Openstax College Physics Instructor Solutions Manual Chapter 27

OpenStax College Physics Instructor Solutions Manual Chapter 27 CHAPTER 27: WAVE OPTICS

27.1 THE WAVE ASPECT OF LIGHT: INTERFERENCE

1. Show that when light passes from air to water, its wavelength decreases to 0.750 times its original value.

Solution       0.750  n n 1.333 . So, the wavelength of light in water is 0.750 times the wavelength in air.

2. Find the range of visible wavelengths of light in crown glass.

Solution 380 nm 760 nm 380 nm 760 nm   to  to  250 nm to 500 nm n n n 1.52 1.52

3. What is the index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum? Identify the likely substance.

Solution   1 n   n    1.49  n n 0.671 Polystyrene

4. Analysis of an interference effect in a clear solid shows that the wavelength of light in the solid is 329 nm. Knowing this light comes from a He-Ne laser and has a wavelength of 633 nm in air, is the substance zircon or diamond?

Solution   633 nm n   n    1.92 ; n  1.92  n n 329 nm Zircon

5. What is the ratio of thicknesses of crown glass and water that would contain the same number of wavelengths of light?

Solution d d d   n n 1.333 c  w  c  c  c  w   0.877 c E d w w  nw nc 1.52

27.3 YOUNG’S DOUBLE SLIT EXPERIMENT OpenStax College Physics Instructor Solutions Manual Chapter 27

6. At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

Solution  450  109 m  d sin   m    sin -1   0.516  -5   5.00  10 m 

7. Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

Solution d sin   m for m  0,1,2,3,...... Using

7 -1 m  -1 35.80  10 m   sin    sin  -4   0.997  d   1.00  10 m 

8. What is the separation between two slits for which 610-nm orange light has its first 30.0 maximum at an angle of ?

7 Solution m 16.1010 m 6 d sin   m  d   1.2210 m sin  sin 30.0

9. Find the distance between two slits that produces the first minimum for 410-nm 45.0 violet light at an angle of .

Solution  1  d sin   m      2  m  1 2 0  1 2 4.10  107 m d      2.90  107 m  0 .290 μm sin  sin (45.0°)

10. 30.0 Calculate the wavelength of light that has its third minimum at an angle of 3.00 μm when falling on double slits separated by . Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution  1  d sin   m     2  d sin  3.00  106 m sin 30.0        6.00  107 m  600 nm (m  1 2) 2.5

11. 2.00 μm What is the wavelength of light falling on double slits separated by if the 60.0 third-order maximum is at an angle of ?

Solution d sin 2.00  106 m sin 60.0 d sin   m        5.77  107 m  577 nm m 3

12. At what angle is the fourth-order maximum for the situation Exercise 27.6?

Solution 7 -1 m  -1 44.50  10 m d sin   m    sin    sin  -5   2.06  d   5.00  10 m 

13. What is the highest-order maximum for 400-nm light falling on double slits separated 25.0 μm by ?

Solution d sin   m sin   1 Looking at , we notice that the highest order occurs when , so d 2.50105 m m    62.5.  4.0010-7 m the highest order is: Since m must be an integer, the highest order is then m = 62.

14. 1.20 μm Find the largest wavelength of light falling on double slits separated by for which there is a first-order maximum. Is this in the visible part of the spectrum?

Solution d sin  d sin   m     d sin  m

6 max  d  1.2010 m  1200 nm Thus, (not visible)

15. What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution 2 d sin   m  2  d  ;   90. sin  dmin  2  1440 nm  1.44 μm Thus,

16. (a) What is the smallest separation between two slits that will produce a second- order maximum for any visible light? (b) For all visible light?

Solution d sin   m  2    380 nm  760 nm. (a)

dmin  2min  2380 nm  760 nm So, d  2760 nm 1520 nm (b)

17. (a) If the first-order maximum for pure-wavelength light falling on a double slit is at 10.0 an angle of , at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

Solution sin 1 1 d sin 1  , d sin 2  2   sin  2 2    sin -1 2 sin    sin -12 sin 10.0 20.3 (a) 2 1

1 sin 1 -1 d sin 1  ,d sin  1     2   1  sin 0.5 sin 10 4.98 2 2 2 sin  1 2 (b)

sin 1 1 1 1 d sin max  mmax     mmax    5.76 sin max mmax sin 1 sin 10 (c) The highest order is m = 6.

18. x Figure 27.56 shows a double slit located a distance from a screen, with the y d distance from the center of the screen given by . When the distance between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, sin    for small angles (where , with in radians), the distance between fringes is y  x / d given by . OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution y  m1 ym+ 1 ym d  m

sin   tan   in radians  For small angles

d sin m  m and d sin m1  m 1 For two adjacent fringes we have Subtracting these equations gives:

d sin θm1  sin θm   m 1 mλ

dθm1 θ m   λ

ym  ym1 ym  tan θm  θ m  d     x  x x  y x d    y  x d

19. Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 27.56.

Solution x 3.00 m 6.33  107 m y      2.37  102 m  2.37 cm d 8.00  10-5 m

20. Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 27.56).

Solution x y 1.20  105 m 7.50  103 m y     d      4.50  107 m  450 nm d x 2.00 m

27.4 MULTIPLE SLIT DIFFRACTION

21. A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum be for 520-nm-wavelength green light? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution 0.0100 m d   5.00 106 m 2000

    5.20  107 m  d sin       sin -1  sin -1   5.97    -6   d   5.00  10 m  Therefore, since

22. Find the angle for the third-order maximum for 580-nm-wavelength yellow light falling on a diffraction grating having 1500 lines per centimeter.

Solution 0.0100 m d   6.667106 m. 1500

7 -1 3  -1 35.80 10 m d sin   m    sin    sin  -6   1.51  d   6.667 10 m  Since

23. How many lines per centimeter are there on a diffraction grating that gives a first- 25.0 order maximum for 470-nm blue light at an angle of ?

Solution m 0.0100 m d sin   m  d   sin  N 0.0100 m sin 0.0100 msin 25.0 N    8.99103 m 14.70 107 m Therefore,

24. What is the distance between lines on a diffraction grating that produces a second- 60.0 order maximum for 760-nm red light at an angle of ?

Solution m 2 7.60  107 m d sin   m  d      1.76  106 m sin  sin 60.0

25. 45.0 Calculate the wavelength of light that has its second-order maximum at when falling on a diffraction grating that has 5000 lines per centimeter. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution The second order maximum is constructive interference, so for diffraction gratings d sin   m for m  0,1,2,3,... we use the equation where the second order m  2 maximum has . Next, we need to determine the slit separation by using the fact that there are 5000 lines per centimeter: 1 1 m d    2.00  106 m 5000 slits/cm 100 cm

  45.0 Since , we can determine the wavelength of the light:

6 d sin  2.00  10 msin 45.0 3     7.07  10 m  707 nm m 2

26. An electric current through hydrogen gas produces several distinct wavelengths of visible light. What are the wavelengths of the hydrogen spectrum, if they form first- 24.2 25.7 29.1 41.0 order maxima at angles of , , , and when projected on a diffraction grating having 10,000 lines per centimeter? Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics.

Solution 0.0100 m d  1.00106 m. So, d sin   m  , 10,000 6 7 a  1.0010 m sin 24.2  4.09910 m  410 nm 6 7 b  1.0010 m sin 25.7  4.33710 m  434 nm 6 7 c  1.0010 m sin 29.1  4.86310 m  486 nm 6 7 d  1.0010 m  sin 41.0  6.56110 m  656 nm

27. (a) What do the four angles in the above problem become if a 5000-line-per- centimeter diffraction grating is used? (b) Using this grating, what would the angles be for the second-order maxima? (c) Discuss the relationship between integral reductions in lines per centimeter and the new angles of various order maxima. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution 0.0100 m d   2.00 106 m ; d sin   m   5000     4.099 107 m    sin 1 1  sin 1    11.8 ; 1,a    6   d   2.0010 m   4.337 107 m    sin 1    12.5 ; 1,b  -6   2.00 10 m   4.863107 m    sin 1    14.1 ; 1,c  -6   2.00 10 m   6.561107 m    sin 1    19.2 1,d  2.00 10-6 m  (a)  

1  2  d sin   m  2    sin 1    d  7 1 24.099 10 m  2,a  sin  -6   24.2 ;  2.00 10 m  7 1 24.337 10 m  2,b  sin  6   25.7 ;  2.00 10 m  7 1 24.86310 m  2,c  sin  -6   29.1 ;  2.00 10 m  7 1 26.56110 m  2,d  sin  6   41.0 2.0010 m (b)  

x (c) Decreasing the number of lines per centimeter by a factor of means that the

x angle for the -order maximum is the same as the original angle for the first- order maximum.

28. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution d sin   . d The maximum number of lines corresponds to the smallest , for the   90  d sin 90    d   longest wavelength in the visible spectrum at . So, d d  780 nm the smallest which can accommodate the entire visible spectrum is .

102 m 102 m 102 m d   N    1.28205  104  12,821  12,800 N d 7.80  10-7 m

29. The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it 36.093 36.129 produces two first-order maxima at and when projected on a 10,000 line per centimeter diffraction grating. What are the two wavelengths to an accuracy of 0.1 nm?

Solution 0.0100 m d  1.00106 m, so   d sin   10,000 6 7 a  1.00010 msin 36.093  5.89110 m  589.1nm 6 7 b  1.00010 msin 36.129  5.89610 m  589.6nm

30. What is the spacing between structures in a feather that acts as a reflection grating, 30.0 given that they produce a first-order maximum for 525-nm light at a angle?

Solution m 1 5.25  107 m d sin   m  d      1.05  106 m sin  sin 30.0

31. Structures on a bird feather act like a reflection grating having 8000 lines per centimeter. What is the angle of the first-order maximum for 600-nm light?

Solution 0.0100 m d   1.25106 m. So, 8000 7 1  m  1 16.0010 m d sin   m    sin    sin  -6   28.7  d   1.2510 m 

32. An opal such as that shown in Figure 27.17 acts like a reflection grating with rows 8 μm separated by about . If the opal is illuminated normally, (a) at what angle will red light be seen and (b) at what angle will blue light be seen? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution d sin   m Constructive interference of light from rows in crystal is given by . The 17.00107  sin    0.088 8106    5.0 angle that red light (700 nm) will be seen is , or , d  8m   460 nm   3.3 where spacing . For blue light, , and so .

33. At what angle does a diffraction grating produces a second-order maximum for light 20.0 having a first-order maximum at ?

Solution sin 1 1 d sin 1   ; d sin  2  2    sin  2 2 1 1  2  sin 2 sin 1   sin 2sin 20.0  43.2

34. Show that a diffraction grating cannot produce a second-order maximum for a given 30.0 wavelength of light unless the first-order maximum is at an angle less than .

Solution sin 2 1 The largest possible second order occurs when . Using the equation

d sin m  m, we see that the value for the slit separation and wavelength are the same for the first and second order maximums, so that: sin 1 1  d sin 1   d sin 2  2 sin2 2 and , so that:

sin 2 Now, since we know the maximum value for , we can solve for the maximum

1 value for : 1 1  1   1  1  sin  sin 2  1,max  sin    30.0  2   2  so that:

35. If a diffraction grating produces a first-order maximum for the shortest wavelength of 30.0 visible light at , at what angle will the first-order maximum be for the longest wavelength of visible light? OpenStax College Physics Instructor Solutions Manual Chapter 27

7 Solution 1 3.8010 m 7 d sin 1  1  d    7.6010 m ; sin 1 sin 30.0  7.60107 m  d sin     sin 1   90.0 2 2  -7   7.6010 m 

36. (a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light. (b) Would such a grating be useful for ultraviolet spectra? (c) For infrared spectra?

Solution   380 nm, d sin   . , d (a) For a given the smallest corresponds to the largest

 380 nm 0.0100 m sin  or sin  1  d    3.80 107 m  sin  1 N 0.0100 m  N   26,315  26,300 lines/cm 3.80 10-7 m

 sin   1.   380 nm d (b) For UV, and so Therefore, yes it would be useful for UV.   760 nm and sin  1 (c) For IR, , so no it would not be useful for IR.

37. (a) Show that a 30,000-line-per-centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a first- order maximum? (c) What is the greatest number of lines per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution (a) First we need to calculate the slit separation: 1 line 1 line 1 m d     3.333  107 m  333.3 nm. N 30,000 lines/cm 100 cm

d sin   m Next, using the equation , we see that the longest wavelength will be sin  1 and m 1 d    333.3 nm for , so in that case, , which is not visible. (b) From part (a), we know that the longest wavelength is equal to the slit separation, or 333 nm. (c) To get the largest number of lines per cm and still produce a complete spectrum, we want the smallest slit separation that allows the longest wavelength of visible

max  760 nm light to produce a second order maximum, so (see Example 27.3). m  2 and sin   1, For there to be a second order spectrum, so 6 d  2max  2760 nm  1.52 10 m

Now, using the technique in step (a), only in reverse: 1 line 1 line 1 m N     6.58  103 lines/cm d 1.52  10-6 m 100 cm

38. A He–Ne laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.50 m from the CD, and the first fringe is 0.600 m from the central maximum, what is the spacing of grooves on the CD?

Solution y (0.600 m) d sin θ  m,sin θ    0.4. x 1.5m   6.3310 7 m For the He–Ne laser . m 6.33107 m d    1.5810 6 m. sin θ 0.4 Therefore

39. The analysis shown in the figure below also applies to diffraction gratings with lines d separated by a distance . What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 600-nm light, if the screen is 1.50 m away? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution 102 m x 1.50 m 6.00  107 m d   8.00  105 m y      1.13  102 m 125 d 8.00  10-5 m , so that

40. Unreasonable Results Red light of wavelength of 700 nm falls on a double slit separated by 400 nm. (a) At what angle is the first-order maximum in the diffraction pattern? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Solution   700 nm  d sin   m    sin 1  sin 1  D  400 nm    sin 1 1.75 (a) (b) The sine of a real angle cannot be greater than one.   d (c) Assuming that you will get a diffraction pattern if is unreasonable.

41. Unreasonable Results (a) What visible wavelength has its fourth-order maximum at 25.0 an angle of when projected on a 25,000-line-per-centimeter diffraction grating? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Solution d sin   m, for m  0,1,2,3,4,... (a) For diffraction gratings, we use the equation m  4 where the fourth order maximum has . We first need to determine the slit

separation by using the fact that there are 25,000 lines per centimeter: 1 1 m d    4.00  107 m 25,000 lines /cm 100 cm

  25.0 So, since , we can determine the wavelength of the light:

d sin  4.00  107 m sin 25.0       4.226  108 m  42.3 nm m 4

(b) This wavelength is not in the visible spectrum. (c) The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light. OpenStax College Physics Instructor Solutions Manual Chapter 27

27.5 SINGLE SLIT DIFFRACTION

43. (a) At what angle is the first minimum for 550-nm light falling on a single slit of width 1.00 μm ? (b) Will there be a second minimum?

Solution 7 1 m  1 15.5010 m Dsin   m    sin    sin  -6   33.4  D   1.0010 m  (a) For m  2,sin  1 No (b)

44. 2.00 - μm (a) Calculate the angle at which a -wide slit produces its first minimum for 410-nm violet light. (b) Where is the first minimum for 700-nm red light?

Solution 7 1 m  1 14.1010 m Dsin   m   sin    sin  -6  11.8  D   2.0010 m  (a) 7 1 17.0010 m   sin  -6   20.5  2.0010 m  (b)

45. (a) How wide is a single slit that produces its first minimum for 633-nm light at an 28.0 angle of ? (b) At what angle will the second minimum be?

Solution 7 m 16.3310 m 6 D sin   m  D      1.3510 m sin   sin 28.0  (a) 7 1  m  1 26.3310 m   sin    sin  -6   69.9  D   1.34810 m  (b)

46. 60.0 (a) What is the width of a single slit that produces its first minimum at for 600- 62.0 nm light? (b) Find the wavelength of light that has its first minimum at . OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution  6.00107 m Dsin   m  D    6.928107 m  693 nm sin  sin 60.0 (a)   Dsin   6.92810 7 m sin 62.0  6.117 10 7 m  612 nm (b)

47. 48.6 Find the wavelength of light that has its third minimum at an angle of when it 3.00 μm falls on a single slit of width .

Solution D sin  3.00  106 msin 48.6 D sin   m      7.50  107 m  750 nm m 3

48. 36.9 Calculate the wavelength of light that produces its first minimum at an angle of 1.00 μm when falling on a single slit of width .

Solution Dsin   m, Since 6 D sin  1.00  10 msin 36.9 7     6.004  10 m  600 nm m 1

49. (a) Sodium vapor light averaging 589 nm in wavelength falls on a single slit of width 7.50 μm . At what angle does it produces its second minimum? (b) What is the highest-order minimum produced?

Solution 7 1 m  1 25.8910 m Dsin   m   sin    sin  -6   9.04  D   7.5010 m  (a) D sin 90 7.50 106 m m    12.73  Highest order  12  5.89 10-7 m (b)

50. (a) Find the angle of the third diffraction minimum for 633-nm light falling on a slit of 20.0 μm 85.0 width . (b) What slit width would place this minimum at ? Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution 7 1 m  1 36.3310 m Dsin   m   sin    sin  -5   5.45  D   2.0010 m  (a) m 3 6.33107 m D     1.91106 m 1.91 μm sin  sin 85.0 (b)

51. (a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and 589.6 nm, when they fall upon a single slit of width 2.00 μm . (b) What is the distance between these minima if the diffraction pattern falls on a screen 1.00 m from the slit? (c) Discuss the ease or difficulty of measuring such a distance.

Solution     5.891107 m  Dsin   m    sin 1 1  sin 1  17.13056 and 1 1 1    -6   D   2.0010 m   5.896107 m    sin 1  17.14554 so that    0.0150 2  2.0010-6 m  2 1 (a)  

y  x tan   1.00 m tan0.0150  2.62 10 4 m  0.262 mm (b) (c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable.

52.  (a) What is the minimum width of a single slit (in multiples of ) that will produce a  first minimum for a wavelength ? (b) What is its minimum width if it produces 50 minima? (c) 1000 minima?

Solution m Dsin   m  D   m (for   90)  D  1.00 sin  (a) D  50.0  (b) D  1000 (c) OpenStax College Physics Instructor Solutions Manual Chapter 27

53. 14.5 (a) If a single slit produces a first minimum at , at what angle is the second-order minimum? (b) What is the angle of the third-order minimum? (c) Is there a fourth-order minimum? (d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).

Solution sin 1 1 1 1 Dsin  m  m     2  sin 2sin 1   sin 2sin 14.5  30.1 sin  2 2 (a) 1 1 3  sin 3sin 1   sin 3sin 14.5 48.7 (b) 1 1 θ4  sin 4sin 14.5  sin 1.0015  No (c)

21  214.5  29, 2 1  30.05 14.5  15.56 Thus, 29  2 15.56  31.1 (d)   

54. A double slit produces a diffraction pattern that is a combination of single and double slit interference. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern. (This will greatly reduce the intensity of the fifth maximum.)

Solution D d The problem is asking us to find the ratio of to . For the single slit, using the Dsin   n n 1 equation , we have . For the double slit, using the equation

d sin   m m  5 (because we have a maximum), we have . Dividing the single slit

equation by double slit equation, where the angle and wavelength are the same D n 1 D     0.200. d m 5 d gives: So, the slit separation is five times the slit width.

55. Integrated Concepts A water break at the entrance to a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves of 20.0-m wavelength approach the opening straight on. At what angle to the incident direction are the boats inside the harbor most protected against wave action?

Solution We are looking for the first minimum for single slit diffraction because the 50.0 m Dsin   m, where m 1 wide opening acts as a single slit. Using the equation , we OpenStax College Physics Instructor Solutions Manual Chapter 27

can determine the angle for first minimum:  m  120.0 m   sin 1    sin 1  23.58  23.6  D   50.0 m 

Since the main peak for single slit diffraction is the main problem, a boat in the harbor at an angle greater than this first diffraction minimum will feel smaller waves. At the second minimum, the boat will not be affected by the waves at all:  m  220.0 m   sin 1   sin 1  53.13  53.1  D   50.0 m 

56. Integrated Concepts An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Outside the door, on a line perpendicular to the opening in the door, a jet engine makes a 600-Hz sound. At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is 0.800 m wide and the speed of sound is 340 m/s?

Solution v 340 m s v  f      0.5667 m f 600 Hz

We are looking for the first minimum for single slit diffraction:  m  10.5667 m   sin 1   sin 1  45.1  D   0.800 m 

27.6 LIMITS OF RESOLUTION: THE RAYLEIGH CRITERION

57. The 300-m-diameter Arecibo radio telescope pictured in Figure 27.28 detects radio waves with a 4.00 cm average wavelength. (a) What is the angle between two just- resolvable point sources for this telescope? (b) How close together could these point sources be at the 2 million light year distance of the Andromeda galaxy?

Solution 1.22 1.22 4.00102 m      1.63104 rad D 300 m (a) y tan      y  x, so that y  2.00106 ly1.63104 rad  3.25 ly x (b)

58. Assuming the angular resolution found for the Hubble Telescope in Example 27.5, what is the smallest detail that could be observed on the Moon? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution y tan   y  x tan  x, so that y  3.82108 m2.80107 rad  107 m x

59. Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading of a flashlight beam that is originally 5.00 cm in diameter with an average wavelength of 600 nm.

Solution 1.22 1.22 6.00 107 m       1.46 105 rad D 0.0500 m

60. (a) What is the minimum angular spread of a 633-nm wavelength He-Ne laser beam that is originally 1.00 mm in diameter? (b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be? (c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.) Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics.

Solution 1.22 1.22 6.33107 m       7.72 104 rad D 1.00 10-3 m (a)  y D Y

(b) x D  2y, y The diameter of the spot will be where is the width of the spread of the spot edge due to diffraction. y  x  1.50104 m7.723104 rad11.5845 m Y  D  2y  1.00103 m 211.5845 m  23.2 m

D (c) We can ignore since it is so small. Y  2y  2x  23.82  108 m7.723  104 rad 5.90  105 m  590 km OpenStax College Physics Instructor Solutions Manual Chapter 27

61. A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be projected onto a satellite or the Moon. (a) If this is done with the Mount Wilson telescope, producing a 2.54-m-diameter beam of 633-nm light, what is the minimum angular spread of the beam? (b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the Moon, assuming a 3.84108 m lunar distance of ?

Solution 1.22 1.22 6.33  107 m       3.04  107 rad D 2.54 m (a) d  D  2y  D  2x  2.54 m  23.84  108 m3.04  107 rad  2.54 m  233 m  235 m (b)

62. The limit to the eye’s acuity is actually related to diffraction by the pupil. (a) What is the angle between two just-resolvable points of light for a 3.00-mm-diameter pupil, assuming an average wavelength of 550 nm? (b) Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.30 m apart? (c) What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye? (d) How does your answer to (c) compare to details you normally observe in everyday circumstances?

Solution   5.50  107 m    1.22  1.22    2.237 104 rad  2.24  104 rad   -3  D  3.00  10 m  (a) s r  (b) The distance between two objects, a distance away, separated by an angle

s 1.30 m r    5.811  103 m  5.81 km s  r  2.237  10-4 rad is , so: (c) Using the same equation as in part (b): s  r  0.800 m2.237  104 rad  1.790  104 m  0.179 mm

(d) Holding a ruler at arm’s length, you can easily see the millimeter divisions, so you can resolve details 1.0 mm apart. Therefore, you probably can resolve details 0.2 mm apart at arm’s length. OpenStax College Physics Instructor Solutions Manual Chapter 27

63. What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.00 km on the Moon some 384,000 km away? Assume an average wavelength of 550 nm for the light received.

Solution 5.00 km   so that 384,000 km 1.22 1.22 5.50  107 m 384,000 km D      5.15  102 m  5.15 cm  5.00 km

64. You are told not to shoot until you see the whites of their eyes. If the eyes are separated by 6.5 cm and the diameter of your pupil is 5.0 mm, at what distance can you resolve the two eyes using light of wavelength 555 nm?

Solution Apply the Rayleigh criterion to the eyes:  s sD 0.065 m0.0050 m θ  1.22   R    480 m D R 1.22 1.225.55107 m .

65. (a) The planet Pluto and its Moon Charon are separated by 19,600 km. Neglecting atmospheric effects, should the 5.08-m-diameter Mount Palomar telescope be able to 4.50109 km resolve these bodies when they are from Earth? Assume an average wavelength of 550 nm. (b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. What are the reasons for this?

Solution 1.22 1.22 5.50  107 m       1.321  107 rad D 5.08 m (a) The telescope should be able to discern a separation at: y  x  4.50  109 km1.321  10 7 rad  594 km  19,600 km

Yes. The telescope should easily be able to discern a separation. (b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations.

66. The headlights of a car are 1.3 m apart. What is the maximum distance at which the eye can resolve these two headlights? Take the pupil diameter to be 0.40 cm. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution (5.55107 m) 1.3 m θ  1.22  4.010-3 m R From , we can calculate the maximum distance at R  7.71km which lights can be resolved: .

67. When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh’s criterion. Take the pupil of the eye to be 3.0 mm and the distance from the paper to the eye of 35 cm; find the minimum separation of two dots such that they cannot be resolved. How many dots per inch (dpi) does this correspond to?

Solution (5.55107 m) s θ  1.22   s  7.9105 m  7.9 10-3 cm 3.010-3 m 0.35 m

2.54cm  7.9103 cm   370 dpi 1in Resolution in dpi . These are specs found on most printers today.

68. Unreasonable Results An amateur astronomer wants to build a telescope with a diffraction limit that will allow him to see if there are people on the moons of Jupiter. (a) What diameter mirror is needed to be able to see 1.00 m detail on a Jovian Moon 7.50108 km at a distance of from Earth? The wavelength of light averages 600 nm. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Solution  y  y 1 m 1.22   tan 1     1.333  1012 rad   x  x 7.50  1011 m D 7 1.22 1.226.00  10 m 5  D   -12  5.49  10 m  549 km (a)  1.333  10

(b) This is an unreasonably large telescope. (c) It is unreasonable to assume you could build a telescope large enough to see such detail. OpenStax College Physics Instructor Solutions Manual Chapter 27

27.7 THIN FILM INTERFERENCE

70. A soap bubble is 100 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength and color of visible light is most constructively reflected, assuming the same index of refraction as water?

Solution   2t  n     4 nt (one phase change), 2 2 n   41.331.00107 m 5.32107 m  532 nm (green)

71. An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?

Solution   2t  n  (one phase change), so   4 tn 2 2 n   41.20 107 m1.40  6.72 107 m  672 nm (red)

72. Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.

Solution    2t  n  (one phase change), so t  2 2 n 4n 4.70107 m t     8.39108 m  83.9 nm 41.40

73. Find the minimum thickness of a soap bubble that appears red when illuminated by white light perpendicular to its surface. Take the wavelength to be 680 nm, and assume the same index of refraction as water.

Solution The minimum thickness will occur when there is one phase change, so for light   2t  n  2 2n incident perpendicularly, constructive interference first occurs when . So, using the index of refraction for water from Table 25.1:  6.80  107 m t    1.278  107 m  128 nm 4n 41.33 OpenStax College Physics Instructor Solutions Manual Chapter 27

74. n  1.33 A film of soapy water ( ) on top of a plastic cutting board has a thickness of 233 nm. What color is most strongly reflected if it is illuminated perpendicular to its surface?

Solution  2t    (two phase changes), so n n   2tn  22.33  107 m1.33  6.20  107 m  620 nm (orange)

75. n  1.33 What are the three smallest non-zero thicknesses of soapy water ( ) on Plexiglas if it appears green (constructively reflecting 520-nm light) when illuminated perpendicularly by white light? Explicitly show how you follow the steps in Problem- Solving Strategies for Wave Optics.

Solution   3 2t   , 2 , 3   t  , , . Therefore, n n n 2n n 2n  520 nm  520 nm 3 3520 nm   195 nm ;   391 nm ; and   586 nm 2n 21.33 n 1.33 2n 21.33

76. Suppose you have a lens system that is to be used primarily for 700-nm red light. What is the second thinnest coating of fluorite (magnesium fluoride) that would be non-reflective for this wavelength?

Solution 3 3 2t  n  (two phase changes), so 2 2n 3 3 7.00  107 m t      3.80  107 m  380 nm 4n 41.38

77. (a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one- fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water. (b) Discuss the fragility of the film considering the thickness found. OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution (a) It can be as thick as one-fourth of the longest visible wavelength in water, so that   760 nm t  n    143 nm 4 4n 41.33

(b) This thickness is rather small compared to the diameter of a soap bubble, so it is rather fragile when it appears dark.

78. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.

Solution The path length must be less than one-fourth of the shortest visible wavelength in oil. The thickness of the oil is half the path length, so it must be less than one-eighth of the shortest visible wavelength in oil. If we take 380 nm to be the shortest visible  3.80107 m t  n   3.39108 m  33.9 nm 8 81.40 wavelength in air,

79. Figure 27.34 shows two glass slides illuminated by pure-wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a 0.100-mm-diameter hair at the other end, forming a wedge of air. (a) How far apart are the dark bands, if the slides are 7.50 cm long and 589-nm light is used? (b) Is there any difference if the slides are made from crown or flint glass? Explain.

Solution

Two adjacent dark bands will have thickness differing by one wavelength, i.e., 1.00104 m  hair diameter   tan -1   0.076394.   d2  d1, and tan     slide length  0.075 m  or

x tan  d2  d1  , So, since we see that  5.89107 m x    4.418104 m  0.442 mm tan tan0.076394

(b) The material makeup of the slides is irrelevant because it is the path difference in the air between the slides that gives rise to interference. OpenStax College Physics Instructor Solutions Manual Chapter 27

80. Figure 27.34 shows two 7.50-cm-long glass slides illuminated by pure 589-nm wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on some debris at the other end, forming a wedge of air. How thick is the debris, if the dark bands are 1.00 mm apart?

Solution D d 2  d 1 x 7.5 cm

D tan  7.50 cm

d2  d1   Two dark bands will have thickness differing by one wavelength, i.e., d  d  D tan  2 1   x x 7.50 cm 2 7 7.50 cm 7.5010 m5.8910 m 5 D    4.4210 m x 10-3 m

81. 45 Repeat Exercise 27.70, but take the light to be incident at a angle.

Solution 2t   4nt  n     sin 45.0 2 2n sin 45.0 (one phase change) 41.331.00107 m      7.52107 m  752 nm (red) sin 45 So,

82. 45 Repeat Exercise 27.71, but take the light to be incident at a angle.

Solution 2t    n  sin 45 2 2n For one phase change: , so 4tn 41.20  107 m 1.40       9.50  107 m  950 nm (infrared) sin 45.0 sin 45.0

2t 3 3  n  sin 45 2 2n For three phase changes: , so OpenStax College Physics Instructor Solutions Manual Chapter 27

7 4tn 41.20  10 m1.40 7     3.17  10 m  317 nm ultraviole t 3 sin 45.0 3 sin 45.0

Therefore, the oil film will appear black, since the reflected light is not in the visible part of the spectrum.

83. Unreasonable Results To save money on making military aircraft invisible to radar, an inventor decides to coat them with a non-reflective material having an index of refraction of 1.20, which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers. (a) What thickness should the coating be to inhibit the reflection of 4.00-cm wavelength radar? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Solution n (a) Assuming for the plane is greater than 1.20 then there are two phase changes:

1     4.00102 m  2t     t    0.833 cm 2  n  4n 41.2

(b) It is too thick, and the plane would be too heavy. (c) It is unreasonable to think the layer of material could be any thickness when used on a real aircraft.

27.8 POLARIZATION

84. What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half?

1 1 Solution  2   I   1  2  2 1    1   I  I 0 cos   cos    cos    45.0  I   2    0    

85. 45.0 The angle between the axes of two polarizing filters is . By how much does the second filter reduce the intensity of the light coming through the first?

Solution I 2 2 1  0.500 I  I 0 cos   I 0 cos 45  I 0 2 I 0 therefore, OpenStax College Physics Instructor Solutions Manual Chapter 27

86. 150 W / m2 If you have completely polarized light of intensity , what will its intensity 89.0 be after passing through a polarizing filter with its axis at an angle to the light’s polarization direction?

2 Solution I  I0 cos  Using the equation : 2 2 2 2 2 2 I  I 0 cos   150 W m cos 89.0  4.57  10 W m  45.7 m W m

87. What angle would the axis of a polarizing filter need to make with the direction of 1.00 kW/m 2 10.0 W/m 2 polarized light of intensity to reduce the intensity to ?

1 1 Solution  2  2 2  I   10.0 W m   I  I cos2   cos1     cos1     84.3 0  I   1.00  103 W m2    0     

88. At the end of Example 27.8, it was stated that the intensity of polarized light is 90.0% reduced to of its original value by passing through a polarizing filter with its 18.4 axis at an angle of to the direction of polarization. Verify this statement.

Solution 2 I 2 2 I  I 0 cos    cos   cos 18.4  0.9004  90.04 %  90.0 % I 0

89. 45 Show that if you have three polarizing filters, with the second at an angle of to 90.0 the first and the third at an angle of to the first, the intensity of light passed by 25.0% the first will be reduced to of its value. (This is in contrast to having only the first and third, which reduces the intensity to zero, so that placing the second between them increases the intensity of the transmitted light.)

Solution

I0 I1 I2

2 2 2 2 I1  I 0 cos 1 , I 2  I1 cos  2  I 0 cos 1 cos  2 OpenStax College Physics Instructor Solutions Manual Chapter 27

2 2 I2  I0 cos 45 cos 45  0.25I0

90. I Prove that, if is the intensity of light transmitted by two polarizing filters with axes  I 90.0  at an angle and is the intensity when the axes are at an angle , then

I  I  I0 , the original intensity. (Hint: Use the trigonometric identities cos(90.0  )  sin  cos2   sin 2  1 and .)

Solution 2 2 I  I0 cos , I  I0 cos 90   2 2 2 2 2 2 I  I  I0 cos   I0 cos90    I0 cos   I0 sin    I0 cos   sin    I0

91. At what angle will light reflected from diamond be completely polarized?

Solution n2 2.42 tanb    2.42 1 n1 1 b  tan 2.42  67.6 . Therefore,

92. What is Brewster’s angle for light traveling in water that is reflected from crown glass?

Solution n2 tanb  n1 n2 n1 Using the equation , where is for crown glass and is for water (see

 n   1.52  1  2  1  b  tan    tan    48.75  48.8  n1  1.333 Table 25.1). Brewster’s angle is

93. A scuba diver sees light reflected from the water’s surface. At what angle will this light be completely polarized?

Solution  n  1.333  1  2  1  b  tan    tan    53.1  n1   1.00 

94. At what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank? OpenStax College Physics Instructor Solutions Manual Chapter 27

Solution  n  1.333 1  2  1  b  tan    tan    41.2  n1   1.52 

95. 55.6 Light reflected at from a window is completely polarized. What is the window’s index of refraction and the likely substance of which it is made?

Solution n2 tanb   n2  n1 tan b  1.00tan55.6  1.46 n1

The substance is likely fused quartz.

96. 62.5 (a) Light reflected at from a gemstone in a ring is completely polarized. Can the gem be a diamond? (b) At what angle would the light be completely polarized if the gem was in water?

Solution n2 tan  b   n2  n1 tan  b  1.00tan62.5  1.921  1.92 n1 (a) The gem is not a diamond (it is zircon).

1 1.921  b  tan    55.2 1.333 (b)

97. b If is Brewster’s angle for light reflected from the top of an interface between two

b substances, and is Brewster’s angle for light reflected from below, prove that

b b  90.0 .

Solution n2 n1 tanb  , tanb   tanb  cotb n1 n2 . The equation can be solved with the

cot  tan90  b  90 b  b b  90 trigonometric identity:

98. Integrated Concepts If a polarizing filter reduces the intensity of polarized light to 50.0% of its original value, by how much are the electric and magnetic fields reduced? OpenStax College Physics Instructor Solutions Manual Chapter 27

2 2 Solution I  B  I1  aB1 a where is some constant.

2  B  2 2  2  I 2  aB2  0.500I1  0.500aB1     0.500, so that B2  0.500B1  0.707B1  B1 

99. Integrated Concepts Suppose you put on two pairs of Polaroid sunglasses with their 15.0 axes at an angle of . How much longer will it take the light to deposit a given amount of energy in your eye compared with a single pair of sunglasses? Assume the lenses are clear except for their polarizing characteristics.

Solution 2 I 2 t2 1 I  I 0 cos    cos 15.0  0.933    1.072  t2  1.072t1 I 0 t1 0.933

100. 1.00 kW / m2 Integrated Concepts (a) On a day when the intensity of sunlight is , a circular lens 0.200 m in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle 20.0 100% of . Assuming the sunlight is unpolarized and the polarizers are efficient, C / s 80.0% what is the initial rate of heating of the water in , assuming it is absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain.

Solution (a) After passing through the two polarizers the intensity is: 2 2 2 2 I  I 0 cos   1.00 k W m cos 20.0  883W m

This intensity is focused on the lens, producing a total power hitting the water of: P  IA  883W m2  0.100 m2  27.7 W

Since the water absorbs only 80.0% of the heat, the power transferred to the

Pabsorbed  0.800P  0.80027.7 W  22.2 W water is: 22.2 J of heat are transferred to the water and beaker every second. Finally, we can get an expression for the change in temperature of the water and beaker: Q Q  mw cw T  mAlcAlT  T  mwcw  mAlcAl

Q For the rate of heating, replace with the power transferred: OpenStax College Physics Instructor Solutions Manual Chapter 27

T 22.2 W   1.69102 C/s t 0.250 kg4186 J Kg C  0.300 kg900 J Kg C

(b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.