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
n 
 0.750  .
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
n 
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
n 



380 nm 760 nm 380 nm 760 nm
to

to
 250 nm to 500 nm
n
n
1.52
1.52

n
n

1

 1.49  Polystyrene
n 0.671
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
n 

n
n
 633 nm

 1.92 ; n  1.92  Zircon
n 329 nm
5.
What is the ratio of thicknesses of crown glass and water that would contain the
same number of wavelengths of light?
Solution
dc
c

dw
E

d c c
 nc nw 1.333




 0.877
d w w  nw nc
1.52
27.3 YOUNG’S DOUBLE SLIT EXPERIMENT
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 
  0.516
d sin   m    sin 
-5
 5.00  10 m 
-1
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
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
Using d sin   m for m  0,1,2,3,......

7
m
 m 
-1  3 5.80  10
  sin 
-4
 d 
 1.00  10 m
  sin -1 
  0.997


8.
What is the separation between two slits for which 610-nm orange light has its first
maximum at an angle of 30.0 ?
Solution
d sin   m  d 
9.
Find the distance between two slits that produces the first minimum for 410-nm violet
light at an angle of 45.0 .
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 )


m 1 6.10 10 7 m

 1.22 10 6 m
sin 
sin 30.0


10.
Calculate the wavelength of light that has its third minimum at an angle of 30.0
when falling on double slits separated by 3.00 μm . Explicitly show how you follow
the steps in Problem-Solving Strategies for Wave Optics.
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.
What is the wavelength of light falling on double slits separated by 2.00 μm if the
third-order maximum is at an angle of 60.0 ?
Solution
d sin   m   
12.
At what angle is the fourth-order maximum for the situation Exercise 27.6?
Solution


d sin 
2.00  106 m sin 60.0

 5.77  107 m  577 nm
m
3

7
m
 m 
-1  4 4.50  10
d sin   m    sin -1 

sin


-5
 d 
 5.00  10 m
  2.06


OpenStax College Physics
Instructor Solutions Manual
Chapter 27
13.
What is the highest-order maximum for 400-nm light falling on double slits separated
by 25.0 μm ?
Solution
Looking at d sin   m , we notice that the highest order occurs when sin   1 , so
d 2.50  10 5 m
the highest order is: m  
 62.5. Since m must be an integer, the
 4.00  10 -7 m
highest order is then m = 62.
14.
Find the largest wavelength of light falling on double slits separated by 1.20 μm for
which there is a first-order maximum. Is this in the visible part of the spectrum?
Solution
d sin   m   
Thus, max
d sin 
 d sin 
m
 d  1.20  10 6 m  1200 nm (not visible)
15.
What is the smallest separation between two slits that will produce a second-order
maximum for 720-nm red light?
Solution
d sin   m  2  d 
16.
(a) What is the smallest separation between two slits that will produce a secondorder maximum for any visible light? (b) For all visible light?
Solution
(a) d sin   m  2    380 nm  760 nm.
So, dmin  2min  2380 nm   760 nm
(b) d  2760 nm   1520 nm
17.
(a) If the first-order maximum for pure-wavelength light falling on a double slit is at
an angle of 10.0 , 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
(a) d sin 1   , d sin  2  2 
2
;   90. Thus, d min  2  1440 nm  1.44 μm
sin 
sin 1 1

sin  2 2
  2  sin -1 2 sin 1   sin -1 2 sin 10.0  20.3
sin 1
1
 2   1  sin -1 0.5 sin 10  4.98
(b) d sin 1   , d sin  1   
2
2
2
sin  1
2
sin 1
1
1
1

 mmax 

 5.76
sin  max mmax
sin 1 sin 10
The highest order is m = 6.
(c) d sin  max  mmax  
OpenStax College Physics
18.
Instructor Solutions Manual
Chapter 27
Figure 27.56 shows a double slit located a distance x from a screen, with the
distance from the center of the screen given by y . When the distance d between the
slits is relatively large, there will be numerous bright spots, called fringes. Show that,
for small angles (where sin    , with  in radians), the distance between fringes is
given by y  x / d .
Solution
y
 m1
d
ym + 1
ym
m
x
For small angles sin   tan    in radians 
For two adjacent fringes we have d sin  m  m and d sin  m1  m  1
Subtracting these equations gives:
d sin θ m 1  sin θ m   m  1  mλ
d θ m 1 θ m   λ
ym
y 
y
θ m  d  m 1  m   
x
x 
 x
y
x
d
   y 
x
d
tan θ m 
19.
Using the result of the problem above, calculate the distance between fringes for 633nm 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
-5
d
8.00  10 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
Solution
d
Instructor Solutions Manual
Chapter 27
0.0100 m
 5.00  10 6 m
2000
 5.20  10 7 m 

  5.97
Therefore, since d sin       sin -1    sin -1 
-6
d 
 5.00  10 m 
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
d
0.0100 m
 6.667  10 6 m.
1500


 3 5.80  10 7 m 
 3 
Since d sin   m    sin -1    sin -1 
  1.51
-6
 d 
 6.667  10 m 
23.
How many lines per centimeter are there on a diffraction grating that gives a firstorder maximum for 470-nm blue light at an angle of 25.0 ?
Solution
m
0.0100 m

sin 
N
0.0100 m sin   0.0100 msin 25.0  8.99  103
Therefore, N 
1 4.70  107 m
m
d sin   m  d 


24.
What is the distance between lines on a diffraction grating that produces a secondorder maximum for 760-nm red light at an angle of 60.0 ?
Solution
d sin   m  d 
25.
Calculate the wavelength of light that has its second-order maximum at 45.0 when
falling on a diffraction grating that has 5000 lines per centimeter.
Solution
The second order maximum is constructive interference, so for diffraction gratings
we use the equation d sin   m for m  0,1,2,3,... where the second order
maximum has m  2 . Next, we need to determine the slit separation by using the
fact that there are 5000 lines per centimeter:
1
1m
d

 2.00  10 6 m
5000 slits/cm 100 cm
Since   45.0 , we can determine the wavelength of the light:
d sin  2.00  10 6 m sin 45.0


 7.07  10 3 m  707 nm
m
2


m 2 7.60  10 7 m

 1.76  10 6 m
sin 
sin 60.0
OpenStax College Physics
26.
Solution
Instructor Solutions Manual
Chapter 27
An electric current through hydrogen gas produces several distinct wavelengths of
visible light. What are the wavelengths of the hydrogen spectrum, if they form firstorder maxima at angles of 24.2 , 25.7 , 29.1 , and 41.0 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.
d
0.0100 m
 1.00 10 6 m. So, d sin   m   ,
10,000
a  1.00 10 6 m  sin 24.2  4.099 10 7 m  410 nm
b  1.00 10 6 m  sin 25.7  4.337 10 7 m  434 nm
c  1.00 10 6 m  sin 29.1  4.863 10 7 m  486 nm
d  1.00 10 6 m  sin 41.0  6.56110 7 m  656 nm
27.
Solution
(a) What do the four angles in the above problem become if a 5000-line-percentimeter 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.
(a)
0.0100 m
 2.00  10 6 m ; d sin   m  
5000
 4.099  10 7 m 
 
  11.8 ;
 sin 1  1   sin 1 
6
d 
 2.00  10 m 
d
1,a
 4.337  10 7 m 
  12.5 ;
-6
 2.00  10 m 
1,b  sin 1 
 4.863  10 7 m 
  14.1 ;
-6
 2.00  10 m 
1,c  sin 1 
 6.561  10 7 m 
  19.2
-6
 2.00  10 m 
1,d  sin 1 
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
1
2
(b) d sin   m  2    sin 1  
 d 
7
 2 4.099  10 m 
 2,a  sin 1 
  24.2 ;
-6
 2.00  10 m 






 2  4.337  10 7 m 
  25.7 ;
6
 2.00  10 m 
 2,b  sin 1 
 2  4.863  10 7 m 
  29.1 ;
-6
 2.00  10 m 
 2,c  sin 1 
 2  6.561  10 7 m 
  41.0
6
 2.00  10 m 
(c) Decreasing the number of lines per centimeter by a factor of x means that the
angle for the x -order maximum is the same as the original angle for the firstorder maximum.
 2,d  sin 1 
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?
Solution
d sin   . The maximum number of lines corresponds to the smallest d , for the
longest wavelength in the visible spectrum at   90  d sin 90    d   . So,
the smallest d which can accommodate the entire visible spectrum is d  780 nm .
10 2 m
10 2 m
10 2 m
d
N

 1.28205  10 4  12,821  12,800
-7
N
d
7.80  10 m
29.
The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it
produces two first-order maxima at 36.093 and 36.129 when projected on a
10,000 line per centimeter diffraction grating. What are the two wavelengths to an
accuracy of 0.1 nm?
Solution
d
0.0100 m
 1.00 10 6 m, so   d sin  
10,000
a  1.000 10 6 m sin 36.093  5.89110 7 m  589.1 nm
b  1.000 10 6 m sin 36.129  5.896 10 7 m  589.6 nm
30.
What is the spacing between structures in a feather that acts as a reflection grating,
given that they produce a first-order maximum for 525-nm light at a 30.0 angle?
Solution
d sin   m  d 


m 1 5.25  10 7 m

 1.05  10 6 m
sin 
sin 30.0
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
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
 1.25  10 6 m. So,
8000
7
m
1  m 
1  1 6.00  10
d sin   m    sin 
  sin 
  28.7
-6
 d 
 1.25  10 m 
d


32.
An opal such as that shown in Figure 27.17 acts like a reflection grating with rows
separated by about 8 μm . 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?
Solution
Constructive interference of light from rows in crystal is given by d sin   m . The
17.00 10 7 
angle that red light (700 nm) will be seen is sin  
8 106   0.088 , or   5.0 ,
where spacing d  8m . For blue light,   460 nm , and so   3.3 .
33.
At what angle does a diffraction grating produces a second-order maximum for light
having a first-order maximum at 20.0 ?
Solution
d sin 1   ; d sin  2  2 
sin 1 1
 
sin  2 2
 2  sin 1 2 sin 1   sin 1 2sin 20.0  43.2
34.
Show that a diffraction grating cannot produce a second-order maximum for a given
wavelength of light unless the first-order maximum is at an angle less than 30.0 .
Solution
The largest possible second order occurs when sin 2  1 . 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   and d sin 2  2 , so that:

sin  2 2
Now, since we know the maximum value for sin  2 , we can solve for the maximum
value for 1 :
1

1  sin  sin  2 
2

35.
1
so that: 1,max
1
 sin  
2
1
 30.0
If a diffraction grating produces a first-order maximum for the shortest wavelength of
visible light at 30.0 , at what angle will the first-order maximum be for the longest
wavelength of visible light?
OpenStax College Physics
Solution
Instructor Solutions Manual
d sin 1  1  d 
Chapter 27
1
3.80  10 7 m

 7.60  10 7 m ;
sin 1
sin 30.0
 7.60  10 7 m 
  90.0
d sin  2  2  sin 
-7
7
.
60

10
m


1
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
(a)   380 nm, d sin   . For a given  , the smallest d 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

(b) For UV,   380 nm and so sin    1. Therefore, yes it would be useful for UV.
d
(c) For IR,   760 nm and sin   1 , 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?
Solution
(a) First we need to calculate the slit separation:
1 line
1 line
1m
d


 3.333  10 7 m  333.3 nm.
N
30,000 lines/cm 100 cm
Next, using the equation d sin   m , we see that the longest wavelength will be
for sin   1 and m  1, so in that case, d    333.3 nm , 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
light to produce a second order maximum, so max  760 nm (see Example 27.3).
For there to be a second order spectrum, m  2 and sin   1, so
d  2max  2760 nm   1.52  10 6 m
Now, using the technique in step (a), only in reverse:
1 line
1 line
1m
N


 6.58  10 3 lines/cm
-6
d
1.52  10 m 100 cm
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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
d sin θ  m , sin θ 
Therefore d 
39.
Solution
y (0.600 m)

 0.4. For the He–Ne laser   6.33  10 7 m .
x
1.5 m
6.33  10 7 m
m

 1.58  10 6 m.
sin θ
0.4
The analysis shown in the figure below also applies to diffraction gratings with lines
separated by a distance d . 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?
d


x 1.50 m  6.00  10 7 m
10 2 m
 8.00  10 5 m , so that y 

 1.13  10 2 m
-5
125
d
8.00  10 m
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
(a) d sin   m    sin 1

 700 nm 
 sin 1 

D
 400 nm 
  sin 1 1.75
(b) The sine of a real angle cannot be greater than one.
(c) Assuming that you will get a diffraction pattern if   d is unreasonable.
41.
Unreasonable Results (a) What visible wavelength has its fourth-order maximum at
an angle of 25.0 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
(a) For diffraction gratings, we use the equation d sin   m , for m  0,1,2,3,4,...
where the fourth order maximum has m  4 . We first need to determine the slit
separation by using the fact that there are 25,000 lines per centimeter:
1
1m
d

 4.00  10 7 m
25,000 lines /cm 100 cm
So, since   25.0 , 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
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Chapter 27
(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.
27.5 SINGLE SLIT DIFFRACTION
43.
Solution
44.
Solution
(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?

7
m
 m 
1  1 5.50 10

sin
(a) D sin   m    sin 1 


-6
 D 
 1.00 10 m
(b) For m  2, sin   1  No
Solution

Solution

7
m
 m 
1  1 4.10 10
(a) D sin   m    sin 1 
  sin 
  11.8
-6
 D 
 2.00 10 m 
7
m
1  1 7.00 10
(b)   sin 
  20.5
-6
 2.00 10 m 

(a) How wide is a single slit that produces its first minimum for 633-nm light at an
angle of 28.0 ? (b) At what angle will the second minimum be?


m  1 6.33  10 7 m 

 1.35  10 6 m

sin  
sin 28.0

7
m
 m 
1  2 6.33 10
(b)   sin 1 
  sin 
  69.9
-6
 D 
 1.348 10 m 
(a) D sin   m  D 

46.


(a) Calculate the angle at which a 2.00 - μm -wide slit produces its first minimum for
410-nm violet light. (b) Where is the first minimum for 700-nm red light?

45.
  33.4

(a) What is the width of a single slit that produces its first minimum at 60.0 for 600nm light? (b) Find the wavelength of light that has its first minimum at 62.0 .

6.00 10 7 m

 6.928 10 7 m  693 nm
(a) D sin   m  D 
sin 
sin 60.0
7
(b)   D sin   6.928  10 m sin 62.0  6.117  10 7 m  612 nm
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Chapter 27
47.
Find the wavelength of light that has its third minimum at an angle of 48.6 when it
falls on a single slit of width 3.00 μm .
Solution
D sin   m   
48.
Calculate the wavelength of light that produces its first minimum at an angle of 36.9
when falling on a single slit of width 1.00 μm .
Solution
Since D sin   m ,

49.
Solution
Solution
(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?

7
m
 m 
1  2 5.89 10
(a) D sin   m    sin 
  sin 
-6
 D 
 7.50 10 m
1
D sin 90


  9.04


7.50  10 6 m
 12.73  Highest order  12
5.89  10 -7 m
(a) Find the angle of the third diffraction minimum for 633-nm light falling on a slit of
width 20.0 μm . (b) What slit width would place this minimum at 85.0 ? Explicitly
show how you follow the steps in Problem-Solving Strategies for Wave Optics.

7
m
 m 
1  3 6.33 10
(a) D sin   m    sin 1 
  sin 
-5
 D 
 2.00 10 m
(b) D 
51.

D sin  1.00  10 6 m sin 36.9

 6.004  10 7 m  600 nm
m
1
(b) m 
50.

D sin  3.00  10 6 m sin 48.6

 7.50  10 7 m  750 nm
m
3


m 3 6.33 10 7 m

 1.9110 6 m  1.91 μm
sin 
sin 85.0
  5.45


(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.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 27
 5.89110 7 m 

(a) D sin 1  m1  1  sin 1  1   sin 1 
  17.13056 and
-6
D
 2.00 10 m 
 5.896 10 7 m 
  17.14554 so that  2  1  0.0150
-6
 2.00 10 m 
 2  sin 1 
(b) y  x tan   1.00 m tan 0.0150  2.62  10 4 m  0.262 mm
(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
(a) D sin   m  D 
(b) D  50.0 
(c) D  1000
m
 m (for   90)  D  1.00
sin 
53.
(a) If a single slit produces a first minimum at 14.5 , 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
(a) D sin  m  m 
sin 1 1
   2  sin 1 2 sin 1   sin 1 2sin 14.5  30.1
sin  2 2
(b)  3  sin 1 3 sin 1   sin 1 3sin 14.5  48.7
(c) θ4  sin 1 4 sin 14.5  sin 1 1.0015  No
(d) 21  214.5  29, 2  1  30.05  14.5  15.56
Thus, 29  215.56  31.1
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
The problem is asking us to find the ratio of D to d . For the single slit, using the
equation D sin   n , we have n  1 . For the double slit, using the equation
d sin   m (because we have a maximum), we have m  5 . Dividing the single slit
equation by double slit equation, where the angle and wavelength are the same
D n 1
D
    0.200. So, the slit separation is five times the slit width.
gives:
d m 5
d
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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
wide opening acts as a single slit. Using the equation D sin   m , where m  1 , we
can determine the angle for first minimum:
 m 
1  120.0 m  
  sin 1 
 23.58  23.6
  sin 
 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 
1  2 20.0 m  
  sin 1 
 53.13  53.1
  sin 
 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

 0.5667 m
f
600 Hz
We are looking for the first minimum for single slit diffraction:
 m 
1  10.5667 m  
  sin 1 
 45.1
  sin 
 D 
 0.800 m 
v  f   
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 justresolvable 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
(a)  
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?


1.22 1.22 4.00  102 m

 1.63  104 rad
D
300 m
y
(b) tan      y  x , so that y  2.00  10 6 ly 1.63  10  4 rad   3.25 ly
x
OpenStax College Physics
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
Chapter 27


Solution
tan 
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

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
(a)  
y
 y  x tan  x , so that y  3.82  10 8 m 2.80  10 7 rad  107 m
x


1.22 1.22 6.00  10 7 m

 1.46  10 5 rad
D
0.0500 m


1.22 1.22 6.33  10 7 m

 7.72  10 4 rad
-3
D
1.00  10 m

(b)
y
D
Y

y
x
The diameter of the spot will be D  2 y, where y is the width of the spread of
the spot edge due to diffraction.
y  x  1.50 10 4 m 7.723 10 4 rad  11.5845 m





Y  D  2 y  1.00 10 m  211.5845 m  23.2 m
(c) We can ignore D since it is so small.
Y  2 y  2 x  23.82  108 m7.723  10 4 rad   5.90  105 m  590 km
61.
3
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
8
lunar distance of 3.84  10 m ?
OpenStax College Physics
Solution
Instructor Solutions Manual

Chapter 27

1.22 1.22 6.33  10 7 m

 3.04  10 7 rad
D
2.54 m
(b) d  D  2 y  D  2 x  2.54 m   2 3.84  10 8 m 3.04  10 7 rad
(a)  



 2.54 m  233 m  235 m
62.
Solution
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?
 5.50  107 m 
  2.237  104 rad  2.24  104 rad
 1.22
-3
D
 3.00  10 m 
(b) The distance s between two objects, a distance r away, separated by an angle 
(a)   1.22

1.30 m
 5.811  10 3 m  5.81 km
-4
 2.237  10 rad
(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.
is s  r , so: r 
s

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

D
5.00 km
so that
384,000 km
1.22


1.225.50  10 7 m 384,000 km 

 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:
0.065 m0.0050 m  480 m .
 s
sD
θ  1.22   R 

D R
1.22
1.22 5.55  10 7 m


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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
resolve these bodies when they are 4.50 109 km 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
(a)  
D
5.08 m
The telescope should be able to discern a separation at:
y  x  4.50  10 9 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.
Solution

From θ  1.22

(5.55  10 7 m) 1.3 m

, we can calculate the maximum distance at
4.0  10-3 m
R
which lights can be resolved: R  7.71 km .
67.
Solution
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?
θ  1.22
(5.55  10 7 m)
s

 s  7.9  10 5 m  7.9  10-3 cm
-3
3.0  10 m
0.35 m
Resolution in dpi  7.9  10 3 cm 
2.54 cm
 370 dpi . These are specs found on most
1in
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
at a distance of 7.50  108 km from Earth? The wavelength of light averages 600 nm.
(b) What is unreasonable about this result? (c) Which assumptions are unreasonable
or inconsistent?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 27
1m
1.22
(a)   tan 1  y   y 
 1.333  10 12 rad 
11
D
 x  x 7.50  10 m
1.22 1.22 6.00  10 7 m 
D

 5.49  105 m  549 km

1.333  10 -12
(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 
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 

   4 nt (one phase change),
2 2n
  41.331.00  10 7 m   5.32  10 7 m  532 nm (green)
n

(one phase change), so   4 tn
2 2n
  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 2n
4n
7
4.70  10 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


incident perpendicularly, constructive interference first occurs when 2t  n 
.
2 2n
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
41.33
4n
74.
A film of soapy water ( n  1.33 ) 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?
OpenStax College Physics
Instructor Solutions Manual
Chapter 27

Solution
2t  n 
75.
What are the three smallest non-zero thicknesses of soapy water ( n  1.33 ) 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 ProblemSolving Strategies for Wave Optics.
Solution
(two phase changes), so
n
  2tn  22.33  10 7 m1.33  6.20  10 7 m  620 nm (orange)
2t  n , 2n , 3 n  t 
  3
, , . Therefore,
2n n 2n

520 nm
 520 nm
3 3520 nm 

 195 nm ; 
 391 nm ; and

 586 nm
21.33
2n 21.33
n
1.33
2n
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
2t 
3 n 3

( two phase changes), so
2
2n
3 3 7.00  10 7 m
t

 3.80  10 7 m  380 nm
41.38
4n


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 onefourth 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.
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.
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
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
wavelength in air, t  n 
 3.39 10 8 m  33.9 nm
8
81.40
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
(a)
Two adjacent dark bands will have thickness differing by one wavelength, i.e.,
 1.00  104 m 
hair diameter
  0.076394.
  d 2  d1 , and tan  
or   tan -1 
slide length
 0.075 m 
So, since x tan   d2  d1  , we see that

5.89  107 m
 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.
x
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?
OpenStax College Physics
Instructor Solutions Manual
Solution

D
d2
d1
Chapter 27
x
7.5 cm
D
7.50 cm
Two dark bands will have thickness differing by one wavelength, i.e., d2  d1  
d  d1 
D
tan   2
 
x
x 7.50 cm
7.50 cm   7.50  10 2 m 5.89  10 7 m  4.42  10 5 m
D
x
10 -3 m
tan  



81.
Repeat Exercise 27.70, but take the light to be incident at a 45 angle.
Solution

2t

4nt
(one phase change)
 n 
 
sin 45.0 2 2n
sin 45.0
41.33 1.00  107 m  7.52  107 m  752 nm (red)
So,  
sin 45

82.
Solution

Repeat Exercise 27.71, but take the light to be incident at a 45 angle.

2t

, so
 n 
sin 45 2 2n

4tn
41.20  10 7 m 1.40


 9.50  10 7 m  950 nm (infrared)
sin 45.0
sin 45.0
3
2t
3
 n 
For three phase changes:
, so
sin 45
2
2n
4 1.20  10 7 m 1.40  3.17  10 7 m  317 nm ultraviole t 
4tn


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.
For one phase change:

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?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 27
(a) Assuming n 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?
Solution
 I  12 
 1  12 

1
I  I 0 cos   cos     cos     45.0
 I 0  
 2  


1
2
85.
The angle between the axes of two polarizing filters is 45.0 . By how much does the
second filter reduce the intensity of the light coming through the first?
Solution
I  I 0 cos 2   I 0 cos 2 45 
1
I
I 0 therefore,
 0.500
2
I0
86.
2
If you have completely polarized light of intensity 150 W / m , what will its intensity
be after passing through a polarizing filter with its axis at an 89.0 angle to the
light’s polarization direction?
Solution
Using the equation I  I 0 cos 2  :


I  I 0 cos 2   150 W m 2 cos 2 89.0  4.57  10 2 W m 2  45.7 m W m 2
87.
What angle would the axis of a polarizing filter need to make with the direction of
polarized light of intensity 1.00 kW/m 2 to reduce the intensity to 10.0 W/m 2 ?
Solution
  12 
 10.0 W m 2  12 
I
   84.3
I  I 0 cos 2   cos 1     cos 1 
 I 0  
 1.00  103 W m 2  




88.
At the end of Example 27.8, it was stated that the intensity of polarized light is
reduced to 90.0% of its original value by passing through a polarizing filter with its
axis at an angle of 18.4 to the direction of polarization. Verify this statement.
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
Solution
I  I 0 cos 2  
89.
Show that if you have three polarizing filters, with the second at an angle of 45 to
the first and the third at an angle of 90.0 to the first, the intensity of light passed by
the first will be reduced to 25.0% 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.)
I
 cos 2   cos 2 18.4  0.9004  90.04 %  90.0 %
I0
Solution
I0
I1
I2


I 1  I 0 cos 2 1 , I 2  I 1 cos 2  2  I 0 cos 2 1 cos 2  2
I 2  I 0 cos 45 cos 45  0.25I 0
2
90.
2
Prove that, if I is the intensity of light transmitted by two polarizing filters with axes
at an angle  and I  is the intensity when the axes are at an angle 90.0   , then
I  I   I 0 , the original intensity. (Hint: Use the trigonometric identities
cos(90.0   )  sin  and cos 2   sin 2   1 .)
Solution
I  I 0 cos 2  , I   I 0 cos 2 90   


I  I   I 0 cos 2   I 0 cos90     I 0 cos 2   I 0 sin    I 0 cos 2   sin 2   I 0
2
2
91.
At what angle will light reflected from diamond be completely polarized?
Solution
tan  b 
92.
What is Brewster’s angle for light traveling in water that is reflected from crown
glass?
Solution
Using the equation tan  b 
n2 2.42

 2.42 . Therefore,  b  tan 1 2.42  67.6
n1
1
n2
, where n2 is for crown glass and n1 is for water (see
n1
n 
 1.52 
Table 25.1). Brewster’s angle is  b  tan 1  2   tan 1 
  48.75  48.8
 1.333 
 n1 
93.
A scuba diver sees light reflected from the water’s surface. At what angle will this
light be completely polarized?
OpenStax College Physics
Solution
94.
Solution
Instructor Solutions Manual
Chapter 27
 n2 
 1.333 
  tan 1 
  53.1
 1.00 
 n1 
 b  tan 1 
At what angle is light inside crown glass completely polarized when reflected from
water, as in a fish tank?
 n2 
 1.333 
  tan 1 
  41.2
n
1
.
52


 1
 b  tan 1 
95.
Light reflected at 55.6 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
 n2  n1 tan  b  1.00 tan 55.6  1.46
n1
The substance is likely fused quartz.
96.
(a) Light reflected at 62.5 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
(a) tan  b 
97.
If  b is Brewster’s angle for light reflected from the top of an interface between two
substances, and  b is Brewster’s angle for light reflected from below, prove that
 b   b  90.0 .
Solution
n2
n
, tan  b  1  tan  b  cot  b . The equation can be solved with the
n1
n2
trigonometric identity: cot   tan 90      b  90   b   b   b  90
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?
Solution
I  B 2  I1  aB1 where a is some constant.
tan  b 
n2
 n2  n1 tan  b  1.00 tan 62.5  1.921  1.92
n1
The gem is not a diamond (it is zircon).
 1.921 
(b)  b  tan 1 
  55.2
 1.333 
tan  b 
2
2
B 
I 2  aB  0.500 I 1  0.500aB   2   0.500, so that B2  0.500 B1  0.707 B1
 B1 
2
2
2
1
OpenStax College Physics
Instructor Solutions Manual
Chapter 27
99.
Integrated Concepts Suppose you put on two pairs of Polaroid sunglasses with their
axes at an angle of 15.0 . 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
I  I 0 cos 2  
100.
Integrated Concepts (a) On a day when the intensity of sunlight is 1.00 kW / m2 , 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
of 20.0 . Assuming the sunlight is unpolarized and the polarizers are 100% efficient,
what is the initial rate of heating of the water in C / s , assuming it is 80.0%
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:
I  I 0 cos 2   1.00 k W m 2 cos 2 20.0  883 W m 2
This intensity is focused on the lens, producing a total power hitting the water of:
2
P  IA  883 W m 2  0.100 m   27.7 W
Since the water absorbs only 80.0% of the heat, the power transferred to the
water is: Pabsorbed  0.800P  0.80027.7 W   22.2 W
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 
mw cw  mAlcAl
For the rate of heating, replace Q with the power transferred:
T
22.2 W

 1.69  10 2 C/s
0.250 kg 4186 J Kg  C  0.300 kg 900 J Kg  C
t
(b) Yes, the polarizing filters get hot because they absorb some of the lost energy
from the sunlight.
t
I
1
 cos 2 15.0  0.933  2 
 1.072  t 2  1.072t1
I0
t1 0.933
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