Assg10 Answers Mastering Physics

11/1/2015 Assignment 10

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4032261 1/26

Assignment 10Due: 11:59pm on Sunday, November 1, 2015

You will receive no credit for items you complete after the assignment is due. Grading Policy

± FM Radio Interference

You are listening to the FM radio in your car. As you come to a stop at a traffic light, you notice that the radio signal is fuzzy. By pulling up a short distance,you can make the reception clear again. In this problem, we work through a simple model of what is happening.Our model is that the radio waves are taking two paths to your radio antenna:

the direct route from the transmitteran indirect route via reflection off a building

Because the two paths have different lengths, they can constructively or destructively interfere. Assume that the transmitter is very far away, and that thebuilding is at a 45degree angle from the path to the transmitter.

Point A in the figure is where you originally stopped, and point B is where the station is completelyclear again. Finally, assume that the signal is at its worst at point A, and at its clearest at point B.

Part A

What is the distance between points A and B?

Express your answer in wavelengths, as a fraction.

Hint 1. What is the pathlength difference at point A?

Since we know that the waves traveling along the two paths interfere destructively at point A, we know something about the difference in the

d

N



lengths of those two paths. What is the difference between the two path lengths, in integer multiples of the wavelength?

ANSWER:

Hint 2. What is the pathlength difference at point B?

Since we know that the waves traveling along the two paths interfere constructively at point B, we know something about the difference in thelengths of those two paths. What is the difference between the two path lengths, in integer multiples of the wavelength?

ANSWER:

Hint 3. What is the path length of reflected waves?

Consider the lengths of the paths taken by the two reflected waves. Note that the path from the transmitter to the building is larger for one wave,while the path from the building to the antenna is larger for the other. What is the difference in length between the path of the reflected wave fromthe transmitter to A, and the path of the reflected wave from the transmitter to B, in integer multiples of the wavelength?

ANSWER:

N

0N2N+1

2N

N + 12

N

0N2N+1

2N

N + 12

N



ANSWER:

Correct

Part B

Your FM station has a frequency of megahertz. The speed of light is about meters per second. What is the distance between points Aand B?

Express your answer in meters, to two significant figures.

ANSWER:

Correct

Problem 26.02

A person with a radiowave receiver starts out equidistant from two FM radio transmitters and that are 11.0 apart, each one emitting inphase radiowaves at 92.0 . She then walks so that she always remains 50.0 from transmitter . (See the figure.) Limit your solution to the cases where

.

0N2N+1

2N

N + 12

= 0.500 wavelengths d

100 3.00 × 108 d

= 1.5 d m

A B mMHz m B

50.0 m ≤ x ≤ 65.0 m



Part A

For what values of will she find the radio signal to be maximally enhanced?

Enter your answers in increasing order separating them with commas.

ANSWER:

Correct

Part B

For what values of will she find the radio signal to be cancelled?

Enter your answers in increasing order separating them with commas.

ANSWER:

x

50.0,53.3,56.5,59.8,63.0 m

x

51.6,54.9,58.2,61.4,64.7 m



Correct

Introduction to TwoSource Interference

Learning Goal:

To gain an understanding of constructive and destructive interference.

Consider two sinusoidal waves (1 and 2) of identical wavelength , period , and maximum amplitude . A snapshot of one of these waves taken at a certaintime is displayed in the figure below. Let and represent the displacement of eachwave at position at time . If these waves were to be in the same location ( ) at the same time,they would interfere with one another. This would result in a single wave with a displacement given by

.This equation states that at time the displacement of the resulting wave at position is thealgebraic sum of the displacements of the waves 1 and 2 at position at time . When themaximum displacement of the resulting wave is less than the amplitude of the original waves, thatis, when , the waves are said to interfere destructively because the result is smaller thaneither of the individual waves. Similarly, when , the waves are said to interfereconstructively because the resulting wave is larger than either of the individual waves. Notice that

.

Part A

To further explore what this equation means, consider four sets of identical waves that move in the +x direction. A photo is taken of each wave at time and is displayed in the figures below.

Rank these sets of waves on the basis of the maximum amplitude of the wave that results from the interference of the two waves in each set.

Rank from largest amplitude on the left to smallest amplitude on the right. To rank items as equivalent, overlap them.

ANSWER:

λ T A(x, t)y1 (x, t)y2

x t xy(x, t)

y(x, t) = (x, t) + (x, t)y1 y2t y(x, t) x

x t

< Aymax> Aymax

0 ≤ ≤ 2Aymax

t



Correct

When identical waves interfere, the amplitude of the resulting wave depends on the relative phase of the two waves. As illustrated by the set ofwaves labeled A, when the peak of one wave aligns with the peak of the second wave, the waves are in phase and produce a wave with the largestpossible amplitude. When the peak of one wave aligns with the trough of the other wave, as illustrated in Set C, the waves are out of phase by and produce a wave with the smallest possible amplitude, zero!

Part B

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except that one of them isshifted relative to the other in the pair by the distance shown:

A. B. C. D. E. F. G.

Identify which of the seven pairs will interfere constructively and which will interfere destructively. Each letter represents a pair of waves.

Enter the letters of the pairs that correspond to constructive interference in alphabetical order and the letters of the pairs that correspond topairs that interfere destructively in alphabetical order separated by a comma. For example if pairs A, B and D interfere constructively and pairsC and F interfere destructively enter ABD,CF.

λ/2

−(1/2)λ2λ−5λ(3/2)λ0(17/2)λ(6/2)λ



ANSWER:

CorrectDo you notice a pattern? When the phase difference between two identical waves can be written as , where , thewaves will interfere constructively. When the phase difference can be expressed as , where , the waves willinterfere destructively.

Consider what water waves look like when you throw a rock into a lake. These waves start at the point where the rock entered the water and travel out in alldirections. When viewed from above, these waves can be drawn as shown, where the solid linesrepresent wave peaks and troughs are located halfway between adjacent peaks.

Part C

Now look at the waves emitted from two identical sources (e.g., two identical rocks that fall into a lake at the same time). The sources emit identicalwaves at the exact same time.

Identify whether the waves interfere constructively or destructively at each point A to D.

For points A to D enter either c for constructive or d for destructive interference. For example if constructive interference occurs at points A, Cand D, and destructive interference occurs at B, enter cdcc.

BCEG,ADF

λmC = 0, ±1, ±2, ±3, …mC(λ/2)mD = ±1, ±3, ±5, …mD



Hint 1. How to approach the problem

Recall that constructive interference occurs when the two waves are in phase when they interfere, so that the peak (or trough) of one wave alignswith the peak (or trough) of the other wave. Destructive interference occurs when waves are out of phase so that the peak of one wave alignswith the trough of the other wave.

Study the picture to find where each type of interference occurs.

ANSWER:

Correct

Each wave travels a distance or from its source to reach Point B. Since the distance between consecutive peaks is equal to , from the picture youcan see that Point B is away from Source 1 and away from Source 2. The pathlength difference, , is the difference in the distance each wavetravels to reach Point B:

.

λ/2

ccdd

d1 d2 λ2λ 3λ ΔdB

Δ = − = 2λ − 3λ = −1λdB d1 d2



Part D

What are the pathlength differences at Points A, C, and D (respectively, , , and )?

Enter your answers numerically in terms of separated by commas. For example, if thepathlength differences at Points A, C, and D are , , and , respectively, enter4,.5,1.

ANSWER:

ΔdA ΔdC ΔdD

λ4λ λ/2 λ

, , = 0,1.5,0.5 , , ΔdA ΔdC ΔdD λ λ λ



Correct

Knowing the pathlength difference helps to confirm what you found in Part C. When the pathlength difference is , where , the waves interfere constructively. When the pathlength difference is , where ,

the waves interfere destructively.

Part E

What are the pathlength differences at Points L to P?

Enter your answers numerically in terms of separated by commas. For example, if thepathlength differences at Points L, M, N, O, and P are , , , , and ,respectively, enter 5,2,1.5,1,6.

ANSWER:

Correct

Every point along the line connecting Points L to P corresponds to a pathlength difference . This means that at every point along this line,waves from the two sources interfere constructively.

The figure below shows two other lines of constructive interference: One corresponds to a pathlength difference , and the othercorresponds to . It should make sense that the line halfway between the two sources corresponds to a pathlength difference of zero, sinceany point on this line is equally far from each source. Notice the symmetry about the line of the and the lines.

λmC= 0, ±1, ±2, ±3, …mC (λ/2)mD = ±1, ±3, ±5, …mD

λ5λ 2λ λ3

2λ 6λ

, , , , = 1,1,1,1,1 , , , , ΔdL ΔdM ΔdN ΔdO ΔdP λ λ λ λ λ

Δd = λ

Δd = −λΔd = 0

Δd = 0 Δd = λ Δd = −λ



A similar figure can be drawn for the lines of destructive interference. Notice that the pattern of lines is still symmetric about the line halfway betweenthe two sources; however, the lines along which destructive interference occurs fall midway between adjacent lines of constructive interference.

Constructive and Destructive Interference Conceptual Question

Two sources of coherent radio waves broadcasting in phase are located as shown below. Each grid square is 0.5 square, and the radio sources broadcastat .

mλ = 2.0 m



Part A

At Point A is the interference between the two sources constructive or destructive?

Hint 1. Pathlength difference

Since the two sources emit radio waves in phase, the only possible phase difference between the waves at various points is due to the differentdistances the waves have traveled to reach those points. The difference in the distances traveled by the two waves from source to point of interestis termed the pathlength difference.If the pathlength difference is an integer multiple of the wavelength of the waves, one wave will pass through an integer number of completecycles more than the other wave, placing the two waves back in perfect synchronization, resulting in constructive interference. If the pathlengthdifference is a halfinteger multiple of the wavelength, one wave will be onehalf of a cycle, or 180 , out of phase, resulting in destructiveinterference.

Hint 2. Find the pathlength difference

What is the distance from the left source to Point A, ? What is the distance from the right source to Point A, ?

Enter the distances in meters separated by a comma.

ANSWER:

degrees

dA,left dA,right



ANSWER:

Correct

Part B

At Point B is the interference between the two sources constructive or destructive?

Hint 1. Find the pathlength difference

What is the distance from the left source to Point B, ? What is the distance from the right source to Point B, ?

Enter the distances in meters separated by a comma.

ANSWER:

ANSWER:

Correct

, = 3,3 , dA,left dA,right m m

constructivedestructive

dB,left dB,right

, = 1.5,4.5 , dB,left dB,right m m




Part C

At Point C is the interference between the two sources constructive or destructive?

ANSWER:

Correct

Part D

At Point D is the interference between the two sources constructive or destructive?

ANSWER:

Correct

± Fringes from Different Interfering Wavelengths

Coherent light with wavelength 602 passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 fromthe slits. The firstorder bright fringe is a distance of 4.84 from the center of the central bright fringe.

Part A

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

Express your answer in micrometers (not in nanometers).



nm mmm




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

Hint 2. Interference pattern equation

The equation for the constructive interference fringes from two slits projected on ascreen is

,

where is the distance between the two slits, is the wavelength of light, and isthe angle between the constructive peak and the centerline. Note that

.

For the destructive interference pattern, one can use the equation

,

where all the variables are the same as for the case 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" destructive interference fringe. The correct way tolook at the situation is to use the main equation for destructive interference,

,

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:

d

d sin( ) = mλθm

d λ θm

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



Correct

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

Antireflective Coating

A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). The index of refraction of the polystyrene is 1.49, and theindex of refraction of the fabulite is 2.409.

Part A

What is the minimum thickness of film required? Assume that the wavelength of the light in air is 450 nanometers.

Express your answer in nanometers.


1.20 μm

d sin(θ) λλ 2λ



An antireflective coating works because destructive interference occurs between the light that is reflected off the surface of the coating and thelight that is reflected off the coating/substrate interface. As a result, there is a specific thickness of coating in which a certain wavelength ofreflected light will experience destructive interference.

Hint 2. What is the phase shift?

Which statement accurately describes the phase shift of the light reflected off the coating?

Hint 1. Phase shift of reflection off different materials

Recall that if light is reflected off a surface that has a higher index of refraction than that in which the light ray is being propagated, there willbe a halfcycle phase shift. If the light reflects off a surface with lower index of refraction than that in which it is being propagated, it will notexperience a phase shift.

ANSWER:

Hint 3. Find the wavelength of light in the coating

The light reflected off the top of the coating has no phase shift, while the light that reflects off the bottom has a halfwave phase shift.The light reflected off the top of the coating has a halfwave phase shift, and the light that reflects off the bottom also has a halfwavephase shift.The light reflected off the top of the coating has a halfwave phase shift, while the light that reflects off the bottom has no phase shift.The light reflected off the top of the coating has no phase shift, and the light that reflects off the bottom also has no phase shift.

λ



Find the wavelength of the light as it propagates in the antireflective coating.

Express your answer in nanometers.

Hint 1. Speed of light in a material

Recall that the speed of light in a material with index of refraction is given by

,

where is the speed of light in the material and is the speed of light in vacuum.

Hint 2. Relation between speed and frequency

The speed of a wave is related to its frequency and wavelength via the equation . The frequency of the light cannot changeas it passes into a material (or else there would be discontinuites over time), so the wavelength of the light must decrease by a factor of theindex of refraction, relative to the wavelength in vacuum, in order for the correct velocity of light in the material to be obtained.

ANSWER:

ANSWER:

Correct

Problem 26.22

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge, so that a very thinwedge of air is formed. The plates are illuminated at normal incidence by 548 light from a mercuryvapor lamp. Interference fringes are formed, with 16.0

.

λcoating

n

v = cn

v c

v f λ v = λf

= 302 λcoating nm

75.5 nm

nmfringes/centimeter



Part A

Find the angle of the wedge.

ANSWER:

All attempts used; correct answer displayed

Problem 26.32

Light of wavelength 585 falls on a slit 0.0666 wide.

Part A

On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central brightspot? Solve this problem without calculating all the angles! (Hint: What is the largest that can be? What does this tell you is the largest that canbe?)

ANSWER:

Correct

Part B

At what angle will the dark fringe that is most distant from the central bright fringe occur?

ANSWER:

= 2.51×10−2 θ ∘

nm mm

sinθ m

= 226 N fringes

= 83.0 |θ| ∘



Correct

Problem 26.40

Light of wavelength 631 passes through a diffraction grating having 299 .

Part A

What is the total number of bright spots (indicating complete constructive interference) that will occur on a large distant screen? Solve this problemwithout finding the angles. (Hint: What is the largest that can be? What does this imply for the largest value of ?)

Express your answer as an integer.

ANSWER:

All attempts used; correct answer displayed

Part B

What is the angle of the bright spot farthest from the center?

ANSWER:

Answer Requested

Problem 26.44

Electromagnetic waves of wavelength 0.169 fall on a crystal surface. As the angle of incidence is gradually increased, starting at 0 , you find that the

nm lines/mm

sin θ m

11

= 70.6 θ ∘

nm ∘

∘



first strong interference maximum occurs when the beam makes an angle of 21.4 with the surface of the crystal planes in the Bragg reflection.

Part A

What is the distance between the crystal planes?

ANSWER:

Answer Requested

Part B

At what other angles will interference maxima occur?

ANSWER:

Answer Requested

± Resolving Distant Objects

Consider a telescope with a small circular aperture of diameter 2.0 centimeters.

Part A

If two point sources of light are being imaged by this telescope, what is the maximum wavelength at which the two can be resolved if their angularseparation is radians?

Express your answer in nanometers to two significant figures.


∘

= 0.232 d nm

= 46.9 θ ∘

λ3.0 × 10−5



You are given the angular separation between two point sources and the diameter of the circular aperture that they travel through. You want toknow at what wavelength you will just satisfy the Rayleigh criterion. You already know the angular separation of the two centers of the diffractionpatterns (i.e., the angle between the two sources). Therefore, you can put that angle and the known diameter into the equation for Rayleigh'scriterion and then solve for .

Hint 2. Rayleigh's criterion

Recall that Rayleigh's criterion for resolvability says that two point sources are just resolved when the center of one's diffraction pattern coincideswith the first minimum of the other's diffraction pattern.

The angle to the first minimum from the center of a diffraction pattern made by a circular aperture of diameter is given by the equation

,

where is the wavelength of the light used to make the pattern.

ANSWER:

Correct

Part B

In this applet, change the settings to match the situation in Part A. Notice how the diffraction pattern looks when the two sources are at the resolutionlimit (i.e., just resolved). Now change the wavelength to 650 nanometers and the aperture diameter to 1.5 centimeters. Use the applet to roughlydetermine at what angular separation the two points are just resolved.

Express your answer in radians to two significant figures.

ANSWER:

Correct

λ

θ1 D

sin( ) = 1.22θ1λD

λ

= 490 λ nm

5.3×10−5 rad

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4032261



Part C

Calculate the angular separation at which two point sources of wavelength 600 nanometers are just resolved when viewed through a circular apertureof diameter 1.5 centimeters. Use the applet to check whether your answer is reasonable.

Express your answer in radians to three significant figures.

ANSWER:

Correct

Rayleigh's criterion is not a hard cutoff between resolvable and unresolvable. If you move the slider back and forth between radians and radians, it is unlikely that you will exclaim, "Wow! Point sources separated by radians are resolved so well!" However,

having a standard to decide whether a measurement is valid is quite important. Rayleigh's criterion serves this purpose by giving a lower bound forvalidity below which measurements from two different point sources are not generally considered reliable.

Problem 26.49: Resolution of the eye, I.

Even if the lenses of our eyes functioned perfectly, our vision would still be limited due to diffraction of light at the pupil.

Part A

Using Rayleigh's criterion, what is the smallest object a person can see clearly at his near point of 22.5 with a pupil 2.25 in diameter and light ofwavelength 492 ? (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.)

ANSWER:

Correct

Part B

Based upon your answer, does it seem that diffraction plays a significant role in limiting our visual acuity?

θ1

= 4.880×10−5 θ1 rad

4.8 × 10−5

5.0 × 10−5 5.0 × 10−5

cm mmnm

= 6.00×10−2 y mm



ANSWER:

Correct

Problem 26.54: Coating eyeglass lenses.

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye.

Part A

If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, what minimum thickness of film is neededon the lenses to cancel light of wavelength 548 reflected toward the eye at normal incidence?

ANSWER:

Answer Requested

Part B

Will any other wavelengths of visible light be cancelled in the reflected light?

ANSWER:

Yes.No.

nm

= 95.7 t nm



Correct

Part C

Will any other wavelengths of visible light be enhanced in the reflect light?

ANSWER:

Correct

Score Summary:Your score on this assignment is 69.8%.You received 90.72 out of a possible total of 130 points.

There are three visible wavelengths in air for which there is destructive interference.The only visible wavelength in air for which there is destructive interference is 550 .There are two visible wavelengths in air for which there is destructive interference.There are no visible wavelengths in air for which there is destructive interference.

nm

There are three visible wavelengths in air for which there is constructive interference.The only visible wavelength in air for which there is constructive interference is 550 .There are two other wavelengths in air for which there is construcive interference.There are no visible wavelengths in air for which there is constructive interference.

nm

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