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Homework 7Due: 11:59pm on Sunday, May 8, 2011

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

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Where Is the Image?

Learning Goal: To understand how to use ray tracing to determine image location and size for a convexlens.

Ray tracing is one way to determine the location and size of an image formed by a lens. Ray tracing allowsyou to follow the path of a few principal rays originating from a point on the imaged object. The thin lensapproximation simplifies ray tracing by ignoring details of light propagation inside the lens.

A converging lens focuses a set of light rays entering the lens on one side parallel to its axis to a single focalpoint on the opposite side of the lens. If the principal rays emitted from a single point on an object allintersect at a single point after passing through a converging lens, then this intersection point helps indetermining the location and size of a real image of the object. The most useful principal rays for converginglenses are the following:The P ray is directed parallel to the axis of the lens. It emerges from the lens directed toward the focal pointon the side of the lens opposite the object.The M ray is directed toward the midpoint of the lens. It emerges from the lens unchanged in direction.The F ray is directed toward the focal point on the same side of the lens as the object. It emerges from thelens directed parallel to the lens axis.

Part A

Trace the P ray and M ray from the tip of the object for the converging lens shown (use the label for the

segment of the P ray on the same side as the object and for the segment on the opposite side). The

points labeled F are the focal points of the lens. Be sure to extend the M ray enough to intersect the P ray.

Hint A.1 Path of the P ray

Hint not displayed

Hint A.2 Path of the M ray

Hint not displayed

Draw the vectors for the incident rays starting from the tip of the object. The location andorientation of the vectors will be graded.

ANSWER:

View All attempts used; correct answer displayed

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

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Part B

If the focal length (the distance from the lens to either focal point F) of the lens is , which of the following

is true of the horizontal distance from the lens to the image?

Hint B.1 Location of the image formed by the lens

Hint not displayed

ANSWER:

Correct

Part C

Now add the F ray to your diagram (use the label for the segment of the F ray on the same side as the

object and for the segment on the opposite side).

Hint C.1 Path of the F ray

Hint not displayed

Draw the vectors starting from the tip of the object. The location and orientation of the vectors willbe graded.

ANSWER:


As you can see, all three principal rays pass through the same intersection point. In fact, all rays froma single point on the object will pass through a single point on the opposite side of a converging lens.Therefore you can use the intersection point of any two rays to find the location of the image; theprincipal rays are usually the easiest ones to use.

Part D

What happens to the image formed by the same lens if the distance between the object and the focal pointis reduced? Assume that the object remains to the left of the focal point.


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Hint D.1 How to approach the problem

Hint not displayed

ANSWER: The image becomes smaller.

The image is unchanged.

The image becomes larger.

The image disappears.

Correct

The situation described so far is typical of convex lenses. Whenever the object is farther from the lensthan the focal point, the lens creates a real image that is inverted. If the distance between the objectand the focal point is reduced, the size of the image is increased. This is basically what happens in acamera. When you zoom in, you are actually moving the lens closer to the object, reducing the distancebetween the object and the focal point of the lens. As a result, the size of the image on the filmincreases.

Part E

Now consider the case in which the object is between the lens and the focal point. Trace the P ray (use thelabel for the segment of the P ray on the same side as the object and for the segment on the

opposite side) and the M ray.

Draw the vectors for the incident rays starting from the tip of the object. The location andorientation of the vectors will be graded.

ANSWER:

View Correct

In this case the P and M rays diverge. The lens now forms a virtual image that can be found byextending the diverging rays backward, as shown in the figure.


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A convex lens can act as both a converging lens and a diverging lens. It forms a real image when theobject is located at a distance from the lens greater than the distance of the focal point, while it formsa virtual image when the object is between the focal point and the lens.

A Simple Magnifier

Part A

To view an enlarged upright image of an object through a simple magnifier, where must the object be located?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Angular magnification

Hint not displayed

ANSWER:

Correct

Part B

Two people, Micah and Lyra, with different near points are equally close to an object. Both inspect the objectthrough the same magnifier by holding the lens close to the eye. Micah's near point is located farther away fromhis eye than Lyra's near point is located relative to her eye. Micah will experience a larger magnification forwhich of the following reasons?

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Find the object angular size when the object is at the near point

Hint not displayed


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Check all that apply.

ANSWER: When the object is located at his near point, the angular size of the object issmaller for Micah than for Lyra.

Micah sees a larger image than Lyra sees.

Micah can see the image clearly from a larger distance than Lyra can.

Lyra can see the image clearly from a larger distance than Micah can.

The angular size of the image relative to the angular size of the object at the nearpoint is greater for Micah than for Lyra.

Correct

The angular magnification produced by a magnifier depends not only on the focal length of the magnifier,but also on the near point of the observer. Since the average person has a near point 25 from the eye,the angular magnification is usually expressed as

,

but this expression changes for different near-point distances.

An Antique TelescopeIn a two-lens system, the image produced by one lens acts as the object for the next lens. This simpleprinciple finds applications in many optical instruments, including some of common use such as themicroscope and the telescope.

Part A

On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath theinstrument says that the telescope has a magnification of 20 and consists of two converging lenses, theobjective and the eyepiece, fixed at either end of a tube 60.0 long. Assuming that this telescope wouldallow an observer to view a lunar crater in focus with a completely relaxed eye, what is the focal length

of the eyepiece?Note that to view the crater with a completely relaxed eye, the eyepiece must form its image at infinity.


Hint not displayed

Hint A.2 Magnification of a telescope

Hint not displayed

Hint A.3 Find an expression for the image distance for the objective

Hint not displayed

Hint A.4 Find an expression for the object distance for the eyepiece

Hint not displayed


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Hint A.5 Find an expression for the length of the telescope

Hint not displayed

Express your answer in centimeters.

ANSWER: = 2.86Correct

Multiple Optics with Plane Mirror - Virtual ObjectsA plane mirror is located at the origin. A converging lens with focal length 5.00 is located at .

An object is placed at .

Part A

What is the location of the final image, as seen by an observer looking toward the mirror through the lens?

Hint A.1 Multiple Optics

Hint not displayed

Hint A.2 Find the location of the image due to the lens

Hint not displayed

Hint A.3 Find the location of the image formed by the plane mirror

Hint not displayed

Hint A.4 The image formed by the lens (after reflection)

Hint not displayed

Hint A.5 Find the distance from the lens to the object created by the mirror

Hint not displayed

Express your answer in meters, to three significant figures.

ANSWER: = 3.22

Correct

Part B

Is the image in Part A real or virtual?

ANSWER: real


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virtual

Correct

Part C

What is the magnitude of the magnification of the image found in Part A?

Hint C.1 Magnification in multiple optics

Hint not displayed

Express your answer to three significant figures.

ANSWER: = 0.111

Correct

Part D

Is the image in Part A upright or inverted?

ANSWER: upright

inverted

Correct

Reflection and Refraction Ranking TaskA ray of light is incident onto the interface between material 1 and material 2.

Part A


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Given the indices of refraction and of material 1 and material 2, respectively, rank these scenarios onthe basis of the phase shift in the refracted ray.

Hint A.1 Distinguish between reflection and refraction

The light that reflects off of the interface does so in a similar manner to light reflecting from a mirror. Thelight that refracts through the interface has its path bent by the differing indices of refraction of the twomaterials. A phase shift occurs in only one of these two processes. Which one?

ANSWER: reflection

refraction

Correct

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:


Part B

Rank these scenarios on the basis of the phase shift in the reflected ray.

Hint B.1 Phase change upon reflection

Hint not displayed


ANSWER:

View Correct

Reflection Vector Drawing


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Vittorio needs to clean the dusty mirror hanging on the wall before he can use it, but he would like to clean aslittle of the mirror as possible.

Part A

Vittorio would like to be able to see the logo on his shirt. Draw the incident and reflected rays showing thelight from the logo reflecting off the mirror into his eyes. The rays should meet at the point on the mirror thatneeds cleaning.

Hint A.1 The law of reflection

Hint not displayed

Hint A.2 Equal angles and equal distances

Hint not displayed

Draw the vectors, for the reflected and incident ray, starting and ending at the surface of the mirrorrespectively. The location and orientation of the vectors will be graded.

ANSWER:

View Correct

Part B

Vittorio would like to be able to check whether his shoelaces are tied. Draw the incident and reflected raysshowing the light from his shoelaces reflecting off the mirror into his eyes. Again, the rays should meet atthe point on the mirror that needs cleaning.(See the hints from Part A if you need help.)

Draw the vectors, for the reflected and incident ray, starting and ending at the surface of the mirrorrespectively. The location and orientation of the vectors will be graded.

ANSWER:

View Correct

± Image Size in a Mirror


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A guy named Joe, who is 1.6 meters tall, enters a room in which someone has placed a large convex mirrorwith radius of curvature equal to 30 meters. The mirror has been cut in half, so that the axis of the mirror is

at ground level. As Joe enters the room, he is 5meters in front of the mirror, but he is looking the otherway, so he fails to see it. When he turns around, he isstartled by his own image in the mirror.

Part A

How far away does the image appear to Joe?

Hint A.1 Find the focal length

Hint not displayed

Hint A.2 Finding the distance from Joe to his image

Hint not displayed

Express your answer in meters, to three significant figures or as a fraction.

ANSWER: 8.75Correct

Part B

What is the image height that Joe sees in the mirror?

Hint B.1 Finding the image height

Hint not displayed


ANSWER: = 1.2

Correct

Joe is so startled by his image that he falls forward. (Assume that his feet stay at the same position.)


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Part C

Now what is the length (i.e., the distance from head to toe) of Joe's image?

Hint C.1 Finding the image's length

Hint not displayed


ANSWER: 0.9783Correct

How Deep Is the Goldfish?A tank whose bottom is a mirror is filled with water to a depth of 20.9 . A small fish floats motionless 8.00

under the surface of the water.

Part A

What is the apparent depth of the fish when viewed at normal incidence to the water?

Hint A.1 Use the object-image relation

Hint not displayed

Hint A.2 Radius of curvature of the surface

Hint not displayed

Express your answer in centimeters. Use 1.33 for the index of refraction of water.

ANSWER: 6.02Correct

Part B

What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normalincidence?

Hint B.1 Follow the path of light


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Hint not displayed

Express your answer in centimeters. Use 1.33 for the index of refraction of water.

ANSWER: 25.4Correct

± Shutter Speeds with Different LensesCamera A has a lens with an aperture diameter of 8.40 . It photographs an object using the correctexposure time of 3.33×10−2 .

Part A

What exposure time should be used with camera B in photographing the same object with the same film ifthis camera has a lens with an aperture diameter of 22.8 ? The lenses of cameras A and B have thesame focal length.


Hint not displayed

Hint A.2 Time is of essence

Hint not displayed

Express your answer in seconds to three significant figures.

ANSWER: 4.52×10−3

Correct

Nearsightedness and FarsightednessA person with normal vision can focus on objects as close as a few centimeters from the eye up to objectsinfinitely far away. There exist, however, certain conditions under which the range of vision is not so extended.For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while afarsighted person cannot focus on objects closer than a certain point (the near point). Note that even thoughthe presence of a near point is common to everyone, a farsighted person has a near point that is much fartherfrom the eye than the near point of a person with normal vision.Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In thiscase, the eye converges the light coming from the image formed by the corrective lens rather than from theobject itself.

Part A

When glasses (or contact lenses) are used to correct nearsightedness, where should the corrective lensform an image of an object located at infinity in order for the eye to form a clear image of that object?

Hint A.1 Range of vision in nearsightedness


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Hint not displayed

ANSWER: The lens should form the image at the near point.

The lens should form the image at the far point.

The lens should form the image at a point closer to the eye than the nearpoint.

The lens should form the image at a point farther from the eye than the farpoint.

Correct

This effect is achieved by the use of a diverging lens, as shown in the figure.

Part B

If a nearsighted person has a far point that is 3.50 from the eye, what is the focal length of the

contact lenses that the person would need to see an object at infinity clearly?


Hint not displayed

Hint B.2 Find the object distance

Hint not displayed

Hint B.3 Find the image distance

Hint not displayed

Express your answer in meters.

ANSWER: = -3.50Correct


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Part C

When glasses (or contact lenses) are used to correct farsightedness, where should the corrective lens forman image of an object located between the eye and the near point in order for the eye to form a clear imageof that object?

Hint C.1 Range of vision in farsightedness

Hint not displayed

ANSWER: The lens should form the image at the near point.

The lens should form the image at the far point.

The lens should form the image at a point closer to the eye than the nearpoint.

The lens should form the image at a point farther from the eye than the farpoint.

Correct

This effect is achieved by the use of a converging lens, as shown in the figure.

Part D

If a farsighted person has a near point that is 0.600 from the eye, what is the focal length of the

contact lenses that the person would need to be able to read a book held at 0.350 from the person'seyes?

Hint D.1 How to approach the problem

Hint not displayed

Hint D.2 Find the object distance

Hint not displayed


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Hint D.3 Find the image distance

Hint not displayed



Double Slit with ReflectionsA radar tower sends out a signal of wavelength . It is

meters tall, and it stands on the edge of the ocean.A weather balloon is released from a boat that is adistance out to sea. The balloon floats up to an

altitude . In this problem, assume that the boat and

balloon are so far away from the radar tower that thesmall angle approximation holds.

Part A

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

Hint A.1 Small-angle approximation

Hint not displayed

Hint A.2 Interference with reflections

Hint not displayed

Hint A.3 Double slit approximation

Hint not displayed


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Express your answer in terms of , , and .

ANSWER: =

Correct

Part B

What is the maximum wavelength that will interfere constructively?


ANSWER: =

Correct

Interference of Two Radio WavesTwo coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves withwavelength 6.00 meters. Consider points along the line connecting the two sources.

Part A

At what distance from source A is there constructive interference between points A and B?


Hint not displayed

Hint A.2 An equation for constructive interference

Hint not displayed


ANSWER: 2.5Correct

Part B

At what distances from source A is there destructive interference between points A and B?


Hint not displayed

Hint B.2 An equation for destructive interference


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Hint not displayed

Note that there will be two separate interference fringes between point A and point B. Enter youranswers in ascending order separated by a comma.

ANSWER: 1,4Correct

Two-Slit InterferenceAs 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. (Itcontains the soul of quantum mechanics only after you swallow the preposterous notion that particles inmotion are described by a wave equation rather than the laws of Newtonian mechanics.)In this problem, you will look at a classic wave interference problem involving electromagnetic waves. Young'sdouble-slit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one ofthe most elegant experiments in physics (because it demonstrates the important concept of interference sosimply). For the purposes of this problem, we assume that two long parallel slits extending along the z axis(out of the plane shown in the figure) are separted by a distance . They are illuminated coherently, that is, in

phase, by light with a wavelength , for example by a laser beam polarized in the z direction. (Lacking a

laser, Young used an intense source diffracted by a slit to produce 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 sumof the electric fields emanating from each of the two slits. Hence, the relative phase of these electric fields atsome observation point determines whether they add in phase (constructively) or out or phase

(destructively).To refresh your memory about traveling waves, theelectric field that is incident on the double slits

from the left is a function of and . Let us assume

that it has amplitude . We will also assume a

cosine trigonometric function with the arbitrary phaseset 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 the angular frequency ( ) and is the

wave number defined by . The phase increases by each time the distance increases by .

Moreover, the phase is constant for an observer moving in the positive x direction at the speed of light, i.e.,for whom .

Part A

Now consider the electric field observed at a point that is far from the two slits, say at a distance

from the midpoint of the segment connecting the


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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 slitsto point are not equal; hence the waves will be

out of phase due to the longer distance traveled bythe wave from one slit relative to the other.Calculate the phase of the wave from

the lower slit that arrives at point .

Hint A.1 Definition of phase

Hint not displayed

Hint A.2 Calculate the extra distance

Hint not displayed

Hint A.3 Find the phase of the lower wave

Hint not displayed

Express your answer in terms of , , , , , , and constants like .

ANSWER: =

Answer Requested

Part B

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

Hint B.1 Find the difference in distances

The distance from the upper slit to point is slightly shorter than . Find the distance by which the

path is shorter. Be sure to exploit the fact that . This means that the path from the upper slit to

point is essentially parallel to the path from the middle of the slits to point . Hence all the path

difference occurs in the vicinity of the slits.

Express your answer in terms of and .

ANSWER: = Answer not displayed

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


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ANSWER: =

Answer Requested

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 from midway between the slits is

, now find the magnitude of the combined field at due to the two slits. You may ignore variations

in the maximum amplitude and consider only variations in phase of the waves emerging from the slits.

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

Hint not displayed


ANSWER: =

Answer Requested

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 two-slit interference is the dependence of the total intensity at point on the angle .

Find this intensity .

The formula for intensity is


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.

Hint D.1 Amplitude of the electric field

Hint not displayed

Express your answer in terms of , , , and , where . Note: should be

coded as cos(x)^2.

ANSWER: =

Answer Requested

Part E

Two-slit 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 interferencepattern. Find the angular spacing of the interference pattern.

Hint E.1 How to approach the problem

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

difference in phase between two different directions equal to :

.

Note that once you make the substitution for both phases, you will have a relatively simple

expression involving , where .

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

ANSWER: =

Answer Requested

This is a famous formula:

.

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


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± Thin Film (Oil Slick)A scientist notices that an oil slick floating on water when viewed from above has many different rainbowcolors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelengthto be 750 (in air). The index of refraction of water is 1.33.

Part A

The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

Hint A.1 Thin-film interference

Hint not displayed

Hint A.2 Path-length phase difference

Hint not displayed

Hint A.3 Phase shift due to reflections

Hint not displayed

Express your answer in nanometers to three significant figures.

ANSWER: = 313

Correct

Part B

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

Hint B.1 Phase shift due to reflections

Hint not displayed


ANSWER: = 125

Correct

Part C

Now assume that the oil had a thickness of 200 and an index of refraction of 1.5. A diver swimmingunderneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is thelongest wavelength of the light in water that is transmitted most easily to the diver?

Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Determine the wavelength of light in air


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Hint not displayed

Hint C.3 Relationship between wavelength and index of refraction

Hint not displayed


ANSWER: = 451Correct

This problem can also be approached by finding the wavelength with the minimum reflection.Conservation of energy ensures that maximum transmission and minimum reflection occur at the sametime (i.e., if the energy did not reflect, then it must have been transmitted to conserve energy), sofinding the wavelength of minimum reflection must give the same answer as finding the wavelength ofmaximum transmission. In some cases, working the problem one way may be substantially easier, soyou should keep both approaches in mind.

Antireflective CoatingA thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). The indexof refraction of the polystyrene is 1.49, and the index 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 450nanometers.


Hint not displayed

Hint A.2 What is the phase shift?

Hint not displayed

Hint A.3 Find the wavelength of light in the coating


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Hint not displayed

Express your answer in nanometers.

ANSWER: 75.5Correct

± Double Slit 2A laser with wavelength is shining light on a double slit with slit separation 0.400 . This results in an

interference pattern on a screen a distance away from the slits. We wish to shine a second laser, with a

different wavelength, through the same slits.

Part A

What is the wavelength of the second laser that would place its second maximum at the same location

as the fourth minimum of the first laser, if = 0.400 ?

Hint A.1 Second maximum and fourth minimum

Hint not displayed

Hint A.2 Locating interference maxima and minima

Hint not displayed

Express your answer in millimeters.

ANSWER: = 8.75×10−2

Correct

Diffraction and Interference Ranking Task

A monochromatic beam of light is sent through each of the following six optical slides.

Part A

Rank these scenarios on the basis of the angle of the first interference maximum.

Hint A.1 Double-slit interference

Hint not displayed

Hint A.2 Diffraction gratings

Hint not displayed


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Hint A.3 Single-slit interference

Hint not displayed


ANSWER:

View Correct

Single-Slit Diffraction Conceptual QuestionAn experiment is conducted in which red light is diffracted through a single slit.

Then, each of the following alterations to the original experiment is made, one at a time, and the experimentis repeated. After each alteration, the experiment is returned to its original configuration.

AThe slit width is halved.

BThe distance between the slits and the screen ishalved.

CThe slit width is doubled.DA green, rather than red, light source is used.EThe experiment is conducted in a water-filled tank.

F The distance between the slits and the screen isdoubled.

Part A

Which of these alterations decreases the angles at which the diffraction minima appear?

Hint A.1 Minima of single-slit diffraction

Hint not displayed

Hint A.2 Effect of moving the screen

Hint not displayed

Hint A.3 Wavelength of light in water

Hint not displayed

Enter the letters of the correct statements in alphabetical order. For example, if statements A and Bare corrent, enter AB.


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ANSWER: CDECorrect

± Circular Diffraction PatternsMonochromatic light of wavelength nanometers is incident on a small pinhole in a piece of paper. On a

screen meters from the pinhole, you observe the diffraction pattern shown in the figure. You carefully

measure the diameter of the central maximum to be millimeters, as shown in the figure.

Part A

What is the diameter of the pinhole?

Hint A.1 Find the angular separation

Hint not displayed

Hint A.2 Solve for the diameter

Hint not displayed

Express your answer in millimeters, to three significant figures.


Doorway Diffraction

Part A

Sound with frequency 1290 leaves a room through a doorway with a width of 1.14 . At what minimum

angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound?Use 344 for the speed of sound in air and assume that the source and listener are both far enough


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from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.


Hint not displayed

Hint A.2 The equation for "dark" fringes

Hint not displayed

Hint A.3 Find the wavelength of the sound wave

Hint not displayed

Express your answer in radians.

ANSWER: 0.236Correct radians

Light of Differing Wavelengths on a Diffraction GratingVisible light passes through a diffraction grating that has 900 slits per centimeter, and the interference patternis observed on a screen that is 2.26 from the grating.

Part A

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.62 . What is the difference between these wavelengths?


Hint not displayed

Hint A.2 Equation for the intensity maxima

Hint not displayed

Hint A.3 Find the distance between two slits

Hint not displayed

Hint A.4 Finding the difference between wavelengths

Hint not displayed

Hint A.5 Find the angle between the maxima

Hint not displayed



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ANSWER: 1.29×10−8

Correct

± A Diffraction Grating SpectrometerSuppose that you have a reflection diffraction grating with 115 lines per millimeter. Light from a sodiumlamp passes through the grating and is diffracted onto a distant screen.

Part A

Two visible lines in the sodium spectrum have wavelengths 498 and 569 . What is the angularseparation of the first maxima of these spectral lines generated by this diffraction grating?

Hint A.1 Find reflection angle of 498 spectral line

Hint not displayed

Hint A.2 Find reflection angle of 569 line

Hint not displayed

Express your answer in degrees to two significant figures.


Part B

How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59nanometers, which are a well known pair of lines for sodium, in the second order ( )?

Hint B.1 Find the necessary spectral resolving power

Hint not displayed

Hint B.2 Two expressions for resolving power

Hint not displayed

Hint B.3 Relation between and the grating width.

Hint not displayed

Express your answer in millimeters to two significant figures.

ANSWER: 4.3Correct


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Most diffraction gratings are much wider than this. What you are actually finding is the width of the partof the grating that must be illuminated. The parts of the grating that no light shines onto obviously can'taffect the way the light diffracts. To see why illuminating more of the grating gives better resolvingpower, recall that the width of an interference maximum decreases as you add more slits to the screenthat the light passes through. Illuminating more of the grating has the same effect as adding slits to thescreen. If the interference maxima for different wavelengths are narrower, the wavelengths can beresolved more finely.

± Resolving Power of the EyeIf you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arcminute,equal to 1.67×10−2 .

Part A

If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system doesthis correspond? Use Rayleigh's criterion and assume that the wavelength of the light is 530 .

Hint A.1 Definition of Rayleigh's criterion

Recall that Rayleigh's criterion for resolving objects is that two objects are just distinguishable from eachother if the center of one object's diffraction pattern lies on the first minimum of the other object'sdiffraction pattern.

Hint A.2 Diffraction equation for a circular aperature

The diffraction equation for a circular aperature is

,

where is the wavelength of light used, is the diameter of the circular aperture, and is the angle at

which the first minimum lies (relative to a centerline from the aperature). The factor 1.22 comes from thefact that we are using a circular aperture for the diffraction pattern instead of an infinite slit.


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Express your answer in millimeters to three significant figures.

ANSWER: 2.22Correct

Score Summary:Your score on this assignment is 94.9%.You received 94.92 out of a possible total of 100 points.


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