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PHY208FALL2008

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The following three problems concern interference from two sources or along two paths.

Introduction to Two-Source Interference

Description: This problem provides students with a basic introduction to the interference of two identical waves. Theconditions for constructive and destructive interference are discussed in terms of phase difference and path length difference.

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 certain time is displayed in the figure below.Let and represent the displacement of each wave

at position at time . If these waves were to be in the same

location ( ) at the same time, they would interfere with oneanother. 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 the algebraic sum of thedisplacements of the waves 1 and 2 at position at time . When

the maximum displacement of the resulting wave is less than theamplitude of the original waves, that is, when , the

waves are said to interfere destructively because the result issmaller than either of the individual waves. Similarly, when , the waves are said to interfere constructively 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 istaken 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 twowaves in each set.

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

ANSWER:

View

When identical waves interfere, the amplitude of the resulting wave depends on the relative phase of the two waves. Asillustrated by the set of waves labeled A, when the peak of one wave aligns with the peak of the second wave, the wavesare in phase and produce a wave with the largest possible amplitude. When the peak of one wave aligns with the troughof 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!

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Part B

Consider sets of identical waves with the following phase differences:

A.

B.

C.

D.

E.

F.

G.

Identify which sets will interfere constructively and which will interfere destructively.

Enter the letters of the sets corresponding to constructive interference in alphabetical order and the letterscorresponding to sets that interfere destructively in alphabetical order separated by a comma. For example if sets Aand B interfere constructively and sets C and F interfere destructively enter AB,CF.

ANSWER: BCEG,ADF

Do you notice a pattern? When the phase difference between two identical waves can be written as , where

, the waves will interfere constructively. When the phase difference can be expressed as

, where , the waves will interfere destructively.

Consider what water waves look like when you throw a rock into a lake. These waves start at the point where the rockentered the water and travel out in all directions. When viewed from above, these waves can be drawn as shown, where thesolid lines represent wave peaks and troughs are located halfwaybetween 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). Thesources emit identical waves at the exact same time.

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



Hint C.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 (ortrough) of one wave aligns with the peak (or trough) of the other wave. Destructive interference occurs when waves are

out of phase so that the peak of one wave aligns with the trough of the other wave.

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

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

ANSWER: ccdd

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 you can see that Point B is away

from Source 1 and away from Source 2. The path-length

difference, , is the difference in the distance each wave

travels to reach Point B:

.

Part D

What are the path-length differences at Points A, C, and D (respectively, , , and )?

Enter your answers numerically in terms of separated by

commas. For example, if the path-length differences at PointsA, C, and D are , , and , respectively, enter 4,.5,1.

ANSWER: , , = , ,

Knowing the path-length difference helps to confirm what you found in Part C. When the path-length difference is

, where , the waves interfere constructively. When the path-length difference is ,

where , the waves interfere destructively.

Part E

What are the path-length differences at Points L to P?

Enter your answers numerically in terms of separated by

commas. For example, if the path-length differences at PointsL, M, N, O, and P are , , , , and , respectively,

enter 5,2,1.5,1,6.



ANSWER: , , , , = , , , ,

Every point along the line connecting Points L to P corresponds to a path-length 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 path-length difference , and the other corresponds to . It should make sense that the line halfway between the two sources

corresponds to a path-length difference of zero, since any point on this line is equally far from each source. Notice thesymmetry about the line of the and the lines.

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

Constructive Interference

Description: Short quantitative problem on a double-slit experiment and interference pattern. Students should know how tofind the wavelength of light in different materials. Based on Young/Geller Quantitative Analysis 26.1.

The figureshows the interference pattern obtained in a double-slit experimentwith light of wavelength .



Part A

Identify the fringe or fringes that result from the interference of two waves whose phases differ by exactly .

Hint A.1 How to approach the problem

In a double-slit experiment, when two waves reach the screen shifted by an integer number of wavelengths, such as or

, at that point the waves are in phase and interfere constructively. Their amplitudes add up and the resulting band, or

fringe, on the screen is a bright region where light intensity is at a maximum. Note that several bright fringes can be seen inthe interference pattern shown in the figure. Each of them is produced by waves that are shifted by a different (integer)number of wavelengths. Choose those that correspond to a path difference of .

Part A.2 Find which fringes correspond to constructive interference

Which of the fringes shown in the figure correspond to constructive interference?

Hint A.2.a Constructive interference fringes

Recall that, in a double-slit experiment, waves that interfere constructively strike the screen in regions where light intensityis maximum, creating a series of bright bands or fringes.

ANSWER: fringe C onlyfringes A, B, and Cfringes A, B, and Dfringes A, B, and Efringes A, D, and Efringes A, B, D, and E

Which of these fringes correspond to a path difference of and which to a path difference of ?

Hint A.3 Constructive interference pattern

In a double-slit experiment, fringes of maximum intensity are produced by waves whose path difference is an integer numberof wavelengths, such as , , , . The bigger this number, the farther from the central bright band is the

corresponding fringe. For example, if the center of the fringe produced by waves that are shifted by is at 1 from the

center of the central band, the center of the fringe produced by waves shifted by will be at approximately 4 from the

central band. Moreover, the fringes appear symmetrical with respect to the central band. Thus, when the center of the fringeproduced by waves that are shifted by is at 1 from the center of the central band, the center of the fringe produced by

waves shifted by will be located at about the same distance from the center of the central band, but on the opposite side.

Enter the letter(s) indicating the fringe(s) in alphabetical order. For example, if you think that fringes A and C areboth correct, enter AC.

ANSWER: AD

Part B

The same double-slit experiment is then immersed in water (with an index of refraction of 1.33) and repeated. When in thewater, what happens to the interference fringes?

Hint B.1 How to approach the problem

When in water, the wavelength of light changes. Will this affect the location of the interference fringes on the screen? Recallthat in a double-slit experiment the positions of the centers of the bright fringes on the screen can be expressed in terms ofthe difference in path length traveled by the waves.

Part B.2 Find the wavelength of light in water

Consider light with wavelength in air. What is the wavelength of the same light in water, ?

Hint B.2.a Wavelength of light in a material

The wavelength of light is different in different materials. In particular, the wavelength of light in a material of index

of refraction is given by

,

where is the wavelength of the same light in vacuum. This is because in any material the velocity of a wave is given by

the product of frequency and wavelength, . Since the speed of light in a material is less than it is in vacuum, but

its frequency does not change, its wavelength must be less than the wavelength of the same light in vacuum. Note that

the wavelength of light in air is essentially the same as that of light in vacuum.

Express your answer in terms of .



ANSWER: =

Thus, when the experiment is immersed in water, the wavelength of light decreases. How does this affect the locationof the interference fringes?

Part B.3 Find how the centers of the fringes change with the wavelength of light

Let be the position of the center of the mth band (measured from the center of the central band) when the experimentis in air, and let be the same position when the experiment is in water. Which of the following expressions iscorrect?

Hint B.3.a Using proportional reasoning

To solve this problem use proportional reasoning to find a relation between the position of the mth band when theexperiment is in air, , and the same position when the experiment is in water, .

Find the simplest equation that contains these variables and other known quantities from the problem.Write this equation twice, once to describe and again for .You need to write each equation so that all the constants are on one side and your variables are on the other. Sinceyour variable is in this problem, you want to write your equations in the form .To finish the problem you need to compare the two cases presented in the problem. For this question you shouldfind the ratio .

Hint B.3.b Position of the centers of the bright fringes in a double-slit experiment

In a double-slit experiment with light of wavelength , waves whose path difference is (where is an integer)

interfere constructively and create the bright band on the screen. The position of the center of the mth band(measured from the center of the central band) on the screen is given by

,

where and are, respectively, the distance from the screen to the slits and the distance between the slits.

ANSWER:

As you found out, when the experiment is immersed in water, the centers of the bright fringes are closer to the centralband than when the experiment is in air. But does the location of the center of the central band change when thewavelength of light decreases?

Part B.4 Find how the center of the central band changes with the wavelength of light

In a double slit-experiment, the central fringe of the interference pattern corresponds to waves that traveled the same distancefrom the source. If the wavelength of the waves changes, will the position of the center of this fringe change? Assume theslits are horizontal so that the fringe pattern displayed on a screen is oriented vertically..

Hint B.4.a How to approach the question

In a double-slit experiment, the central fringe is created by waves that travel the same distance from the slits, and its centeris always located at the same distance from both slits. Thus, its position changes only if the location of the slits changes,regardless of the properties of the light that shines through the slits.

ANSWER: The position of the center of the central fringe will shift upward.The position of the center of the central fringe will shift downward.The position of the center of the central fringe will not change.

ANSWER: They are more closely spaced than in air by a factor of 1.33.They are more widely spaced than in air by a factor of 1.33.They are spaced the same as in air.They are shifted upward.They are shifted downward.

When the experiment is immersed in water, the wavelength of light decreases because the index of refraction of water ishigher than that of air. Since the positions of the centers of the bright bands depend on the wavelength of light, light witha smaller wavelength will produce interference fringes that are more closely spaced; the higher the index of refraction, themore closely spaced are the fringes.



FM Radio Interference

Description: Calculation involving interference of radio waves and its effect on radio reception.

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 whatis 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 isvery far away, and that the building is at a 45-degree angle from the path to the transmitter.

Point A in the figure is where you originally stopped, and point B iswhere the station is completely clear again. Finally, assume that thesignal 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?

Part A.1 What is the path-length difference at point A?

Since we know that the waves traveling along the two paths interfere destructively at point A, we know something about thedifference in the lengths of those two paths. What is the difference between the two path lengths, in integer multiples of

the wavelength?

ANSWER:

To have destructive interference, the two waves must be shifted relative to one another by half of a wavelength. Integernumbers of wavelengths in the path difference do not change the relative positions of the waves, so the value of is

irrelevant.

Part A.2 What is the path-length difference at point B?

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

the wavelength?

ANSWER:

For constructive interference, the waves must align crest to crest, which corresponds to a path difference of zero. Sinceinteger numbers of wavelengths in the path difference do not change the relative positions of the waves, any integer

number of wavelengths in path difference also gies constructive interference.



Part A.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 islarger for one wave, while the path from the building to the antenna is larger for the other. What is the difference in lengthbetween the path of the reflected wave from the transmitter to A, and the path of the reflected wave from the transmitter toB, in integer multiples of the wavelength?

ANSWER:

Express your answer in wavelengths, as a fraction.

ANSWER: = wavelengths

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 A and B?

Express your answer in meters, to two significant figures.

ANSWER: =

This problem discusses a diffraction grating.

Problem 22.44

Description: For your science fair project you need to design a diffraction grating that will disperse the visible spectrum (400-700 nm) over 30.0 degree(s) in first order. (a) How many lines per millimeter does your grating need? (b) What is the first-order...

For your science fair project you need to design a diffraction grating that will disperse the visible spectrum (400-700 nm) over in first order.

Part A

How many lines per millimeter does your grating need?

ANSWER: lines/mm

Part B

What is the first-order diffraction angle of light from a sodium lamp

ANSWER: =

The following two problems concern thin-film interference.

Thin Film (Oil Slick)

Description: This problem explores thin film interference for both transmission and reflection.

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting offthe surface. She aims a spectrometer at a particular spot and measures the wavelength to be 750 (in air). The index ofrefraction 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

In thin films, there are interference effects because light reflects off the two different surfaces of the film. In this problem, thescientist observes the light that reflects off the air-oil interface and off the oil-water interface. Think about the phasedifference created between these two rays. The phase difference will arise from differences in path length, as well asdifferences that are introduced by certain types of reflection. Recall that if the phase difference between two waves is (a

full wavelength) then the waves interfere constructively, whereas if the phase difference is (half of a wavelength) thewaves interfere destructively.

Hint A.2 Path-length phase difference

The light that reflects off the oil-water interface has to pass through the oil slick, where it will have a different wavelength.The total "extra" distance it travels is twice the thickness of the slick (since the light first moves toward the oil-waterinterface, and then reflects back out into the air).

Hint A.3 Phase shift due to reflections

Recall that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift of radians, whichcorresponds to a shift of half of a wavelength. What used to be a maximum is now a minimum! Be careful, though; if twobeams each reflect off a surface with a higher index of refraction, they will both get a half-wavelength shift, canceling outthat effect.

Express your answer in nanometers to three significant figures.

ANSWER: =

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

Keep in mind that when light reflects off a surface with a higher index of refraction, it gains an extra shift of half of awavelength. What used to be a maximum is now a minimum! Be careful, though; if two beams reflect, they will both get ahalf-wavelength shift, canceling out that effect. Also, reflection off a surface with a lower index of refraction yields no phaseshift.


ANSWER: =

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 oilslick 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 How to approach the problem

For transmission of light, the same rules hold as before, only now one beam travels straight through the oil slick and into thewater, while the other beam reflects twice (once off the oil-water interface and once again off the oil-air interface) beforebeing finally transmitted to the water.

Part C.2 Determine the wavelength of light in air

Find the wavelength of the required light in air.

Express your answer numerically in nanometers.

ANSWER: =

Hint C.3 Relationship between wavelength and index of refraction

There is a simple relationship between the wavelength of light in one medium (with one index of refraction ) and the

wavelength in another medium (with a different index of refraction ):

.


ANSWER: =

This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energyensures 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 thesame 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.

Why Butterfly Wings Shimmer

Description: Basic conceptual and quantitative questions on thin film interference, followed by an application to theiridescence of butterfly wings.

Learning Goal: To understand the concept of thin-film interference and how to apply it.

Thin-film interference is a commonly observed phenomenon. It causes the bright colors in soap bubbles and oil slicks. It alsoleads to the iridescent colors on many insects and bird feathers. In this problem, you will learn how to work with thin-filminterference and see how it creates the dazzling display of a tropical butterfly's wings.

When light is incident on a thin film, some of the light will bereflected at the front surface of the film, and the rest will betransmitted into the film. Some of the transmitted light will bereflected from the back surface of the film. The light reflected fromthe front surface and the light reflected from the back surface willinterfere. Depending upon the thickness of the film, thisinterference may be constructive or destructive. We will bestudying the interference of light normal to the surface of the film.The figure shows the light entering at a small angle to normal onlyfor the purpose of showing the incident and reflected rays.

For this problem, you will only be concerned with the geometric aspects of thin-film interference, so ignore phase shiftscaused by reflection from a medium with higher index of refraction. (Because of the structure of a butterfly's wings, such phaseshifts do not contribute much to what you actually see when you look at the butterfly.)

Part A

Assume that light is incident normal to the surface of a film of thickness . How much farther does the light reflected from the

back surface travel than the light reflected from the front surface?


ANSWER:

Part B

For constructive interference to occur, the difference between the two paths must be an integer multiple of the wavelength ofthe light (as is true in any interference problem), i.e. the general criterion for constructive interference is

,

where is a positive integer. This is usually stated in the slightly more explicit form

.

Given the thickness of the film, , what is the longest wavelength that can exhibit constructive interference?


ANSWER: =

Part C

If you have a thin film of thickness 300 , what is the third-longest wavelength of light that exhibits

constructive interference with the reflected light?

Note that this corresponds to .


ANSWER: =



Part D

The criterion for destructive interference is very similar to the criterion for constructive interference. For destructiveinterference to occur, the difference between the two paths must be some integer number of wavelengths plus half awavelength:

,

or

,

where is a nonnegative integer. What is the second-longest wavelength that will not be visible (i.e., will have

strong destructive interference for the reflected waves) when reflected from a film of thickness 300 ?

Note that the longest wavelength corresponds to for destructive interference. This is why the notation used for the

second-longest wavelength is instead of .


ANSWER: =

The blue morpho butterfly lives in tropical rainforests and can have a wingspan greater than 15 cm. The brilliant blue color ofits wings is a result of thin-film interference. A pigment wouldnot produce such vibrant, pure colors. What cannot be conveyedby a picture is that the colors vary with the viewing angle, whichcauses the shimmering iridescence of the actual butterfly.

The scales of the butterfly's wings consist of two thin layers ofkeratin (a transparent substance with index of refraction greaterthan one), separated by a 200- gap filled with air.

Part E

What is the longest wavelength of light that will exhibit constructive interference at normal incidence? The keratin layers

are thin enough that you can think of them as representing the surfaces of a 200- "film" of air.

Express your answer in nanometers to two significant figures.

ANSWER: =

This wavelength lies near the cutoff between visible (violet) and ultraviolet light. The shorter wavelengths that arestrongly reflected (corresponding to in ) will not be relevant to what humans see when they look at

the butterfly.

To understand why the color changes with viewing angle, try drawing a diagram of light incident on a thin film at a largeangle. The distance the light travels within the film will be increased. However, at large angles the light reflected fromthe front surface will actually have to travel farther to an observer outside of the film than the light reflected from theback surface. The increased distance outside of the material for the front surface reflection actually makes the net path-length difference smaller than it would be for normal incidence. As viewing angle increases, the largest wavelength thatexperiences constructive interference gets shorter. Thus, although the butterfly is blue at normal incidence, at large angleof incidence no particular wavelength of visible light is short enough to be strongly reflected.

Part F

The wavelength that appears in the interference equations given in Parts B and D represents the wavelength of light within

the medium of the film. So far we have assumed that the medium composing the film is air, but many thin-film problems willinvolve films with an index of refraction different from that of air.

Suppose that the butterfly gets wet, thus filling the gaps between the keratin sheets with water ( ). What wavelength

in air will be strongly reflected now?



Hint F.1 How to approach the problem

In Part E, you found the wavelength that exhibits constructive interference; represents the wavelength of the light

within the medium of the film (the medium being water, in this case). Now find the wavelength of light in air that, once inwater, has a wavelength of .

Hint F.2 Relating the wavelength in air to the wavelength in water

Recall that the wavelength within a medium is related to the wavelength in vacuum (or air, to the accuracy we are

working with) by the equation

,

where is the index of refraction of the medium. Be careful not to confuse this with the arbitrary positive integer in theequations introduced earlier in this problem.

Express your answer in nanometers to two significant figures.

ANSWER: =

Part G

Several thin films are stacked together in each butterfly wing scale. How would these multiple layers of thin films affect thelight reflected by the butterfly's wings?

Hint G.1 A picture of the situation

The figure shows a model of the layers in the scales on thebutterfly's wing. Make a similar drawing so that you can drawin the rays of light to help you visualize what is occurring inthis situation. Keep in mind that when light strikes a transparentboundary, some of the light is reflected and some is transmitted.

ANSWER: There would be no change, because all of the 400- light would be reflected by the first scale, and sothe light is unaffected by the remaining scales.More 400- light would be reflected, because some of the light transmitted through the first layer couldbe reflected by the second layer, light transmitted by the second layer could be reflected by the third, etc.Less 400- light would be reflected, because light reflected from the second layer can interferedestructively with light reflected by the first layer.There would be no change, because the interference effects would all occur at the first layer.More 400- light would be reflected, because the effective gap would be three times as wide.Less 400- light would be reflected, because the effective gap would be three times as wide.

Layering of thin films is used to make wavelength-specific mirrors, used widely in laser applications, that are far morereflective than metal mirrors. Layered films are also used to make antireflective coatings for camera lenses.

These last three problems are about diffraction.

Problem 22.17

Description: The second minimum in the diffraction pattern of a a-wide slit occurs at theta. (a) What is the wavelength of thelight?

The second minimum in the diffraction pattern of a 0.110 -wide slit occurs at 0.480 .

Part A

What is the wavelength of the light?

ANSWER: = nm



Understanding Circular-Aperture Diffraction

Description: The diffraction pattern formed by a circular aperture is described, followed by basic calculations for the locationsof dark rings. Rayleigh's criterion is introduced and used.

Learning Goal: To use the formulas for the locations of the dark bands and understand Rayleigh's criterion of resolvability.

An important diffraction pattern in many situations is diffraction from a circular aperture. A circular aperture is relatively easy tomake: all that you need is a pin and something opaque to poke the pin through. The figure shows a typical pattern. It consists ofa bright central disk, called the Airy disk, surrounded by concentricrings of dark and light.

While the mathematics required to derive the equations for circular-aperture diffraction is quite complex, the derived equations arerelatively easy to use. One set of equations gives the angular radiiof the dark rings, while the other gives the angular radii of the lightrings. The equations are the following:

,

,

where is the wavelength of light striking the aperture, is the

diameter of the aperture, and is the angle between a line normal to the screen and a line from the center of the aperture to the

point of observation. There are more alternating rings farther from the center, but they are so faint that they are not generally ofpractical interest.

Consider light from a helium-neon laser ( nanometers) striking a pinhole with a diameter of 0.220 .

Part A

At what angle to the normal would the first dark ring be observed?

Express your answer in degrees, to three significant figures.

ANSWER: =

Part B

Suppose that the light from the pinhole projects onto a screen meters away. What is the radius of the first dark ring on

that screen? Notice that the angle from Part A is small enough that .

Hint B.1 How to find the radius

You already know the angular radius, from Part A. To find the radius of the ring as projected onto the screen, construct aright triangle with the radius as one leg, the normal line from the center of the aperture as the second leg, and the line fromthe end of your radius to the center of the aperture as the hypotenuse.

Hint B.2 Finding the radius: trigonometry

From simple trigonometry, you should see that the radius is multiplied by meters. Use the small-angle

approximation given in the problem and the formula for from the introduction to find the radius.

Express your answer in millimeters, to three significant figures.

ANSWER: =

Part C

The first dark ring forms the boundary for the bright Airy disk at the center of the diffraction pattern. What is the area of the

Airy disk on the screen from Part B?

Express your answer in , to three significant figures.

ANSWER: =



Diffraction due to a circular aperture is important in astronomy. Since a telescope has a circular aperture of finite size, stars arenot imaged as points, but rather as diffraction patterns. Two distinct points are said to be just resolved (i.e., have the smallestseparation for which you can confidently tell that there are two points instead of just one) when the center of one point'sdiffraction pattern is found in the first dark ring of the other point's diffraction pattern. This is called Rayleigh's criterion forresolvability.

Consider a telescope with an aperture of diameter 1.03 .

Part D

What is the angular radius of the first dark ring for a point source being imaged by this telescope? Use nanometers for

the wavelength, since this is near the average for visible light.

Express your answer in degrees, to three significant figures.

ANSWER: =

Part E

Two stars in a certain binary star system have angular separation of degrees when viewed from earth. Can they be

resolved with the telescope described above?

ANSWER: yesno

Resolving Power of the Eye

Description: Use the measured resolving power of a person's eye to calculate the diameter of the pupil of the eye by assumingthat the resolving power is diffraction limited.

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? UseRayleigh's criterion and assume that the wavelength of the light is 570 .

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 centerof 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 circularaperture for the diffraction pattern instead of an infinite slit.



Express your answer in millimeters to three significant figures.

ANSWER:

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