I1

Part A

At what fraction of the distance from the central maximum to the first minimum is the intensity ?

ANSWER:

Correct

Problem 22.71

Light consisting of two nearly equal wavelengths and , where , is incident on a diffraction grating.The slit separation of the grating is .

Part A

Find the angular separation of these two wavelengths in the th order.

Express your answer in terms of the given quantities.

ANSWER:

Correct

Part B

Sodium atoms emit light at 589.0 and 589.6 . What is the firstorder angular separations (in degrees) ofthese two wavelengths for a 600 grating?


ANSWER:

Correct

Part C

What is the secondorder angular separations (in degrees) of these two wavelengths for a 600 grating?


I1

= 0.667y/y1

λ + Δλ λ Δλ ≪ λd

m

= ΔθΔλ

−( )dm

2λ2√

nm nmlines/mm

= 2.2×10−2 Δθ ∘

lines/mm

ANSWER:

Correct

Score Summary:Your score on this assignment is 102%.You received 7.17 out of a possible total of 9 points, plus 2 points of extra credit.

= 5.8×10−2 Δθ ∘

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Mastering Physics HW 8 Ch 22 - Wave Optics