Chapter 28 Physical Optics: Interference and Diffractionphysicsweb.phy.uic.edu/105/JC/28_LectureOutline.pdf · • Young’s Two-Slit Experiment • Interference in Reflected Waves

$Page 1: Chapter 28 Physical Optics: Interference and Diffractionphysicsweb.phy.uic.edu/105/JC/28_LectureOutline.pdf · • Young’s Two-Slit Experiment • Interference in Reflected Waves$
Chapter 28Physical Optics:

Interference and Diffraction

Units of Chapter 28• Superposition and Interference

• Young’s Two-Slit Experiment• Interference in Reflected Waves

• Diffraction• Resolution

• Diffraction Gratings

28-1 Superposition and InterferenceIf two waves occupy the same space, their amplitudes add at each point. They may interfere either constructively or destructively.

28-1 Superposition and Interference

Interference is only noticeable if the light sources are monochromatic (so all the light has the same wavelength) and coherent (different sources maintain the same phase relationship over space and time).

If this is true, interference will be constructive where the two waves are in phase, and destructive where they are out of phase.


In this illustration, interference will be constructive where the path lengths differ by an integral number of wavelengths,

and destructive where they differ by a half-odd integral number of wavelengths.


To summarize, the two path lengths and will interfere constructively or destructively according to the following:

28-2 Young’s Two-Slit Experiment

In this experiment, the original light source need not be coherent; it becomes so after passing through the very narrow slits.


If light consists of particles, the final screen should show two thin stripes, one corresponding to each slit. However, if light is a wave, each slit serves as a new source of “wavelets,” as shown, and the final screen will show the effects of interference. This is called Huygens’s principle.

28-2 Young’s Two-Slit ExperimentAs the pattern on the screen shows, the light on the screen has alternating light and dark fringes, corresponding to constructive and destructive interference.The path difference is given by:

Therefore, the condition for bright fringes (constructive interference) is:


The dark fringes are between the bright fringes; the condition for dark fringes is:


This diagram illustrates the numbering of the fringes.

28-4 Diffraction

A wave passing through a small opening will diffract, as shown. This means that, after the opening, there are waves traveling in directions other than the direction of the original wave.

28-4 Diffraction

Diffraction is why we can hear sound even though we are not in a straight line from the source – sound waves will diffract around doors, corners, and other barriers.

The amount of diffraction depends on the wavelength, which is why we can hear around corners but not see around them.

28-4 Diffraction

To investigate the diffraction of light, we consider what happens when light passes through a very narrow slit. As the figure indicates, what we see on the screen is a single-slit diffraction pattern.

28-4 DiffractionThis pattern is due to the difference in path length from different parts of the opening.

The first dark fringe occurs when:

28-4 Diffraction

The second dark fringe occurs when:

28-4 Diffraction

In general, then, we have for the dark fringes in a single-slit interference pattern:

The positive and negative values of m account for the symmetry of the pattern around the center.Diffraction fringes can be observed by holding your finger and thumb very close together (it helps not to be too farsighted!)

28-5 Resolution

Diffraction through a small circular aperture results in a circular pattern of fringes. This limits our ability to distinguish one object from another when they are very close.The location of the first dark fringe determines the size of the central spot:

28-5 Resolution

Rayleigh’s criterion relates the size of the central spot to the limit at which two objects can be distinguished:If the first dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sources responsible for the patterns will appear to be a single source.

The size of the spot increases with wavelength, and decreases with the size of the aperture.

28-5 Resolution

On the left, there appears to be a single source; on the right, two sources can be clearly resolved.

21

22

Chapter 28: Physical Optics: Interference and Diffraction James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

28 – 2

2. Picture the Problem: Coherent, in phase, 26.0-m wavelength waves interfere at various observation points. Strategy: Subtract the distances from each source to find the difference in path length. Divide this distance by the

wavelength. If the result is an integer value, then the waves constructively interfere. If the result is a half-integer value, the waves destructively interfere.

Solution: 1. (a) Calculate the difference in path lengths: 2 1 221 m 91.0 m 130 md d d' � �

2. Divide the difference by the wavelength: 130 m 5.0

26.0 mdO'

3. Because the path difference is 5.0 wavelengths, the waves interfere constructively at the observation point. 4. (b) Calculate the difference in path lengths: 2 1 135 m 44.0 m 91 md d d' � �

5. Divide the difference by the wavelength: 91 m 3.5

26.0 mdO'

6. Because the path difference is 3.5 wavelengths, the waves interfere destructively at the observation point. Insight: Whenever the path length difference is an integer the waves will interfere constructively. When the path length

difference is a half-integer value the waves will interfere destructively. 3. Picture the Problem: Two coherent waves interfere at an observation point 161 meters from one source and 295 meters

from the other source. Strategy: The longest wavelength that will give constructive interference has a length equal to the path length

difference. Solution: Calculate the path length difference: 295 m 161 m 134 mO ' � A

Insight: The longest wavelength that will give destructive interference is 268 m. 4. Picture the Problem: The figure shows a car traveling at 17 m/s

perpendicular to the line connecting two radio transmitters. The car picks up a maximum signal when it is at point A.

Strategy: Set the wavelength equal to the difference in distances between the two towers. Use the Pythagorean theorem to calculate the distance between the car and each tower as a function of time. Set the difference in distance length equal to half a wavelength (for destructive interference) and solve for the time.

Solution: 1. (a) Find the difference in path lengths: 450 m 150 m 300 mO ' � A

2. (b) Find the distance to the far tower as a function of time: � �2 2 2

1 450 m v t �A

3. Find the distance to the near tower as a function of time: � �2 2 2

2 150 m v t �A

4. Set the difference equal to half a wavelength: � � � �

set2 22 2 2 21 2 450 m 150 m 150 m

2v t v t O

� � � � A A

5. Rearrange the expression, square both sides, and evaluate:

� � � �

� � � � � � � �

� �

2 22 22 2 2 2

2 2 22 2 2 2

2 2 2

450 m 150 m 150 m

450 m 150 m 2 300 m 150 m

150 m

v t v t

v t v t

v t

ª º ª º� � �« » « »¬ ¼ ¬ ¼

ª º� � �¬ ¼ª º� �¬ ¼

6. Rearrange the equation and square both sides again: � � � � � �

22 222 2 2

4 2 2 2 6 2

450 m 2 150 m150 m

300 m

2.25 10 m 2.756 10 m

v t

v t

ª º�ª º� « »« »¬ ¼ « »¬ ¼u � u



28 – 3

7. Solve for t: 6 2 4 21 2.756 10 m 2.25 10 m 30 s

17 m/st u � u

Insight: As the car moves farther away, the path difference 'A decreases until it is nearly zero and the waves constructively interfere.

5. Picture the Problem: The figure shows two students listening to

two coherent sound sources. Both students hear constructive interference.

Strategy: Since student A is equidistant from both speakers the difference in path lengths from each speaker is zero. For student B to also hear constructive interference (and for the wavelength to be a maximum) the path length difference between him and the two speakers must be equal to one wavelength. Calculate the distance of student B from each speaker. Set the wavelength equal to the difference in distances. Divide the speed of sound by the wavelength to calculate the frequency.

Solution: 1. Calculate the distance to the far speaker: � � � �2 22 1.5 m 3.0 m 3.354 m � A

2. Subtract the distances: 2 1 3.354 1.50 1.854 mO� � A A

3. Divide the speed of sound by the wavelength: 343 m/s 0.18 kHz

1.854 mvfO

Insight: The next higher frequency at which constructive interference occurs is 0.37 kHz.f This frequency corresponds to a path length difference of 2O for the two waves.

6. Picture the Problem: The figure shows two students listening to

two loudspeakers that emit a 185 Hz tone. The tone from each speaker is 180q out of phase with the other speaker.

Strategy: The waves are 180° out of phase when they are emitted, so that the two paths must be different by another 180°, or half a wavelength, in order for student B to hear constructive interference. The path difference should be an integer multiple of a wavelength for student B to observe destructive interference. Calculate the path length difference between student B and the two speakers. Divide this difference by the wavelength ( c fO ) to evaluate the type of interference observed by student B.

Solution: 1. Calculate the distance to the far speaker: � � � �2 22 1.5 m 3.0 m 3.35 m � A

2. Subtract the distances: 2 1 3.35 1.50 1.85 m� � A A

3. Calculate the wavelength: 343 m/s 1.85 m

185 Hzvf

O

4. Divide the distance by the wavelength: 2 1 1.85 m 1.00

1.85 mO�

A A

5. Because the path difference is equal to one wavelength, the student at location B observes a sound minimum.

Insight: Student A will also hear a minimum because he is equidistant from both speakers and the speakers produce tones that are out of phase with each other.

23

24



28 – 5

Solution: 1. Find Moe’s distance to the near speaker: � � � �2 21 1.00 m 0.40 m 3.00 m 3.059 m � � A

2. Find Moe’s distance to the far speaker: � � � �2 22 1.00 m 0.40 m 3.00 m 3.311 m � � A

3. Set the difference equal to a half wavelength: � �12 12 2 3.311 m 3.059 m 0.502 mO O � � � A A

4. Divide the speed by the wavelength: 343 m/s 0.68 kHz

0.502 mvfO

5. Set 32 O' A to find O: � �3 2

2 12 3 3.311 m 3.059 m 0.167 mO O � � � A A


0.167 mvfO

Insight: The path difference between the two speakers is the same for Curly as it is for Moe, so each will hear a sound minimum at the same frequencies.

10. Picture the Problem: The image shows three people standing

3.00 m in front of two speakers that are separated by 0.800 m. The people are each standing 1.00 m apart with the middle person directly in front of the midpoint between the speakers. The speakers produce a single tone sound that is 180q out of phase with the other speaker.

Strategy: Moe and Curly will hear maximum sound intensity when the difference in their distances from each speaker is a half-integral wavelength. Use the Pythagorean theorem to determine their distances from the speakers. Set 1

2 O' A for

the lowest frequency and 32 O' A for the second lowest

frequency and solve for the wavelength. Then divide the speed of sound by the wavelength to calculate the frequency.

Solution: 1. (a) The sound travels the same distance from each speaker to reach Larry. The two sound waves will destructively interfere and Larry will hear a minimum sound intensity because the speakers are 180q out of phase.

Solution: 1. Calculate the distance to the close speaker: � � � �2 21 1.00 m 0.40 m 3.00 m 3.059 m � � A

2. Calculate the distance to the far speaker: � � � �2 22 1.00 m 0.40 m 3.00 m 3.311 m � � A


4. Divide the speed by the wavelength:

343 m/s 0.68 kHz0.502 m

vfO

5. Set 32 O' A to find O: � �3 2

2 12 3 3.311 m 3.059 m 0.167 mO O � � � A A


0.167 mvfO




28 – 5

Solution: 1. Find Moe’s distance to the near speaker: � � � �2 21 1.00 m 0.40 m 3.00 m 3.059 m � � A

2. Find Moe’s distance to the far speaker: � � � �2 22 1.00 m 0.40 m 3.00 m 3.311 m � � A



0.502 mvfO

5. Set 32 O' A to find O: � �3 2

2 12 3 3.311 m 3.059 m 0.167 mO O � � � A A


0.167 mvfO


10. Picture the Problem: The image shows three people standing

3.00 m in front of two speakers that are separated by 0.800 m. The people are each standing 1.00 m apart with the middle person directly in front of the midpoint between the speakers. The speakers produce a single tone sound that is 180q out of phase with the other speaker.

Strategy: Moe and Curly will hear maximum sound intensity when the difference in their distances from each speaker is a half-integral wavelength. Use the Pythagorean theorem to determine their distances from the speakers. Set 1

2 O' A for

the lowest frequency and 32 O' A for the second lowest

frequency and solve for the wavelength. Then divide the speed of sound by the wavelength to calculate the frequency.

Solution: 1. (a) The sound travels the same distance from each speaker to reach Larry. The two sound waves will destructively interfere and Larry will hear a minimum sound intensity because the speakers are 180q out of phase.

Solution: 1. Calculate the distance to the close speaker: � � � �2 21 1.00 m 0.40 m 3.00 m 3.059 m � � A

2. Calculate the distance to the far speaker: � � � �2 22 1.00 m 0.40 m 3.00 m 3.311 m � � A


4. Divide the speed by the wavelength:

343 m/s 0.68 kHz0.502 m

vfO

5. Set 32 O' A to find O: � �3 2

2 12 3 3.311 m 3.059 m 0.167 mO O � � � A A


0.167 mvfO


25

26



28 – 10

23. Picture the Problem: Light passes through two slits separated by 0.230 mm and creates an interference pattern on a

screen that is 2.50 m away.

Strategy: If the sharp edge of each razor blade is in the center of the blade’s thickness, then the two sharp edges of the blades will be exactly one blade thickness or 0.230 mm apart, and so will the two slits. Use the slit separation together with equation 28-1 to calculate the wavelength, using m = 1 and the angle from equation 28-3.

Solution: 1. Use equation 28-3 to find ș:

31 1 7.15 10 m

tan tan 0.16392.50 m

yL

T�

� � u q

2. Solve equation 28-1 for O�: � � � �30.230 10 m sin 0.1639sin

658 nm1

dmTO

�u q

Insight: Because the angle is small, we could have used the relation tan sinT T T| | to calculate the wavelength:

� � � ��

3 30.230 10 m 7.15 10 m658 nm

1 2.5 m

d ymL

O� �u u

24. Picture the Problem: The figure shows interference pattern

for diffracted light on a screen that is 1.20 m from two slits that are separated by 135 µm. The separation distance between the minima m and (m + 4) is 23.0 mm.

Strategy: Because 23.0 mm is much smaller than the 1.20 m screen distance, use the small angle approximations tan sin .T T T| | Use equation 28-2 to calculate the angles for the minima of order m and (m + 4). Use equation 28-3 to calculate the location of the minima. Set the difference equal to the 23.0 mm and solve for the wavelength.

Solution: 1. (a) Calculate the angles for minima of order m and (m + 4):

� �

� � � �

1m m 2

71m+4 2 2

sin

4

md

m md d

OT T

O OT

| �

� � �

2. Use equation 28-3 to find the locations of the minima, and set the distance between them equal to 23.0 mm:

� �

� � � �

m 4 m m+4 m m+4 m

set7 12 2

tan tan

4 23.0 mm

y y L L L

L m m Ld d d

T T T T

O O O� � � �

ª º � � � « »¬ ¼

3. Solve for the wavelength:

� � � ��

-60.023 m 135 10 m23.0 mm647 nm

4 4 1.2 m

dL

Ou

4. (b) Equation 28-2 shows that the wavelength is proportional to the diffraction angle. Therefore, increasing the frequency (which decreases the wavelength) will cause the bright spots to move closer together.

Insight: If the wavelength decreased to 480 nm, the separation between the four fringes would decrease to 17.1 mm. 25. Picture the Problem: An instructor passes 632.8-nm light through two slits separated by 0.0220 mm in order to

produce fringes that are separated by 2.50 mm on a screen a distance L away.

Strategy: Use equation 28-1 to find the angle to the first maximum, and then insert that angle into tany L T and solve for the distance L to the screen.

Solution: 1. Find the angle to the first maximum:

� �� 1 1 1 632.8 nmsin sin 1.648

0.0220 mm

mdOT � � q

2. Solve equation 28-3 for L:

2.50 cm86.9 cm

tan tan 1.648

yLT

q

Insight: In this case the small angle approximation is appropriate, and the solution could be written as .L yd mO



28 – 10

23. Picture the Problem: Light passes through two slits separated by 0.230 mm and creates an interference pattern on a

screen that is 2.50 m away.

Strategy: If the sharp edge of each razor blade is in the center of the blade’s thickness, then the two sharp edges of the blades will be exactly one blade thickness or 0.230 mm apart, and so will the two slits. Use the slit separation together with equation 28-1 to calculate the wavelength, using m = 1 and the angle from equation 28-3.

Solution: 1. Use equation 28-3 to find ș:

31 1 7.15 10 m

tan tan 0.16392.50 m

yL

T�

� � u q

2. Solve equation 28-1 for O�: � � � �30.230 10 m sin 0.1639sin

658 nm1

dmTO

�u q

Insight: Because the angle is small, we could have used the relation tan sinT T T| | to calculate the wavelength:

� � � ��

3 30.230 10 m 7.15 10 m658 nm

1 2.5 m

d ymL

O� �u u

24. Picture the Problem: The figure shows interference pattern

for diffracted light on a screen that is 1.20 m from two slits that are separated by 135 µm. The separation distance between the minima m and (m + 4) is 23.0 mm.

Strategy: Because 23.0 mm is much smaller than the 1.20 m screen distance, use the small angle approximations tan sin .T T T| | Use equation 28-2 to calculate the angles for the minima of order m and (m + 4). Use equation 28-3 to calculate the location of the minima. Set the difference equal to the 23.0 mm and solve for the wavelength.

Solution: 1. (a) Calculate the angles for minima of order m and (m + 4):

� �

� � � �

1m m 2

71m+4 2 2

sin

4

md

m md d

OT T

O OT

| �

� � �

2. Use equation 28-3 to find the locations of the minima, and set the distance between them equal to 23.0 mm:

� �

� � � �

m 4 m m+4 m m+4 m

set7 12 2

tan tan

4 23.0 mm

y y L L L

L m m Ld d d

T T T T

O O O� � � �

ª º � � � « »¬ ¼

3. Solve for the wavelength:

� � � ��

-60.023 m 135 10 m23.0 mm647 nm

4 4 1.2 m

dL

Ou

4. (b) Equation 28-2 shows that the wavelength is proportional to the diffraction angle. Therefore, increasing the frequency (which decreases the wavelength) will cause the bright spots to move closer together.

Insight: If the wavelength decreased to 480 nm, the separation between the four fringes would decrease to 17.1 mm. 25. Picture the Problem: An instructor passes 632.8-nm light through two slits separated by 0.0220 mm in order to

produce fringes that are separated by 2.50 mm on a screen a distance L away.

Strategy: Use equation 28-1 to find the angle to the first maximum, and then insert that angle into tany L T and solve for the distance L to the screen.

Solution: 1. Find the angle to the first maximum:

� �� 1 1 1 632.8 nmsin sin 1.648

0.0220 mm

mdOT � � q

2. Solve equation 28-3 for L:

2.50 cm86.9 cm

tan tan 1.648

yLT

q

Insight: In this case the small angle approximation is appropriate, and the solution could be written as .L yd mO

eq 28-3 y=L tan theta

27

28

29



28 – 21

51. Picture the Problem: The figure shows the diffraction

pattern of 632.8-nm light that passes through a single slit. The second-order minima are separated by 15.2 cm on a screen that is 1.50 m behind the slit.

Strategy: Calculate the angle to the second-order minima using equation 28-3, where y is half of 15.2 cm. Then insert the angle into equation 28-12 and solve for the slit width.

Solution: 1. (a) Solve for the second-order angle:

� ��

121 12

2 2 2

0.152 mtan tan tan 2.90

1.50 m

yy LL

T T � � � q

2. Solve equation 28-12 for the slit width:

� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q

3. (b) If light with a wavelength of 591 nm is used, the distance indicated in the figure will be less than 15.2 cm because the diffraction angle decreases as the wavelength decreases according to equation 28-12, sin .m WT O

Insight: The separation distance between the two second-order minima for 591-nm light is 14.2 cm. 52. Picture the Problem: The image shows a single slit

diffraction pattern on a screen 1.00 m from the slit. The central maximum is 2y1 = 1.60 cm wide.

Strategy: Calculate the angle of the first minimum using

1 1tany L T , where y1 is half the width of the central maximum. Insert ș1 into equation 28-12 to find the ratio of the wavelength to the slit width. Use that ratio together with

equations 28-12 and 28-3 to find the distance 2y2 between the two second-order minima.

Solution: 1. Calculate the first-order minimum:

� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q

2. Solve equation 28-12 for the ratio of the wavelength to the slit width:

� �1sin 0.4584sin

0.008001W m

TO q

3. Calculate the angle to the second minimum: � �1 1

2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹

4. Use equation 28-3 to find 2y2: � � � �2 22 2 tan 2 1.00 m tan 0.917 3.20 cmy L T q

Insight: The separation between consecutive minima is constant as long as the angles remain small. If the small angle approximation is valid, we can therefore write the distance between the various minima as � �12 2 1.60 cm .my my m

53. Picture the Problem: The resolution of the human eye depends upon the diameter of the pupil.

Strategy: Use the relationship between the resolution of the eye and the pupil diameter D, sin 1.22 DT O (equation 28-14) to answer the conceptual question.

Solution: 1. (a) On a cloudy day your pupil diameter will increase in order to allow more light to hit the retina. This will decrease the angular size of the diffraction pattern created by the pupil and increase the resolution as indicated by equation 28-14. We conclude that your eyes have greater resolution on a dark cloudy day than they do on a bright sunny day.

2. (b) The best explanation is I. Your eyes have greater resolution on a cloudy day because your pupils are open wider to allow more light to enter the eye. Statement II erroneously assumes a smaller pupil would produce a greater resolution.

Insight: Part of the reason that telescope designers strive for the largest mirror diameter (and hence the largest pupil or aperture of the instrument) is to minimize the diffraction pattern it produces and maximize the resolution.



28 – 21





� ��

121 12

2 2 2


1.50 m

yy LL

T T � � � q


� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q








� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q


� �1sin 0.4584sin

0.008001W m

TO q


2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹










28 – 21





� ��

121 12

2 2 2


1.50 m

yy LL

T T � � � q


� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q








� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q


� �1sin 0.4584sin

0.008001W m

TO q


2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹










28 – 21





� ��

121 12

2 2 2


1.50 m

yy LL

T T � � � q


� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q








� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q


� �1sin 0.4584sin

0.008001W m

TO q


2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹










28 – 21





� ��

121 12

2 2 2


1.50 m

yy LL

T T � � � q


� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q








� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q


� �1sin 0.4584sin

0.008001W m

TO q


2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹










28 – 21





� ��

121 12

2 2 2


1.50 m

yy LL

T T � � � q


� �� 2

2

2 632.8 nmsin 25.0 m

sin sin 2.90mm W

WO OT P

T �

q








� ��

121 11

1 1 1

0.0160 mtan tan tan 0.4584

1.00 m

yy LL

T T � � � q


� �1sin 0.4584sin

0.008001W m

TO q


2 sin sin 2 0.00800 0.917mWOT � �§ · qª º¨ ¸ ¬ ¼© ¹








30

31

(a) On a cloudy day your pupil diameter will increase in order to allow more light to hit the retina. This will decrease the angular size of the diffraction pattern created by the pupil and increase the resolution as indicated byequation 28-14. We conclude that your eyes have greater resolution on a than they do on a brightsunny day.2. (b) The best explanation is your eyes have better resolution on a cloudy day because your pupils will expand to allow more light to enter the eye.

32

33



28 – 23

58. Picture the Problem: A camera collects 450-nm light from two people that are 2.0 m tall and are 27 km away.

Strategy: Calculate the minimum angle from the inverse tangent of the separation distance and the distance from the camera. Then use equation 28-14 to calculate the aperture size.

Solution: Combine equations 28-15 and 28-3 together with the small angle approximation to find D: � ��

min min

3 9

tan 1.22

1.22 27 10 m 450 10 m1.227.4 mm

2.0 m

y L L LD

LDy

OT T

O�

§ · | ¨ ¸© ¹

u u

Insight: A Blackbird airplane can resolve the people with a fairly small diameter camera. However, it is very difficult to construct a camera whose resolution is limited only by diffraction.

59. Picture the Problem: The Hubble telescope collects light with a wavelength of 550 nm through a 2.4-m aperture and

orbits at an altitude of 613 km.

Strategy: Use the Rayleigh criterion (equation 28-15) together with equation 28-3 to solve for the minimum separation distance of two objects that can be resolved by the Hubble Space Telescope camera. The small angle approximation is appropriate in this situation.

Solution: Combine equations 28-15 and 28-3 together with the small angle approximation to find y: � �

min min

93

tan 1.22

550 10 m613 10 m 1.22 17 cm

2.4 m

y L L LDOT T

�

§ · | ¨ ¸© ¹

§ ·u u ¨ ¸

© ¹

Insight: Contrary to popular legend, this resolution does not allow the satellite to resolve the writing on a newspaper from space.

60. Picture the Problem: A lens that has an aperture of 120 mm and a focal length of 640 mm produces an image with 540-nm light.

Strategy: The lens will produce an image at its focal point if the object is very distant. Use equation 28-14 to calculate the angle from the central maximum to the first minimum, and then double this angle to calculate the angular width of the central maximum. Set the width of the central maximum of the image equal to twice the distance given by equation 28-3, where L is the focal length.

Solution: 1. (a) Solve equation 28-14 for min :T

76

min min

5.4 10 msin 1.22 1.22 5.5 10 rad

0.12 mDOT T

��u

| u

2. Find the angular width of the central maximum: � �6 5min2 2 5.5 10 rad 1.1 10 radT T � � u u

3. (b) Calculate the width at the focal length.

� �� min min

5 6

2 2 tan 2

2 0.64 m 5.5 10 rad 7.0 10 m 7.0 m

y L LT T

P� �

|

u u

Insight: The finite size of the lens causes the light to diffract, spreading it out so that it does not perfectly focus at the focal point.

Documents

Chapter 28 Physical Optics: Interference and Diffractionphysicsweb.phy.uic.edu/105/JC/28_LectureOutline.pdf · • Young’s Two-Slit Experiment • Interference in Reflected Waves