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MasteringPhysics: Assignment Print View

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http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

Homework 7 Due: 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 [Switch to Standard Assignment View]

Where Is the Image? Learning Goal: To understand how to use ray tracing to determine image location and size for a convex lens. Ray tracing is one way to determine the location and size of an image formed by a lens. Ray tracing allows you to follow the path of a few principal rays originating from a point on the imaged object. The thin lens approximation 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 focal point on the opposite side of the lens. If the principal rays emitted from a single point on an object all intersect at a single point after ing through a converging lens, then this intersection point helps in determining the location and size of a real image of the object. The most useful principal rays for converging lenses are the following: The P ray is directed parallel to the axis of the lens. It emerges from the lens directed toward the focal point on 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 the lens 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 segment of the P ray on the same side as the object and

for the

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 and orientation of the vectors will be graded. ANSWER:

View All attempts used; correct answer displayed

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Part B If the focal length (the distance from the lens to either focal point F) of the lens is is true of the horizontal distance Hint B.1

, which of the following

from the lens to the image?

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 object and Hint C.1

for the segment of the F ray on the same side as the

for the segment on the opposite side). 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 will be graded. ANSWER:


As you can see, all three principal rays through the same intersection point. In fact, all rays from a single point on the object will 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; the principal 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 point is 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 lens than the focal point, the lens creates a real image that is inverted. If the distance between the object and the focal point is reduced, the size of the image is increased. This is basically what happens in a camera. When you zoom in, you are actually moving the lens closer to the object, reducing the distance between the object and the focal point of the lens. As a result, the size of the image on the film increases. Part E Now consider the case in which the object is between the lens and the focal point. Trace the P ray (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) and the M ray. Draw the vectors for the incident rays starting from the tip of the object. The location and orientation 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 by extending 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 the object is located at a distance from the lens greater than the distance of the focal point, while it forms a 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


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 object through the same magnifier by holding the lens close to the eye. Micah's near point is located farther away from his eye than Lyra's near point is located relative to her eye. Micah will experience a larger magnification for which of the following reasons? Hint B.1


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 is smaller 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 near point 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 Telescope In a two-lens system, the image produced by one lens acts as the object for the next lens. This simple principle finds applications in many optical instruments, including some of common use such as the microscope and the telescope. Part A On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath the instrument says that the telescope has a magnification of 20 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 60.0 long. Assuming that this telescope would allow 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 A.1


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:

= 2.86 Correct

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

is located 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 Task A 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 on the 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. The light that refracts through the interface has its path bent by the differing indices of refraction of the two materials. 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


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 as little 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 the light from the logo reflecting off the mirror into his eyes. The rays should meet at the point on the mirror that needs 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 mirror respectively. 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 rays showing the light from his shoelaces reflecting off the mirror into his eyes. Again, the rays should meet at the 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 mirror respectively. 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 mirror with 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 5 meters in front of the mirror, but he is looking the other way, so he fails to see it. When he turns around, he is startled 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:

8.75 Correct

Part B What is the image height Hint B.1

that Joe sees in the mirror?

Finding the image height Hint not displayed


= 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


0.9783 Correct

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:

6.02 Correct

Part B What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence? 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:

25.4 Correct

± Shutter Speeds with Different Lenses Camera A has a lens with an aperture diameter of 8.40

. It photographs an object using the correct

exposure 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 if this camera has a lens with an aperture diameter of 22.8 ? The lenses of cameras A and B have the same focal length. Hint A.1


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 Farsightedness A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely 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 a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from 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 lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself. Part A When glasses (or lenses) are used to correct nearsightedness, where should the corrective lens form 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 near point. The lens should form the image at a point farther from the eye than the far point. 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

lenses that the person would need to see an object at infinity clearly? Hint B.1


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:

= -3.50 Correct

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Part C When glasses (or lenses) are used to correct farsightedness, where should the corrective lens form an image of an object located between the eye and the near point in order for the eye to form a clear image of 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 near point. The lens should form the image at a point farther from the eye than the far point. 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

lenses that the person would need to be able to read a book held at 0.350

of the

from the person's

eyes? Hint D.1


Hint D.2

Find the object distance Hint not displayed

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Hint D.3


Find the image distance Hint not displayed


= 0.840 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 weather balloon is released from a boat that is a distance 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 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? 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 of

,

, and

.

ANSWER: = Correct

Part B What is the maximum wavelength Express your answer in of

that will interfere constructively? ,

, and

.

ANSWER: = Correct

Interference of Two Radio Waves Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 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 A.1


Hint A.2

An equation for constructive interference Hint not displayed


2.5 Correct

Part B At what distances from source A is there destructive interference between points A and B? Hint B.1


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 your answers in ascending order separated by a comma. ANSWER:

1,4 Correct

Two-Slit 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 the soul of quantum mechanics only after you swallow the preposterous notion that particles in motion 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's double-slit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one of the most elegant experiments in physics (because it demonstrates the important concept of interference so simply). 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 sum of the electric fields emanating from each of the two slits. Hence, the relative phase of these electric fields at some observation 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 left is a function of that it has amplitude

and . Let us assume

. We will also assume a

cosine trigonometric function with 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 the angular frequency ( wave number defined by

) and

is the

. 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 that


from the x axis. Here, far means

, a regime sometimes called Fraunhofer

diffraction. The critical point is that the distances from the slits to point are not equal; hence the waves will be out of phase due to the longer distance traveled by the wave from one slit relative to the other. Calculate the phase of the wave from the lower slit that arrives at point

.

Definition of phase

Hint A.1

Hint not displayed Calculate the extra distance

Hint A.2

Hint not displayed Find the phase of the lower wave

Hint A.3

Hint not displayed Express your answer in of

, ,

, , , , and constants like

.

ANSWER: = Answer Requested

Part B Now calculate the phase Hint B.1

of the wave from the upper slit that arrives at point

Find the difference in distances

The distance from the upper slit to point

is slightly shorter than . Find the distance

path is shorter. Be sure to exploit the fact that point

.

by which the

. This means that the path from the upper slit to

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 of ANSWER:

= Answer not displayed


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and .

, ,

, , , , and constants such as

.

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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

due to the two slits. You may ignore variations

at

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

Express your answer in of ANSWER:

,

, and

.

= 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 Find this intensity

on the angle .

.

The formula for intensity is

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.

Hint D.1

Amplitude of the electric field Hint not displayed


, ,

, 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 interference pattern. 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 expression involving

, where


for both phases, you will have a relatively simple .

,

, and any needed constants.


This is a famous formula: . The angular spacing of the interference pattern (in radians) is simply the ratio of the wavelength to the slit 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 rainbow colors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelength to 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 Hint A.1

of the oil slick at that spot?

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 Hint B.1

be now?

Phase shift due to reflections Hint not displayed


= 125 Correct

Part C Now assume that the oil had a thickness of 200

and an index of refraction of 1.5. A diver swimming

underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver? Hint C.1


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


= 451 Correct

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 same time (i.e., if the energy did not reflect, then it must have been transmitted to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the wavelength of maximum transmission. In some cases, working the problem one way may be substantially easier, so you should keep both approaches in mind.

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 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 450 nanometers. Hint A.1


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:

75.5 Correct

± Double Slit 2 A laser with wavelength

is shining light on a double slit with slit separation 0.400

interference pattern on a screen a distance

. This results in an

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 Hint A.1

= 0.400

?

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:

−2 = 8.75×10 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


View Correct

Single-Slit Diffraction Conceptual Question An 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 experiment is repeated. After each alteration, the experiment is returned to its original configuration. A The slit width is halved. The distance between the slits and the screen is B halved. CThe slit width is doubled. DA green, rather than red, light source is used. E The experiment is conducted in a water-filled tank. The distance between the slits and the screen is F doubled. 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 B are corrent, enter AB.

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ANSWER:


CDE Correct

± Circular Diffraction Patterns Monochromatic light of wavelength screen

nanometers is incident on a small pinhole in a piece of paper. On a

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 Hint A.1

of the pinhole?

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. ANSWER:

= 0.120 Correct

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 A.1


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:

0.236 radians Correct

Light of Differing Wavelengths on a Diffraction Grating Visible light es through a diffraction grating that has 900 slits per centimeter, and the interference pattern is 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 A.1


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

Express your answer in meters.

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ANSWER:


1.29×10−8 Correct

± A Diffraction Grating Spectrometer Suppose that you have a reflection diffraction grating with

115 lines per millimeter. Light from a sodium

lamp es through the grating and is diffracted onto a distant screen. Part A Two visible lines in the sodium spectrum have wavelengths 498 separation Hint A.1

and 569

. What is the angular

of the first maxima of these spectral lines generated by this diffraction grating? 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. ANSWER:

= 0.47 Correct

Part B How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers, 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:

4.3 Correct

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Most diffraction gratings are much wider than this. What you are actually finding is the width of the part of the grating that must be illuminated. The parts of the grating that no light shines onto obviously can't affect the way the light diffracts. To see why illuminating more of the grating gives better resolving power, recall that the width of an interference maximum decreases as you add more slits to the screen that the light es through. Illuminating more of the grating has the same effect as adding slits to the screen. If the interference maxima for different wavelengths are narrower, the wavelengths can be resolved more finely.

± Resolving Power of the Eye If 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 does this 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 each other if the center of one object's diffraction pattern lies on the first minimum of the other object's diffraction 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 the fact 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:

2.22 Correct

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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