Chapter 34 The Wave Nature of Light; Interference
65. In order for the two reflected halves of the beam to be 180 out of phase with each other, the
minimum path difference (2t) should be 1 in the plastic. Notice that there is no net phase
2
difference between the two halves of the beam due to reflection, because both halves reflect from the
same material.
2t 1 t 780 nm 126 nm
2 n 4n
4 1.55
66. We determine n for each angle using a spreadsheet. The results are shown below.
N 25 50 75 100 125 150
degree 5.5 6.9 8.6 10.0 11.3 12.5
n 1.75 2.19 2.10 2.07 2.02 1.98
The average value is navg 2.02 . The spreadsheet used for this problem can be found on the Media
Manager, with filename “PSE4_ISM_CH34.XLS,” on tab “Problem 34.66.”
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401
CHAPTER 35: Diffraction and Polarization
Responses to Questions
1. Radio waves have a much longer wavelength than visible light and will diffract around normal-sized
objects (like hills). The wavelengths of visible light are very small and will not diffract around
normal-sized objects.
2. You see a pattern of dark and bright lines parallel to your fingertips in the narrow opening between
your fingers.
3. Light from all points of an extended source produces diffraction patterns, and these many different
diffraction patterns overlap and wash out each other so that no distinct pattern can be easily seen.
When using white light, the diffraction patterns of the different wavelengths will overlap because the
locations of the fringes depend on wavelength. Monochromatic light will produce a more distinct
diffraction pattern.
4. (a) If the slit width is increased, the diffraction pattern will become more compact.
(b) If the wavelength of the light is increased, the diffraction pattern will spread out.
5. (a) A slit width of 50 nm would produce a central maximum so spread out that it would cover the
entire width of the screen. No minimum (and therefore no diffraction pattern) will be seen. The
different wavelengths will all overlap, so the light on the screen will be white. It will also be
dim, compared to the source, because it is spread out.
(b) For the 50,000 nm slit, the central maximum will be very narrow, about a degree in width for
the blue end of the spectrum and about a degree and a half for the red. The diffraction pattern
will not be distinct, because most of the intensity will be in the small central maximum and the
fringes for the different wavelengths of white light will not coincide.
6. (a) If the apparatus is immersed in water, the wavelength of the light will decrease and
n
the diffraction pattern will become more compact.
(b) If the apparatus is placed in a vacuum, the wavelength of the light will increase slightly, and the
diffraction pattern will spread out very slightly.
7. The intensity pattern is actually a function of the form sin x 2 (see equations 35-7 and 35-8). The
x
maxima of this function do not coincide exactly with the maxima of sin2 x. You can think of the
intensity pattern as the combination of a sin2 x function and a 1/x2 function, which forces the
intensity function to zero and shifts the maxima slightly.
8. Similarities: Both have a regular pattern of light and dark fringes. The angular separation of the
fringes is proportional to the wavelength of the light, and inversely proportional to the slit size or slit
separation. Differences: The single slit diffraction maxima decrease in brightness from the center.
Maxima for the double slit interference pattern would be equally bright (ignoring single slit effects)
and are equally spaced.
9. No. D represents the slit width and d the distance between the centers of the slits. It is possible for
the distance between the slit centers to be greater than the width of the slits; it is not possible for the
distance between the slit centers to be less than the width of the slits.
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402
Chapter 35 Diffraction and Polarization
10. (a) Increasing the wavelength, λ, will spread out the diffraction pattern, since the locations of the
minima are given by sin θ = mλ/D. The interference pattern will also spread out; the
interference maxima are given by sin θ = mλ/d. The number of interference fringes in the
central diffraction maximum will not change.
(b) Increasing the slit separation, d, will decrease the spacing between the interference fringes
without changing the diffraction, so more interference maxima will fit in the central maximum
of the diffraction envelope.
(c) Increasing the slit width, D, will decrease the angular width of the diffraction central maximum
without changing the interference fringes, so fewer bright fringes will fit in the central
maximum.
11. Yes. As stated in Section 35-5, “It is not possible to resolve detail of objects smaller than the
wavelength of the radiation being used.”
12. Yes. Diffraction effects will occur for both real and virtual images.
13. A large mirror has better resolution and gathers more light than a small mirror.
14. No. The resolving power of a lens is on the order of the wavelength of the light being used, so it is
not possible to resolve details smaller than the wavelength of the light. Atoms have diameters of
about 10-8 cm and the wavelength of visible light is on the order of 10-5 cm.
15. Violet light would give the best resolution in a microscope, because the wavelengths are shortest.
16. Yes. (See the introduction to Section 35-7.) The analysis for a diffraction grating of many slits is
essentially the same as for Young’s double slit interference. However, the bright maxima of a
multiple-slit grating are much sharper and narrower than those in a double-slit pattern.
17. The answer depends on the slit spacing of the grating being used. If the spacing is small enough,
only the first order will appear so there will not be any overlap. For wider slit spacing there can be
overlap. If there is overlap, it will be the higher orders of the shorter wavelength light overlapping
with lower orders of the longer wavelength light. See, for instance, Example 35-9, which shows the
overlap of the third order blue light with the second order red light.
18. The bright lines will coincide, but those for the grating will be much narrower with wider dark
spaces in between. The grating will produce a much sharper pattern.
19. (a) Violet light will be at the top of the rainbow created by the diffraction grating. Principal
maxima for a diffraction grating are at positions given by sin m . Violet light has a shorter
d
wavelength than red light and so will appear at a smaller angle away from the direction of the
horizontal incident beam.
(b) Red light will appear at the top of the rainbow created by the prism. The index of refraction for
violet light in a given medium is slightly greater than for red light in the same medium, and so
the violet light will bend more and will appear farther from the direction of the horizontal
incident beam.
20. The tiny peaks are produced when light from some but not all of the slits interferes constructively.
The peaks are tiny because light from only some of the slits interferes constructively.
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21. Polarization demonstrates the transverse wave nature of light, and cannot be explained if light is
considered only as particles.
22. Take the sunglasses outside and look up at the sky through them. Rotate the sunglasses (about an
axis perpendicular to the lens) through at least 180. If the sky seems to lighten and darken as you
rotate the sunglasses, then they are polarizing. You could also look at a liquid crystal display or
reflections from the floor while rotating the glasses, or put one pair of glasses on top of the other and
rotate them. If what you see through the glasses changes as you rotate them, then the glasses are
polarizing.
23. Black. If there were no atmosphere, there would be no scattering of the sunlight coming to Earth.
Solutions to Problems
1. We use Eq. 35-1 to calculate the angular distance from the middle of the central peak to the first
minimum. The width of the central peak is twice this angular distance.
sin1 1 sin 1 sin 1 680 109 m 1.067
D D 0.0365 103 m
21 21.067 2.13
2. The angle from the central maximum to the first dark fringe is equal to half the width of the central
maximum. Using this angle and Eq. 35-1, we calculate the wavelength used.
1 1 1 32 16
2 2
sin1
D D sin1 2.60 103 mm sin 16 7.17 104 mm 717 nm
3. The angle to the first maximum is about halfway between the angles to the first and second minima.
We use Eq. 35-2 to calculate the angular distance to the first and second minima. Then we average
these to values to determine the approximate location of the first maximum. Finally, using
trigonometry, we set the linear distance equal to the distance to the screen multiplied by the tangent
of the angle.
D sinm m m sin 1 m
D
1 sin 1 1 580 109 m 8.678 2 sin 1 2 580 109 m 17.774
3.8 106 m
3.8 106 m
= 1 2 8.678 17.774 13.23
2 2
y l tan1 10.0 m tan 13.23 2.35 m
4. (a) We use Eq. 35-2, using m=1,2,3,… to calculate the possible diffraction minima, when the
wavelength is 0.50 cm.
D sinm m m sin 1 m
D
1 sin 1 1 0.50 cm 18.2 2 sin1 2 0.50 cm 38.7
1.6 cm 1.6 cm
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404
Chapter 35 Diffraction and Polarization
3 sin 1 3 0.50 cm 69.6 4 sin 1 4 0.50 cm no solution
1.6 cm 1.6 cm
There are three diffraction minima: 18, 39, and 70.
(b) We repeat the process from part (a) using a wavelength of 1.0 cm.
1 sin 1 11.0 cm 38.7 2 sin 1 2 1.0 cm =no real solution
1.6 cm 1.6 cm
The only diffraction minima is at 39
(c) We repeat the process from part (a) using a wavelength of 3.0 cm.
1 sin 1 1 3.0 cm no real solution
1.6 cm
There are no diffraction minima.
5. The path-length difference between the top and bottom of the
slit for the incident wave is D sin i. The path-length Slit, width aD
difference between the top and bottom of the slit for the
diffracted wave is D sin . When the net path-length difference i
is equal to a multiple of the wavelength, there will be an even
number of segments of the wave having a path-length difference i
of /2. We set the path-length difference equal to m (an integer)
times the wavelength and solve for the angle of the diffraction
minimum.
D sini D sin m
sin sini m , m 1, 2, ...
D
From this equation we see that when = 23.0°, the minima will be symmetrically distributed around
a central maximum at 23.0
6. The angle from the central maximum to the first bright maximum is half the angle between the first
bright maxima on either side of the central maximum. The angle to the first maximum is about
halfway between the angles to the first and second minima. We use Eq. 35-2, setting m 3 2 , to
calculate the slit width, D.
1 1 1 35 17.5
2 2
D sinm m D m 3 2633 nm 3157.6 nm 3.2 m
sin1
sin17.5
7. We use the distance to the screen and half the width of the diffraction maximum to calculate the
angular distance to the first minimum. Then using this angle and Eq. 35-1 we calculate the slit
width. Then using the slit width and the new wavelength we calculate the angle to the first minimum
and the width of the diffraction maximum.
tan1 1 y1 1 tan1 1 y1 tan1 1 0.06 m 0.781
2 2 2
l l 2.20 m
sin1 1 D 1 580 nm 42,537 nm
D sin1 sin 0.781
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
sin2 2 2 sin1 2 sin 1 460 nm 0.620
D D 42,537 nm
y2 2l tan2 22.20 m tan 0.620 0.0476 m 4.8cm
8. (a) There will be no diffraction minima if the angle for the first minimum is greater than 90. We
set the angle in Eq. 35-1 equal to 90 and solve for the slit width.
sin D
D sin 90
(b) For no visible light to exhibit a diffraction minimum, the slit width must be equal to the shortest
visible wavelength.
D min 400 nm .
9. We set the angle to the first minimum equal to half of the separation angle between the dark bands.
We insert this angle into Eq. 35-1 to solve for the slit width.
1 1 55.0 27.5
2 2
sin D 440 nm 953nm
D sin sin 27.5
10. We find the angle to the first minimum using Eq. 35-1. The distance on the screen from the central
maximum is found using the distance to the screen and the tangent of the angle. The width of the
central maximum is twice the distance from the central maximum to the first minimum.
sin1 1 sin 1 sin 1 450 109 m 0.02578
D D 1.0 103 m
y1 l tan1 5.0 m tan 0.02578 0.00225 m
y 2 y1 20.00225 m 0.0045 m 0.45 cm
11. (a) For vertical diffraction we use the height of the slit (1.5 m) as the slit width in Eq. 35-1 to
calculate the angle between the central maximum to the first minimum. The angular separation
of the first minima is equal to twice this angle.
sin1 1 sin 1 sin 1 780 109 m 31.3
D D 1.5 106 m
21 231.3 63
(b) To find the horizontal diffraction we use the width of the slit (3.0 m) in Eq. 35-1.
sin1 1 sin 1 sin 1 780 109 m 15.07
D D 3.0 106 m
21 215.07 30
12. (a) If we consider the slit made up of N wavelets each of amplitude E0, the total amplitude at the
central maximum, where they are all in phase, is NE0. Doubling the size of the slit doubles the
number of wavelets and thus the total amplitude of the electric field. Because the intensity is
proportional to the square of the electric field amplitude, the intensity at the central maximum is
increased by a factor of 4.
I E 2 2E0 2 4E02 4I0
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406
Chapter 35 Diffraction and Polarization
(b) From Eq. 35-1 we see that, for small angles, the width of the central maximum is inversely
proportional to the slit width. Therefore doubling the slit width will cut the area of the central
peak in half. Since the intensity is spread over only half the area, where the intensity is four
times the initial intensity, the average intensity (or energy) over the central maximum has
doubled. This is true for all fringes, so when the slit width is doubled, allowing twice the
energy to pass through the slit, the average energy within each slit will also double, in accord
with the conservation of energy.
13. We use Eq. 35-8 to calculate the intensity, where the angle is found from the displacement from
the central maximum (15 cm) and the distance to the screen.
tan y tan –1 15 cm 31.0
l 25 cm
2 D sin
2 1.0 106 m sin 31.0 4.31rad
750 109 m
I sin 2 2 sin 4.31rad 2 2 0.1498 0.15
I0 2
4.31rad 2
So the light intensity at 15 cm is about 15% of the maximum intensity.
14. (a) The secondary maxima do not occur precisely where sin / 2 is a maximum, that is at
/ 2 (m 1 ) where m 1, 2,3,..., because the diffraction intensity (Eq. 35-7) is the ratio of
2
the sine function and / 2 . Near the maximum of the sine function, the denominator of the
intensity function causes the intensity to decrease more rapidly than the sine function causes it
to increase. This results in the intensity reaching a maximum slightly before the sine function
reaches its maximum.
(b) We set the derivative of Eq. 35-7 with respect to equal to zero to determine the intensity
extrema.
0 dI d Io sin 2 2 2Io sin 2 cos 2 sin 2
d d
2 2 2 2
When the first term in brackets is zero, the intensity is a minimum, so the intensity is a
maximum when the second term in brackets is zero.
0 cos 2 sin 2 2 tan 2
2 2
(c) The first and secondary y 10 y = beta/2
maxima are found where 8 y = tan(beta/2)
these two curves intersect, 6
4 2 4 6 8 10 12 14 16
or 1 8.987 and 2
0 beta
2 15.451. We
0
calculate the percent
difference between these
and the maxima of the sine
curve, 1 3 and
2 5 .
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
1 1 8.987 3 0.0464 4.64%
1 3
1
15.451 5 0.0164 1.64%
5
2
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH35.XLS,” on tab “Problem 35.14.”
15. If the central diffraction peak contains nine fringes, there will be four fringes on each side of the
central peak. Thus the fifth maximum of the double slit must coincide with the first minimum of the
diffraction pattern. We use Eq. 34-2a with m = 5 to find the angle of the fifth interference maximum
and set that angle equal to the first diffraction minimum, given by Eq. 35-1, to solve for the ratio of
the slit separation to slit width.
d sin m sin 5 ; sin 5 d 5D
d D d
16. (a) If the central diffraction peak is to contain seventeen fringes, there will be eight fringes on each
side of the central peak. Thus, the ninth minimum of the double slit must coincide with the first
minimum of the diffraction pattern. We use Eq. 34-2b with m = 8 to find the angle of the ninth
interference minimum and set that angle equal to the first diffraction minimum, given by Eq.
35-1, to solve for the ratio of the slit separation to slit width.
d sin m 1 sin 8 1 8.5
2 2 d
d
sin 8.5 d 8.5D
D d
Therefore, for the first diffraction minimum to be at the ninth interference minimum, the
separation of slits should be 8.5 times the slit width.
(b) If the first diffraction minimum is to occur at the ninth interference maximum, we use Eq. 34-2a
with m = 9 to find the angle of the ninth interference maximum and set that angle equal to the
first diffraction minimum, given by Eq. 35-1, to solve for the ratio of the slit separation to slit
width.
d sin m sin 9 9 ; sin 9 d 9D
d d D d
Therefore, for the first diffraction minimum to be at the ninth interference maximum, the
separation of slits should be 9 times the slit width.
17. Given light with 605 nm passing through double slits with separation d 0.120 mm , we use Eq.
34-2a to find the highest integer m value for the interference fringe that occurs before the angle
90o .
d sin m m 0.120 10–3m sin 90 198
605 10–9m
So, including the m 0 fringe, and the symmetric pattern of interference fringes on each side of
0 , there are potentially a total of 198 198 1 397 fringes. However, since slits have width
a 0.040 mm, the potential interference fringes that coincide with the slits’ diffraction minima will
be absent. Let the diffraction minima be indexed by m 1, 2, 3, etc. We then set the diffraction
angles in Eq. 34-2a and Eq. 35-2 equal to solve for the m values of the absent fringes.
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408
Chapter 35 Diffraction and Polarization
sin m m m d 0.120 mm 3 m 3m
D d m D 0.040 mm
Using m 1, 2, 3, etc., the 66 interference fringes on each side of 0 with m 3, 6, 9, ...,198 will
be absent. Thus the number of fringes on the screen is 397 – 266 265 .
18. In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, there is
the m 0 fringe, along with fringes up to m 6 on each side of 0 . Then, at angle , the m 7
interference fringe coincides with the first diffraction minima. We set this angle in Eq. 34-2a and
35-2 equal to solve for the relationship between the slit width and separation.
sin1 m m d m 7 7 d 7D
D d D m 1
Now, we use these equations again to find the m value at the second diffraction minimum, m 0.
sin2 m m m m d 2 7D 14
D d D D
Thus, the six fringes corresponding to m 8 to m 13 will occur within the first and second
diffraction minima.
19. (a) The angle to each of the maxima of the double slit are given by Eq. 34-2a. The distance of a
fringe on the screen from the center of the pattern is equal to the distance between the slit and
screen multiplied by the tangent of the angle. For small angles, we can set the tangent equal to
the sine of the angle. The slit spacing is found by subtracting the distance between two adjacent
fringes.
sinm m ym l tanm l sinm l m
d d
y
ym1 ym l m 1 l m l 1.0 m 580 109 m 0.019 m 1.9cm
d d
d 0.030 103m
(b) We use Eq. 35-1 to determine the angle between the center and the first minimum. Then by
multiplying the distance to the screen by the tangent of the angle we find the distance from the
center to the first minima. The distance between the two first order diffraction minima is twice
the distance from the center to one of the minima.
sin1 1 sin1 sin1 580 109 m 3.325
D D 0.010 103m
y1 l tan1 1.0 m tan 3.325 0.0581 m
y 2 y1 20.0581m 0.116 m 12cm
20. We set d D in Eqs. 34-4 and 35-6 to show . Replacing with in Eq. 35-9, and using the
double angle formula we show that Eq. 35-9 reduces to Eq. 35-7, with 2 . Finally using Eq.
35-6 again, we show that 2 implies that the new slit width D is simply double the initial slit
width.
2 d sin 2 D sin
I I0 sin 2 2 cos2 2 I0 sin2 2 cos2 2 I0 1 sin2 2 2
2 22 4 22
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I0 sin2 I0 sin2 2 , where 2 .
2 22
2 Dsin 2 2 D sin D 2D
21. Using Eq. 34-2a we determine the angle at which the third-order interference maximum occurs.
Then we use Eq. 35-9 to determine the ratio of the intensity of the third-order maximum, where is
given by Eq. 35-6 and is given by Eq. 34-4.
d sin m sin 1 m sin 1 3 4.301
d 40.0
2 D sin 40.0 5 sin 4.301 1.885 rad
2 2
2 d sin 40.0 sin 4.301 9.424 rad
2 2
sin 2 2 cos 2 sin 1.885 rad 2 cos 9.424 2
2
I Io 2 Io 1.885 rad 0.255Io
22. We use Eq. 34-2a to determine the order of the double slit maximum that corresponds to the same
angle as the first order single slit minimum, from Eq. 35-1. Since this double slit maximum is
darkened, inside the central diffraction peak, there will be the zeroth order fringe and on either side
of the central peak a number of maximum equal to one less than the double slit order. Therefore,
there will be 2(m – 1)+1, or 2m – 1 fringes.
d sin m m d sin d d ; N 2m 1 2 d 1
D D D
(a) We first set the slit separation equal to twice the slit width, d = 2.00 D.
N 2 2.00D 1 3
D
(b) Next we set d = 12.0 D.
N 212.D00D 1 23
(c) For the previous two parts, the ratio of slits had been an integer value. This corresponded to the
single slit minimum overlapping the double slit maximum. Now that d = 4.50 D, the single slit
minimum overlaps a double slit minimum. Therefore, the last order maximum, m = 4, is not
darkened and N = 2m + 1.
N 2m 1 2(4) 1 9
(d) In this case the ratio of the slit separation to slit width is not an integer, nor a half-integer value.
The first order single-slit minimum falls between the seventh order maximum and the seventh
order minimum. Therefore, the seventh order maximum will partially be seen as a fringe.
N 2m 1 2(7) 1 15
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410
Chapter 35 Diffraction and Polarization
23. (a) If D , the central maximum of the diffraction pattern will be very E30
wide. Thus we need consider only the interference between slits. We
E 0
construct a phasor diagram for the interference, with 2 d sin as E20
the phase difference between adjacent slits. The magnitude of the E10
electric fields of the slits will have the same magnitude, E10 E20
E30 E0. From the symmetry of the phasor diagram we see that
. Adding the three electric field vectors yields the net electric
field.
E 0 E10 cos E20 E30 cos E0 1 2 cos
The central peak intensity occurs when = 0. We set the intensity proportional to the square of
the electric field and calculate the ratio of the intensities.
I E20 E02 1 2cos 2 1 2cos 2
I0 E020 E02 1 2cos 02
9
(b) We find the locations of the maxima and minima by setting the first derivative of the intensity
equal to zero.
dI d I0 1 2 cos 2 2I0 1 2 cos 2 sin 0
d d 9 9
This equation is satisfied when either of the terms in parentheses is equal to zero. When
1 2cos 0, the intensity equals zero and is a minimum.
1 2 cos 0 cos1 1 2 , 4 , 8 , 10 ,...
2 3 3 3 3
Maxima occur for sin 0, which also says cos 1.
sin 0 sin1 0 0, , 2 ,3 ,...
When cos 1, the intensity is a principal maximum. When cos 1, the intensity is a
secondary maximum.
I (0) I0 1 2cos 2 I0 1 2cos 02 I0
9 9
I ( ) I0 1 2cos 2 I0 1 212 I0
9
9 9
I (2 ) I0 1 2cos 2 2 I0 1 22 I0
9 9
Thus we see that, since cos alternates between +1 and –1, there is only a single secondary
maximum between each principal maximum.
24. The angular resolution is given by Eq. 35-10.
1.22 1.22 560 109 m 2.69 107 rad 180 3600" 0.055"
D 254 102 m rad 1
25. The angular resolution is given by Eq. 35-10. The distance between the stars is the angular
resolution times the distance to the stars from the Earth.
9.46 1015 m
r 16 ly 1ly 550 109 m
D D
1.22 ; l r 1.22 1.22 1.5 1011 m
0.66 m
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26. We find the angle subtended by the planet by dividing the orbital radius by the distance of the star
to the earth. Then using Eq. 35-10 we calculate the minimum diameter aperture needed to resolve
this angle.
r 1.22
d D
D
1.22d 1.22 550 109 m 4 ly 9.4611015 m/ly 0.17 m 20cm
r 1AU 1.496 1011m/AU
27. We find the angular half-width of the flashlight beam using Eq. 35-10 with D 5 cm and
550 nm. We set the diameter of the beam equal to twice the radius, where the radius of the
beam is equal to the angular half-width multiplied by the distance traveled, 3.84 108 m.
1.22 1.22 550 10–9 m 1.310–5 rad
D 0.050 m
d 2r 2 3.84 108 m 1.310–5 rad 1.0 104 m
28. To find the focal length of the eyepiece we use Eq. 33-7, where the objective focal length is 2.00 m,
’ is the ratio of the minimum resolved distance and 25 cm, and is the ratio of the object on the
moon and the distance to the moon. We ignore the inversion of the image.
fo fe fo fo do l 2.0 m 7.5 km 384,000 km 0.098 m 9.8 cm
fe d N 0.10 mm 250 mm
We use Eq. 35-10 to determine the resolution limit.
1.22 1.22 560 109 m 6.2 106 rad
D 0.11 m
This corresponds to a minimum resolution distance, r 384,000 km 6.2 106 rad 2.4 km ,
which is smaller than the 7.5 km object we wish to observe.
29. We set the resolving power as the focal length of the lens multiplied by the angular resolution, as in
Eq. 35-11. The resolution is the inverse of the resolving power.
1 1.22 1
D
RP( f
f D 25 mm 730 lines/mm
1.22 1.22 560 106 mm
/ 2) f 50.0 mm
1
RP( f /16)
3.0 mm 50.0 mm 88 lines/mm
1.22 560 106 mm
30. We use Eq. 35-13 to calculate the angle for the second order maximum.
d sin
m sin 1 m 2 480 109 m 4.1
d sin1 1.35 105 m
31. We use Eq. 35-13 to calculate the wavelengths from the given angles. The slit separation, d, is the
inverse of the number of lines per cm, N. We assume that 12,000 is good to 3 significant figures.
d sin m sin
Nm
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412
Chapter 35 Diffraction and Polarization
1 sin 28.8 4.01105 cm 401 nm 2 sin 36.7 4.98 105 cm 498 nm
12,000 /cm 12,000 /cm
3 sin 38.6 5.201105 cm 520 nm 4 sin 47.9 6.18 105 cm 618 nm
12,000 /cm 12,000 /cm
32. We use Eq. 35-13 to find the wavelength, where the number of lines, N, is the inverse of the slit
separation, or d=1/N.
d sin m sin sin 26.0 4.17 105cm 420 nm
mN
33500 /cm
33. Because the angle increases with wavelength, to have a complete order we use the largest
wavelength. We set the maximum angle is 90° to determine the largest integer m in Eq. 35-13.
m m sin sin 90 2.1
d sin N
700 109 m 6800 /cm100 cm/m
Thus, two full spectral orders can be seen on each side of the central maximum, and a portion of the
third order.
34. We find the slit separation from Eq. 35-13. Then set the number of lines per centimeter equal to the
inverse of the slit separation, N=1/d.
1 sin sin15.0
d m 650 107 cm
d sin m N 3 1300lines cm
35. Since the same diffraction grating is being used for both wavelengths of light, the slit separation will
be the same. We solve Eq. 35-13 for the slit separation for both wavelengths and set the two
equations equal. The resulting equation is then solved for the unknown wavelength.
d sin m d m11 m22 2 m1 sin2 1 2 sin 20.6 632.8 nm 556 nm
sin1 sin2 m2 sin1 1 sin 53.2
36. We find the first order angles for the maximum and minimum wavelengths using Eq. 35-13, where
the slit separation distance is the inverse of the number of lines per centimeter. Then we set the
distance from the central maximum of the maximum and minimum wavelength equal to the distance
to the screen multiplied by the tangent of the first order angle. The width of the spectrum is the
difference in these distances.
d sin m sin1 m sin1 m N
d
1 sin1 410 107 cm 7800 lines/cm 18.65
2 sin1 750 107 cm 7800 lines/cm 35.80
y y2 y1 l tan2 tan1 2.80 mtan 35.80 tan18.65 1.1m
37. We find the second order angles for the maximum and minimum wavelengths using Eq. 35-13,
where the slit separation distance is the inverse of the number of lines per centimeter. Subtracting
these two angles gives the angular width.
d sin m sin 1 m sin1 m N
d
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1 sin1 2 4.5107 m 6.0 105 /m 32.7
2 sin1 2 7.0 107 m 6.0 105 /m 57.1
2 1 57.1 32.7 24
38. The m 1 brightness maximum for the wavelength of 1200 nm occurs at angle . At this same
angle m 2 , m 3 , etc. brightness maximum will form for other wavelengths. To find these
wavelengths, we use Eq. 35-13, where the right hand side of the equation remains constant, and
solve for the wavelengths of higher order.
d sin m11 mm m m11 1
m m
2 1200 nm 600 nm 3 1200 nm 400 nm 4 1200 nm 300 nm
2 3 4
Higher order maxima will have shorter wavelengths. Therefore in the range 360 nm to 2000 nm, the
only wavelengths that have a maxima at the angle are 600 nm and 400 nm besides the 1200 nm.
39. Because the angle increases with wavelength, we compare the maximum angle for the second order
with the minimum angle for the third order, using Eq. 35-13, by calculating the ratio of the sines for
each angle. Since this ratio is greater than one, the maximum angle for the second order is larger
than the minimum angle for the first order and the spectra overlap.
d sin m sin m ; sin2 22 / d 22 2 700 nm 1.2
d sin3 33 / d 33 3400 nm
To determine which wavelengths overlap, we set this ratio of sines equal to one and solve for the
second order wavelength that overlaps with the shortest wavelength of the third order. We then
repeat this process to find the wavelength of the third order that overlaps with the longest wavelength
of the second order.
sin2 1 22 / d 22 3 2 2,max 2 700 nm 467 nm
sin3 33 / d 33 3 3
2 3 3,min 3 400 nm 600 nm
2 2
Therefore, the wavelengths 600 nm – 700 nm of the second order overlap with the wavelengths 400
nm – 467 nm of the third order. Note that these wavelengths are independent of the slit spacing.
40. We set the diffraction angles as one half the difference between the angles on opposite sides of the
center. Then we solve Eq. 35-13 for the wavelength, with d equal to the inverse of the number of
lines per centimeter.
1 r l 2638 2618 2628 26 28 / 60 26.47
2
2
1 d sin sin sin 26.47 4.618 105 cm 462 nm
N 9650 line/cm
2 2r 2l 4102 4027 4044.5 40 44.5/ 60 40.742
2
2
2 sin 40.742 6.763105 cm 676 nm
9650 line/cm
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414
Chapter 35 Diffraction and Polarization
41. If the spectrometer were immersed in water, the wavelengths calculated in Problem 40 would be
wavelengths in water. To change those wavelengths into wavelengths in air, we must multiply by
the index of refraction.
1 4.618 105cm 1.33 614 nm ; 2 6.763105cm 1.33 899 nm
air air
Note that the second wavelength is not in the visible range.
42. We solve Eq. 35-13 for the slit separation width, d, using the given information. Then setting m=3,
we solve for the angle of the third order maximum.
sin m d m 1589 nm 2074 nm 2.07m
d sin
sin16.5
3 sin 1 m sin 1 3 589 nm 58.4
d 2074 nm
43. We find the angle for each “boundary” color from Eq. 35-13, and then use the fact that the
displacement on the screen is given by tan y , where y is the displacement on the screen from the
L
central maximum, and L is the distance from the grating to the screen.
m 1 1m L tan sin1 m
d 610 lines 103mm d
sin
; d mm 1 6.1105 m ; y L tan
l1 L tan sin1 mred L tan sin1 m violet
d d
0.32 m tan 1 700 109 m tan 1 400 109 m
sin1 6.1105 m sin1 6.1105 m
1 1
0.0706 m 7cm
l2 L tan sin1 mred L tan sin1 m violet
d d
0.32 m tan 2 700 109 m tan 2 400 109 m
sin 1 6.1105 m sin 1 6.1105 m
1 1
0.3464 m 35cm
The second order rainbow is dispersed over a larger distance.
44. (a) Missing orders occur when the angle to the interference maxima (Eq. 34-2a) is equal to the
angle of a diffraction minimum (Eq. 35-2). We set d 2D and show that the even interference
orders are missing.
sin m1 m2 m1 d 2D 2 m1 2m2
d D m2 D D
Since m2 1,2,3,4, ..., all even orders of m1 correspond to the diffraction minima and will be
missing from the interference pattern.
(b) Setting the angle of interference maxima equal to the angle of diffraction minimum, with the
orders equal to integers we determine the relationship between the slit size and separation that
will produce missing orders.
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sin m1 m2 d m1
d D D m2
(c) When d D , all interference maxima will overlap with diffraction minima so that no fringes
will exist. This is expected because if the slit width and separation distance are the same, the
slits will merge into one single opening.
45. (a) Diffraction maxima occur at angles for which the incident
light constructive interferes. That is, when the path length
difference between two rays is equal to an integer number of
wavelengths. Since the light is incident at an angle
relative to the grating, each succeeding higher ray, as shown
in the diagram, travels a distance l1 d sin farther to
reach the grating. After passing through the grating the
higher rays travel a distance to the screen that is again longer
by l2 d sin . By setting the total path length difference
equal to an integer number of wavelengths, we are able to
determine the location of the bright fringes.
l l1 l1 d sin sin m, m 0,1, 2,....
(b) The allows for the incident angle and the diffracted angle to have positive and negative
values.
(c) We insert the given data, with m=1, to solve for the angles .
sin 1 sin m sin1 sin15 550 109 m 0.93 and 32
d 0.01m 5000 lines
46. Using Eq. 35-13 we calculate the maximum order possible for this diffraction grating, by setting the
angle equal to 90. Then we set the resolving power equal to the product of the number of grating
lines and the order, where the resolving power is the wavelength divided by the minimum separation
in wavelengths (Eq. 35-19) and solve for the separation.
sin m m d sin 0.01m 6500linessin 90 2.47 2
d
624 109m
Nm 6500 624 nm cm 2 0.015 nm
Nm
lines/cm 3.18
The resolution is best for the second order, since it is more spread out than the first order.
47. (a) The resolving power is given by Eq. 35-19.
R Nm R1 16,0001 16,000 ; R1 16,0002 32,000
(b) The wavelength resolution is also given by Eq. 35-19.
R Nm
Nm
1 410 nm 2.6 102 nm 26 pm ; 1 410 nm 1.3102 nm 13pm
16, 000 1 32, 000 1
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416
Chapter 35 Diffraction and Polarization
48. (a) We use Eq. 35-13, with the angle equal to 90 to determine the maximum order.
sin m m d sin 1050 nmsin 90 1.81
d
580 nm
Since the order must be an integer number there will only be one principal maximum on either
side of the central maximum. Counting the central maximum and the two other principal
maxima there will be a total of three principal maxima.
(b) We use Eq. 35-17 to calculate the peak width, where the full peak width is double the half-peak
width and the angle to the peak is given by Eq. 35-13.
0 0
2 2 2580 nm 6.4 105 rad 0.0037
Nd cos0 l cos0
1.80 102 m cos0
0
1 sin 1 m sin1 1 580 nm 33.5
d 1050 nm
2 2 580 nm 7.7 105 rad 0.0044
l cos1 1.80 102 m cos33.5
1
49. We use Eq. 35-20, with m = 1.
m 2d sin sin 1 m sin1 1 0.138 nm 14.0
2d 2 0.285 nm
50. We use Eq. 35-20 for X-ray diffraction.
(a) Apply Eq. 35-20 to both orders of diffraction.
m 2d sin m1 sin 1 2 sin 1 m2 sin 1 sin 1 2 sin 26.8 64.4
m2 sin 2 m1 1
(b) Use the first order data.
m 2d sin 2d sin 20.24 nmsin 26.8 0.22 nm
m
1
51. For each diffraction peak, we can measure the angle and count the order. Consider Eq. 35-20.
m 2d sin 2d sin1 ; 2 2d sin2 ; 3 2d sin3
From each equation, all we can find is the ratio 2sin sin2 2 sin 3 . No, we cannot
d 3
separately determine the wavelength or the spacing.
52. Use Eq. 35-21. Since the initial light is unpolarized, the intensity after the first polarizer will be half
the initial intensity. Let the initial intensity be I0.
I1 1 I0 ; I2 I1 cos2 1 I0 cos2 I2 cos2 65 0.089
2 2 I0 2
53. If I0 is the intensity passed by the first Polaroid, the intensity passed by the second will be I0 when
the two axes are parallel. To calculate a reduction to half intensity, we use Eq. 35-21.
I I0 cos2 1 I0 cos2 1 45
2 2
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54. We assume that the light is coming from air to glass, and use Eq. 35-22b.
tanp nglass 1.58 p tan11.58 57.7
55. The light is traveling from water to diamond. We use Eq. 35-22a.
tanp ndiamond 2.42 1.82 p tan11.82 61.2
nwater 1.33
56. The critical angle exists when light passes from a material with a higher index of refraction n1 into
a material with a lower index of refraction n2 . Use Eq. 32-7.
n2 sinC sin 55
n1
To find the Brewster angle, use Eq. 35-22a. If light is passing from high index to low index, we
have the following.
n2 tanp sin 55 p tan1 sin 55 39
n1
If light is passing from low index to high index, we have the following.
n1 tanp 1 p tan 1 1 51
n2 sin 55 sin 55
57. Let the initial intensity of the unpolarized light be I0. The intensity after passing through the first
Polaroid will be I1 1 I0. Then use Eq. 35-21.
2
I2 I1 cos2 1 I0 cos2 cos1 2I2
2 I0
(a) cos1 2I2 cos1 2 35.3
I0 3
(b) cos1 2I2 cos1 2 63.4
I0 10
58. For the first transmission, the angle between the light and the polarizer is 18.0. For the second
transmission, the angle between the light and the polarizer is 36.0. Use Eq. 35-21 twice.
I1 I0 cos2 18.0 ; I2 I1 cos2 36.0 I0 cos2 18.0cos2 36.0 0.592I0
Thus the transmitted intensity is 59.2% of the incoming intensity.
59. First case: the light is coming from water to air. Use Eq. 35-22a.
tanp nair p tan1 nair tan 1 1.00 36.9
nwater nwater 1.33
Second case: for total internal reflection, the light must also be coming from water into air. Use Eq.
32-7.
sinC nair p sin1 nair sin 1 1.00 48.8
nwater nwater 1.33
Third case: the light is coming from air to water. Use Eq. 35-22b.
tanp nwater p tan1 nwater tan11.33 53.1
Note that the two Brewster’s angles add to give 90.0.
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418
Chapter 35 Diffraction and Polarization
60. When plane-polarized light passes through a sheet oriented at an angle , the intensity decreases
according to Eq. 35-21, I I0 cos2 . For 45, cos2 1 . Thus sheets 2 through 6 will each
2
reduce the intensity by a factor of 1 . The first sheet reduces the intensity of the unpolarized incident
2
light by 1 as well. Thus we have the following.
2
I I 1 6 0.016 I0
02
61. We assume vertically polarized light of intensity I0 is incident upon the first polarizer. The angle
between the polarization direction and the polarizer is . After the light passes that first polarizer,
the angle between that light and the next polarizer will be 90 . Apply Eq. 35-21.
I1 I0 cos2 ; I I1 cos2 90 I0 cos2 cos2 90 I0 cos2 sin2
We can also use the trigonometric identity sin cos 1 sin 2 to write the final intensity as
2
I I0 cos2 sin2 1 I0 sin2 2 .
4
dI d 1 I0 sin2 2 1 I0 2 sin 2 cos 2 2 I0 sin 2 cos 2 1 I0 sin 4
d 4 4 2
d
1 I0 sin 4 0 4 , 360 0,45,90
2
Substituting the three angles back into the intensity equation, we see that the angles 0 and 90 both
give minimum intensity. The angle 45 gives the maximum intensity of 1 I0.
4
62. We set the intensity of the beam as the sum of the maximum and minimum intensities. Using Eq.
35-21, we determine the intensity of the beam after it has passed through the polarizer. Since Imin is
polarized perpendicular to Imax and the polarizer is rotated at an angle from the polarization of Imax,
the polarizer is oriented at an angle of 90 from Imin.
I0 Imax Imin
I I0 cos2 Imax cos2 Imin cos2 90 Imax cos2 Imin sin2
We solve the percent polarization equation for Imin and insert the result into our intensity equation.
p I max I min I min 1 p I max
I max I min 1 p
I I max cos2 1 p I max sin 2 I max 1 pcos2 1 p sin 2
p
1 1 p
I max cos2 sin2 p cos2 sin2 I max 1 p cos 2
1 p 1 p
63. Because the width of the pattern is much smaller than the distance to the screen, the angles from the
diffraction pattern for this first order will be small. Thus we may make the approximation that
sin tan . We find the angle to the first minimum from the distances, using half the width of the
full first order pattern. Then we use Eq. 35-2 to find the slit width.
tan 1 min 1 8.20 cm 0.01439 sin1min
2 285cm
D sin m D m 1 415 nm 2.88 104 nm 2.88 105 m
sin
0.01439
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
64. If the original intensity is I0 , the first polarizers will reduce the intensity to one half the initial
intensity, or I1 1 I0. Each subsequent polarizer oriented at an angle to the preceding one will
2
reduce the intensity by cos2 , as given by Eq. 35-21. We set the final intensity equal to one quarter
of the initial intensity, with 10 for each polarizer and solve for the minimum number of
polarizers.
n1 ln 2I I0 ln 2 0.25
ln cos2 ln cos2 10
I
1 I0 cos2 n 1 1 23.6 24 polarizers
2
We round the number of lenses up to the integer number of polarizers, so that the intensity will be
less than 25% of the initial intensity.
65. The lines act like a grating. We assume that we see the first diffractive order, so m = 1. Use Eq. 35-
13.
d sin m d m 1 480 nm 580 nm
sin
sin 56
66. We assume the sound is diffracted when it passes through the doorway, and find the angles of the
minima from Eq. 35-2.
v ; D sin m mv sin1 mv , m 1, 2, 3, ...
f f Df
m 1: sin1 mv sin1 1340m s 27
Df 0.88m850 Hz
m 2: sin1 mv sin 1 2 340 m s 65
Df m850 Hz
0.88
m 3: sin1 mv sin 1 3 340 m s sin11.36 impossible
Df m850 Hz
0.88
Thus the whistle would not be heard clearly at angles of 27° and 65° on either side of the normal.
67. We find the angles for the first order from Eq. 35-13.
1
sin 1 m sin1 1 4.4 107 m 19.5
d
0.01 m 7600
2
sin 1 1 6.8 107 m 31.1
0.01 m 7600
The distances from the central white line on the screen are found using the tangent of the angle and
the distance to the screen.
y1 L tan1 2.5m tan19.5 0.89 m
y2 L tan2 2.5m tan 31.1 1.51m
Subtracting these two distances gives the linear separation of the two lines.
y2 y1 1.51m 0.89 m 0.6 m
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420
Chapter 35 Diffraction and Polarization
68. Because the angle increases with wavelength, to miss a complete order we use the smallest visible
wavelength, 400 nm. The maximum angle is 90°. With these parameters we use Eq. 35-13 to find
the slit separation, d. The inverse of the slit separation gives the number of lines per unit length.
d sin m d m 2400 nm 800 nm
sin
sin 90
1 1 12, 500 lines/cm
d 800 107 cm
69. We find the angles for the two first-order peaks from the distance to the screen and the distances
along the screen to the maxima from the central peak.
tan1 y1 1 tan 1 y1 tan 1 3.32 cm 2.88
l l 66.0 cm
tan2 y2 2 tan 1 y2 tan 1 3.71cm 3.22
l l 66.0 cm
Inserting the wavelength of yellow sodium light and the first order angle into Eq. 35-13, we calculate
the separation of lines. Then, using the separation of lines and the second angle, we calculate the
wavelength of the second source. Finally, we take the inverse of the line separation to determine the
number of lines per centimeter on the grating.
d sin1 m1 d m1 1589 nm 11,720 nm
sin1
sin 2.88
2 d sin1 11,720 nmsin 3.22 658 nm
m
1 1 line 853lines/cm
d 11,720 107 cm
70. We find the angles for the first order from Eq. 35-13, with m = 1. The slit spacing is the inverse of
the lines/cm of the grating.
d 8100 1 cm 1m 1 m ; d sin m sin1 m
lines 100 cm 8.1105 d
sin 1 1 sin1 2 sin 1 656 109 m sin 1 410 109 m 13
d d
8.1 1 m 8.1 1 m
105 105
71. (a) This is very similar to Example 35-6. We use the same notation as in that Example, and solve
for the distance l.
s
l l 1.22 l Ds 6.0 103 m 2.0 m 1.8 104 m 18km
D 1.22
1.22 560 109 m
(b) We use the same data for the eye and the wavelength.
1.22 1.22 560 109 m 1.139 104 rad 180 3600 23
D 6.0 103 m rad 1
Our answer is less than the real resolution, because of atmospheric effects and aberrations in the
eye.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
72. We first find the angular half-width for the first order, using Eq. 35-1, sin . Since this angle is
D
small, we may use the approximation that sin tan. The width from the central maximum to the
first minimum is given by y L tan . That width is then doubled to find the width of the beam,
from the first diffraction minimum on one side to the first diffraction minimum on the other side.
y L tan Lsin
y
2 y 2Lsin 2L 2 3.8 108 m 633109 m 4.8 104 m
D 0.010 m
73. The distance between lines on the diffraction grating is found by solving Eq. 35-13 for d, the grating
spacing. The number of lines per meter is the reciprocal of d.
d m 1 sin sin 21.5 5.79 105 lines m
sin d m
16.328 107 m
74. (a) We calculate the wavelength of the mother’s sound by dividing the speed of sound by the
frequency of her voice. We use Eq. 34-2b to determine the double slit interference minima with
d 3.0 m.
v f 340 m s 400 Hz 0.85 m
sin –1 m 1 sin –1 m 1 0.85 m sin1 0.2833m 1 , m 0,1, 2,...
2 2 2
d 3.0 m
8.1, 25, 45, and 83
We use Eq. 35-2 to determine the angles for destructive interference from single slit diffraction,
with D 1.0 m.
sin –1 m sin –1 m 0.85 m sin1 0.85m, m 1,2,...
D
1.0 m
58
(b) We use the depth and length of the room to determine the angle the sound would need to travel
to reach the son.
tan 1 8.0 m 58
5.0 m
This angle is close to the single slit diffraction minimum, so the son has a good explanation for
not hearing her.
75. We use the Brewster angle, Eq. 35-22b, for light coming from air to water.
tanp n p tan1 n tan11.33 53.1
This is the angle from the normal, as seen in Fig. 35-41, so the angle above the horizontal is the
complement of 90.0 53.1 36.9 .
76. (a) Let the initial unpolarized intensity be I0. The intensity of the polarized light after passing the
first polarizer is I1 1 I0. Apply Eq. 35-21 to find the final intensity.
2
I2 I1 cos2 I1 cos2 90 0 .
(b) Now the third polarizer is inserted. The angle between the first and second polarizers is 66, so
the angle between the second and third polarizers is 24. It is still true that I1 1 I0.
2
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422
Chapter 35 Diffraction and Polarization
I2 I1 cos2 66 1 I0 cos2 66 ; I3 I2 cos2 24 1 I0 cos2 66 cos2 24 0.069
2 2
I3 0.069
I1
(c) The two crossed polarizers, which are now numbers 2 and 3, will still not allow any light to pass
through them if they are consecutive to each other. Thus I3 0.
I1
77. The reduction being investigated is that which occurs when the polarized light passes through the
second Polaroid. Let I1 be the intensity of the light that emerges from the first Polaroid, and I2 be
the intensity of the light after it emerges from the second Polaroid. Use Eq. 35-21.
(a) I2 I1 cos2 0.25I1 cos1 0.25 60
(b) I2 I1 cos2 0.10I1 cos1 0.10 72
(c) I2 I1 cos2 0.010I1 cos1 0.010 84
78. (a) We apply Eq. 35-21 through the successive polarizers. The initial light is unpolarized. Each
polarizer is then rotated 30 from the previous one.
I1 1 I0 ; I2 I1 cos2 2 1 I0 cos2 2 ; I3 I2 cos2 3 1 I0 cos2 2 cos2 3 ;
2 2 2
I4 I3 cos2 4 1 I0 cos2 2 cos2 3 cos2 4 1 I 0 cos2 30 cos2 30 cos2 30 0.21 I 0
2 2
(b) If we remove the second polarizer, then the angle between polarizers # 1 and # 3 is now 60
I1 1 I0 ; I3 I1 cos2 3 1 I0 cos2 3 ;
2 2
I4 I3 cos2 4 1 I0 cos2 3 cos2 4 1 I0 cos2 60 cos2 30 0.094 I0
2 2
The same value would result by removing the third polarizer, because then the angle between
polarizers # 2 and # 4 would be 60 Thus we can decrease the intensity by removing either the
second or third polarizer.
(c) If we remove both the second and third polarizers, we will have two polarizers with their axes
perpendicular, so no light will be transmitted.
79. For the minimum aperture the angle subtended at the lens by the smallest feature is the angular
resolution, given by Eq. 35-10. We let l represent the spatial separation, and r represent the altitude
of the camera above the ground.
1.22 l D 1.22r 1.22 580 109 m 25000 m 0.3538m 0.4 m
D r l 0.05 m
80. Let I0 be the initial intensity. Use Eq. 35-21 for both transmissions of the light.
I1 I0 cos2 1 ; I2 I1 cos2 2 I0 cos2 1 cos2 2 0.25I0
1 cos 1 0.25 cos1 0.25 42
cos2 cos 48
81. We find the spacing from Eq. 35-20.
m 2d sin
d m 2 9.73 1011 m 2.45 1010 m
2 sin 2sin 23.4
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
82. The angles for Bragg scattering are found from Eq. 35-20, for X-ray Crystal
m = 1 and m = 2. If the distance from the crystal to the screen
is l, the radius of the diffraction ring is given by r l tan 2.
2d sin m ; r l tan 2 l tan 2 sin1 m 2
2d
l
r1 l tan 2 sin 1 m
2d Screen
0.12 m tan sin 1 1 0.10 109 m 0.059 m
2 0.22 109 m
2
r2
l tan 2 sin 1 m 0.12 m tan 2 0.10 109 m 0.17 m
2d 2sin1
2 0.22 109 m
83. From Eq. 35-10 we calculate the minimum resolvable separation angle. We then multiply this angle
by the distance between the Earth and Moon to obtain the minimum distance between two objects on
the Moon that the Hubble can resolve.
1.22 1.22 550 109 m 2.796 107 rad
D 2.4 m
l s 3.84 108 m 2.796 107 rad 110 m
84. From Eq. 35-10 we calculate the minimum resolvable separation angle. We then multiply this angle
by the distance between Mars and Earth to obtain the minimum distance between two objects that
can be resolved by a person on Mars
1.22 1.22 550 109 m 1.34 104 rad
D 0.005 m
l s 81010 m 1.34 104 rad 1.07 107 m
Since the minimum resolvable distance is much less than the Earth-Moon distance, a person standing
on Mars could resolve the Earth and Moon as two separate objects without a telescope.
85. The distance x is twice the distance to the first minima. We can write x in terms of the slit width D
using Eq. 35-2, with m = 1. The ratio is small, so we may approximate sin tan .
D
sin ; x 2y 2l tan 2l 2l
D D
When the plate is heated up the slit width increases due to thermal expansion. Eq. 17-1b is used to
determine the new slit width, with the coefficient of thermal expansion, , given in Table 17-1. Each
slit width is used to determine a value for x. Subtracting the two values for x gives the change x.
We use the binomial expansion to simplify the evaluation.
2l 1 1 2l
D0 D0 D0
1 T 1 T
x
x x0 2l D0 2l 1 T 1 1
2l 1 T 1 2l T 2 2.0 m 650 109 m 25 106 C 1 55C
D0 D0
22 106 m
1.7 104 m
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424
Chapter 35 Diffraction and Polarization
86. The tangent of the angle for each order is the distance in the table divided by the distance to the
screen. If we call the distance in the table y and the distance to the screen l, then we have this
relationship.
tan y tan1 y sin 0.14
ll
0.12 sin = 0.01471 m + 0.0001
The relationship between the angle 0.10 R2 = 0.9999
and the wavelength is given by Eq. 0.08
35-2, D sin m, which can be
written as sin m. A plot of 0.06
D 0.04
sin vs. m should have a slope of 0.02
, and so the wavelength can be 0.00
D 0
2 4 6 8 10
determined from the slope and the m
slit width. The graph is shown, and
the slope used to calculate the wavelength.
D
slope slope D 0.01471 4.000 105 m 588.4 nm
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH35.XLS,” on tab “Problem 35.86.”
87. We have N polarizers providing a rotation of 90. Thus, each polarizer must rotate the light by an
angle of N 90 N. As the light passes through each polarizer, the intensity will be reduced by a
factor of cos2 N. Let the original intensity be I0.
I1 I0 cos2 N ; I2 I1 cos2 N I0 cos4 N ; I3 I2 cos2 N I0 cos6 N N [cos(90/N )]2N
IN I0 cosN 2N 0.90I0 cos 90 N 2N 0.90 21 0.8890
We evaluate cos90 N2N for various values of N. A table for a few
values of N is shown here. We see that N = 24 satisfies the criteria, and so 22 0.8938
23 0.8982
N 90 24N 90 24N 3.75. So we need to put 24 polarizers in the 24 0.9022
25 0.9060
path of the original polarized light, each rotated 3.75 from the previous one.
The spreadsheet used for this problem can be found on the Media Manager,
with filename “PSE4_ISM_CH35.XLS,” on tab “Problem 35.87.”
88. (a) The intensity of the diffraction pattern is given by Eqs. 35-6 and 35-7. We want to find the
angle where I 1 I0. Doubling this angle will give the desired .
2
I I sin 2 2 1 I0 sin 2 2 or sin , with 1
2 2 2 2 2
0
This equation must be solved numerically. A spreadsheet was developed to find the non-zero
values of that satisfy sin 0. It is apparent from this expression that there will be no
2
solutions for 2. The only non-zero value is 1.392. Now use Eq. 35-6 to find .
D sin sin1 sin1 2 sin1 1.392 ;
2 D 2 D
D
2 2 sin 1 1.392
D
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(b) For D : 2 sin 1 1.392 2 sin 1 1.392 52.6
D
For D 100 : 2 sin 1 1.392 2 sin1 1.392 0.508
D 100
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH35.XLS,” on tab “Problem 35.88.”
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426
CHAPTER 36: The Special Theory of Relativity
Responses to Questions
1. No. The train is an inertial reference frame, and the laws of physics are the same in all inertial
reference frames, so there is no experiment you can perform inside the train car to determine if you
are moving.
2. The fact that you instinctively think you are moving is consistent with the relativity principle applied
to mechanics. Even though you are at rest relative to the ground, when the car next to you creeps
forward, you are moving backward relative to that car.
3. As long as the railroad car is traveling with a constant velocity, the ball will land back in his hand.
4. The relativity principle refers only to inertial reference frames. Neither the reference frame of the
Earth nor the reference frame of the Sun is inertial. Either reference frame is valid, but the laws of
physics will not be the same in each of the frames.
5. The starlight would pass at c, regardless of your spaceship’s speed. This is consistent with the
second postulate of relativity which states that the speed of light through empty space is independent
of the speed of the source or the observer.
6. It deals with space-time (sometimes called “the fabric of space-time”) and the actual passage of time
in the reference frame, not with the mechanical workings of clocks. Any measurement of time
(heartbeats or decay rates, for instance) would be measured as slower than normal when viewed by
an observer outside the moving reference frame.
7. Time actually passes more slowly in the moving reference frames, according to observers outside
the moving frames.
8. This situation is an example of the “twin paradox” applied to parent-child instead of to twins. This
might be possible if the woman was traveling at high enough speeds during her trip. Time would
have passed more slowly for her and she could have aged less than her son, who stayed on Earth.
(Note that the situations of the woman and son are not symmetric; she must undergo acceleration
during her journey.)
9. No, you would not notice any change in your heartbeat, mass, height, or waistline, because you are
in the inertial frame of the spaceship. Observers on Earth, however, would report that your heartbeat
is slower and your mass greater than if you were at rest with respect to them. Your height and
waistline will depend on your orientation with respect to the motion. If you are “standing up” in the
spaceship such that your height is perpendicular to the direction of travel, then your height would not
change but your waistline would shrink. If you happened to be “lying down” so that your body is
parallel to the direction of motion when the Earth observers peer through the telescope, then you
would appear shorter but your waistline would not change.
10. Yes. However, at a speed of only 90 km/hr, v/c is very small, and therefore γ is very close to one, so
the effects would not be noticeable.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
11. Length contraction and time dilation would not occur. If the speed of light were infinite, v/c would
be zero for all finite values of v, and therefore γ would always be one, resulting in t t0 and
l l0.
12. The effects of special relativity, such as time dilation and length contraction, would be noticeable in
our everyday activities because everyday speeds would no longer be so small compared to the speed
of light. There would be no “absolute time” on which we would all agree, so it would be more
difficult, for instance, to plan to meet friends for lunch at a certain time! In addition, 25 m/s would
be the limiting speed and nothing in the universe would move faster than that.
13. Both the length contraction and time dilation formulas include the term 1 v2 c2 . If c were not
the limiting speed in the universe, then it would be possible to have a situation with v > c. However,
this would result in a negative number under the square root, which gives an imaginary number as a
result, indicating that c must be the limiting speed.
14. Mr. Tompkins appears shrunk in the horizontal direction, since that is the direction of his motion,
and normal size in the vertical direction, perpendicular to his direction of motion. This length
contraction is a result of the fact that, to the people on the sidewalk, Mr. Tompkins is in a moving
frame of reference. If the speed of light were only 20 mi/h, then the amount of contraction, which
depends on γ, would be enough to be noticeable. Therefore, Mr. Tompkins and his bicycle appear
very skinny. (Compare to the chapter-opening figure, which is shown from Mr. Tompkin’s
viewpoint. In this case, Mr. Tompkins sees himself as “normal” but all the objects moving with
respect to him are contracted.)
15. No. The relativistic momentum of the electron is given by p mv mv . At low speeds
1 v2 c2
(compared to c) this reduces to the classical momentum, p = mv. As v approaches c, γ approaches
infinity so there is no upper limit to the electron’s momentum.
16. No. To accelerate a particle with nonzero rest mass up to the speed of light would require an infinite
amount of energy, and so is not possible.
17. No. E = mc² does not conflict with the principle of conservation of energy as long as it is understood
that mass is a form of energy.
18. Yes, mass is a form of energy so technically it is correct to say that a spring has more mass when
compressed. However, the change in mass of the spring is very small and essentially negligible.
19. “Energy can be neither created nor destroyed.” Mass is a form of energy, and mass can be
“destroyed” when it is converted to other forms of energy. The total amount of energy remains
constant.
20. Technically yes, the notion that velocities simply add is wrong. However, at everyday speeds, the
relativistic equations reduce to classical ones, so our ideas about velocity addition are essentially
true for velocities that are low compared to the speed of light.
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428
Chapter 36 The Special Theory of Relativity
Solutions to Problems
1. You measure the contracted length. Find the rest length from Eq. 36-3a.
l0 l 38.2 m 72.5m
1 v2 c2
1 0.8502
2. We find the lifetime at rest from Eq. 36-1a.
t0 t 1 v2 c2 4.76 106s 1 2.70 108 m s 2 2.07 106s
3.00 108 m
s
3. The numerical values and 8
graph were generated in a 7
spreadsheet. The graph is 6
shown also. The spreadsheet 5
used for this problem can be 4
found on the Media Manager, 3
with filename 2
“PSE4_ISM_CH36.XLS,” on 1
tab “Problem 36.3.”
0
0.2 0.4 0.6 0.8 1
v /c
4. The measured distance is the contracted length. Use Eq. 36-3a.
l l0 1 v2 c2 135 ly 1 2.80 108 m s 2 48.5 ly
3.00 108 m
s
5. The speed is determined from the time dilation relationship, Eq. 36-1a.
t0 t 1 v2 c2
vc 1 t0 2 c 1 2.60 108 s 2 0.807c 2.42 108 m s
t 4.40 108 s
6. The speed is determined from the length contraction relationship, Eq. 36-3a.
l l0 1 v2 c2 vc 1 l 2 c 1 35 ly 2 0.78c 2.3108 m s
l0 56 ly
7. The speed is determined from the length contraction relationship, Eq. 36-3a. Then the time is found
from the speed and the contracted distance.
l l0 1 v2 c2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
vc 1 l 2 ; t l l 25ly 25 y c 27 y
l0 v c 0.923
l 2 25ly 2
l0 65ly
c 1 c 1
8. The speed is determined from the length contraction relationship, Eq. 36-3a.
l l0 1 v2 c2 vc 1 l 2 c 1 0.9002 0.436c
l0
9. The change in length is determined from the length contraction relationship, Eq. 36-3a. The speed is
very small compared to the speed of light.
l l0 1 v2 c2
l 1 v2 c2 1 v2 1/ 2 1 1 v2 1 1 11.2 103 m s 2 1 6.97 1010
l0 c2 2 c2 2 3.00 108 m s
So the percent decrease is 6.97 108 % .
10. (a) The measured length is the contracted length. We find the rest length from Eq. 36-3a.
l0 l 4.80 m 7.39 m
1 0.7602
1 v2 c2
Distances perpendicular to the motion do not change, so the rest height is 1.35m .
(b) The time in the spacecraft is the rest time, found from Eq. 36-1a.
t0 t 1 v2 c2 20.0s 1 0.7602 13.0s
(c) To your friend, you moved at the same relative speed: 0.760c .
(d) She would measure the same time dilation: 13.0s .
11. (a) We use Eq. 36-3a for length contraction with the contracted length 99.0% of the rest length.
l l0 1 v2 c2 vc 1 l 2 c 1 0.9902 0.141c
l0
(b) We use Eq. 36-1a for time dilation with the time as measured from a relative moving frame
1.00% greater than the rest time.
t0 t 1 v2 c2 vc 1 t0 2 c 1 1 2 0.140 c
t 1.0100
We see that a speed of 0.14 c results in about a 1% relativistic effect.
12. (a) To an observer on Earth, 18.6 ly is the rest length, so the time will be the distance divided by
the speed.
tEarth l0 18.6 ly 19.58 yr 19.6 yr
v
0.950 c
(b) The time as observed on the spacecraft is shorter. Use Eq. 36-1a.
t0 t 1 v2 c2 19.58 yr 1 0.9502 6.114 yr 6.11yr
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430
Chapter 36 The Special Theory of Relativity
(c) To the spacecraft observer, the distance to the star is contracted. Use Eq. 36-3a.
l l0 1 v2 c2 18.6 ly 1 0.9502 5.808 ly 5.81 ly
(d) To the spacecraft observer, the speed of the spacecraft is their observed distance divided by
their observed time.
v l 5.808 ly 0.950 c
t0
6.114 yr
13. (a) In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 36-1a.
t0 t 1 v2 c2 5.0 yr 1 0.742 3.4 yr
(b) Now we assume the 5.0 years is the time as measured on the Enterprise. Again use Eq. 36-1a.
t0 t 1 v2 c2 t t0 5.0 yr 7.4 yr
1 v2 c2 1 0.742
14. We find the speed of the particle in the lab frame, and use that to find the rest frame lifetime and
distance.
v xlab 1.00 m 2.941108 m s 0.9803c
tlab 3.40 109 s
(a) Find the rest frame lifetime from Eq. 36-1a.
t0 tlab 1 v2 c2 3.40 109s 1 0.98032 6.72 1010s
(b) In its rest frame, the particle will travel the distance given by its speed and the rest lifetime.
x0 vt0 2.941108 m s 6.72 1010s 0.198m
This could also be found from the length contraction relationship: x0 xlab .
1 v2 c2
15. Since the number of particles passing per second is reduced from N to N / 2, a time T0 must have
elapsed in the particles’ rest frame. The time T elapsed in the lab frame will be greater, according to
Eq. 36-1a. The particles moved a distance of 2cT0 in the lab frame during that time.
T0 T 1 v2 c2 T T0 ; v x 2cT0 v 4 c 0.894 c
1 v2 c2 T T0 5
1 v2 c2
16. The dimension along the direction of motion is contracted, and the other two dimensions are
unchanged. Use Eq. 36-3a to find the contracted length.
l l0 1 v2 c2 ; V l l0 2 l0 3 1 v2 c2 2.0 m3 1 0.802 4.8 m3
17. The vertical dimensions of the ship will not change, but the horizontal dimensions will be contracted
according to Eq. 36-3a. The base will be contracted as follows.
lbase l 1 v2 c2 l 1 0.952 0.31l
When at rest, the angle of the sides with respect to the base is given by cos1 0.50l 75.52.
2.0l
The vertical component of lvert 2lsin 2l sin 75.52 1.936l is unchanged. The horizontal
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
component, which is 2l cos 2 l 1 0.50l at rest, will be contracted in the same way as the base.
4
lhorizontal 0.50l 1 v2 c2 0.50l 1 0.952 0.156l
Use the Pythagorean theorem to find the length of the leg.
lleg l l 2 2 0.156l2 1.936l2 1.942l 1.94l
horizontal vert
18. In the Earth frame, the average lifetime of the pion will be dilated according to Eq. 36-1a. The speed
of the pion will be the distance moved in the Earth frame times the dilated time.
v d d 1 v2 c2
t t0
1 1 2 0.95c
3.00 108 m s 2.6 108s
25 m
v c 2 c
ct0
1 d
1
19. We take the positive direction in the direction of the Enterprise. Consider the alien vessel as
reference frame S, and the Earth as reference frame S. The velocity of the Earth relative to the alien
vessel is v 0.60c. The velocity of the Enterprise relative to the Earth is ux 0.90c. Solve for
the velocity of the Enterprise relative to the alien vessel, ux , using Eq. 36-7a.
ux ux v 0.90c 0.60c 0.65c
1 0.60 0.90
1 vux
c2
We could also have made the Enterprise as reference frame S, with v 0.90c, and the velocity of
the alien vessel relative to the Earth as ux 0.60c. The same answer would result.
Choosing the two spacecraft as the two reference frames would also work. Let the alien vessel be
reference frame S, and the Enterprise be reference frame S. Then we have the velocity of the Earth
relative to the alien vessel as ux 0.60c, and the velocity of the Earth relative to the Enterprise as
ux 0.90c. We solve for v, the velocity of the Enterprise relative to the alien vessel.
ux ux v v ux ux .60c 0.90c 0.65c
1 vux 1 uxux 1 0.90c .60c
c2 c2
c2
20. The Galilean transformation is given in Eq. 36-4.
(a) x, y, z x vt, y, z 25m 30 m s3.5s,20 m,0 130 m,20 m,0
(b) x, y, z x vt, y, z 25m 30 m s10.0s,20 m,0 325m,20 m,0
21. (a) The person’s coordinates in S are found using Eq. 36-6, with x 25 m , y 20 m , z 0 , and
t 3.5s. We set v 1.80 108 m/s.
x x vt 25m 1.8 108 m/s 3.5s 820 m
1 v2 c2 1 1.8 108 m/s 2 3.0 108 m/s 2
y y 20 m ; z z 0
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432
Chapter 36 The Special Theory of Relativity
(b) We repeat part (a) using the time t 10.0 s.
x x vt 25m 1.8 108 m/s 10.0 s 2280 m
1 v2 c2 1 1.8 108 m/s 2 3.0 108 m/s 2
y y 20 m ; z z 0
22. We determine the components of her velocity in the S frame using Eq. 36-7, where
ux uy 1.10 108 m/s and v 1.80 108 m/s . Then using trigonometry we combine the
components to determine the magnitude and direction.
ux v 1.10 108 m/s 1.80 108 m/s 2.38 108 m/s
vux / c2 1.80 108 m/s 1.10 108 m/s / 3.00 108 m/s
ux1 2
1
uy 2 2
uy 1 v2 c2 1.10 108 m/s 1 1.8 108 m/s 3.0 108 m/s 2 7.21107 m/s
1 vux / c2 1 1.80 108 m/s 1.10 108 m/s / 3.00 108 m/s
u 2 2
ux2 u 2 2.38 108 m/s 7.21107 m/s 2.49 108 m/s
y
tan1 uy tan 1 7.21107 m/s 16.9
ux 2.38 108 m/s
23. (a) We take the positive direction to be the direction of motion of spaceship 1. Consider spaceship
2 as reference frame S, and the Earth reference frame S. The velocity of the Earth relative to
spaceship 2 is v 0.60c. The velocity of spaceship 1 relative to the Earth is ux 0.60c. Solve
for the velocity of spaceship 1 relative to spaceship 2, ux , using Eq. 36-7a.
ux ux v 0.60c 0.60c 0.88 c
1 0.60 0.60
1 vux
c2
(b) Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship
1 is v 0.60c. The velocity of spaceship 2 relative to the Earth is ux 0.60c. Solve for the
velocity of spaceship 2 relative to spaceship 1, ux , using Eq. 36-7a.
ux ux v 0.60c 0.60c 0.88 c
1 0.600.60
1 vux
c2
As expected, the two relative velocities are the opposite of each other.
24. (a) The Galilean transformation is given in Eq. 36-4.
x x vt x vt 100 m 0.92 3.00 108 m s 1.00 106s 376 m
(b) The Lorentz transformation is given in Eq. 36-6. Note that we are given t, the clock reading in
frame S.
t t vx t t vx
c2 c2
x x vt x v t vx x v ct vx
c2 c c
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
1
1 0.922
100m 0.92 1 0.922 3.00 108 m s 1.00 106s 0.92100 m
316 m
25. (a) We take the positive direction in the direction of the first spaceship. We choose reference frame
S as the Earth, and reference frame S as the first spaceship. So v 0.61c. The speed of the
second spaceship relative to the first spaceship is ux 0.87 c. We use Eq. 36-7a to solve for the
speed of the second spaceship relative to the Earth, u.
ux ux v 0.87c 0.61c 0.97 c
1 0.61 0.87
1 vux
c2
(b) The only difference is now that ux 0.87 c.
ux ux v 0.87c 0.61c 0.55c
1 0.61 0.87
1 vux
c2
The problem asks for the speed, which would be 0.55c
26. We assume that the given speed of 0.90c is relative to the planet that you are approaching. We take
the positive direction in the direction that you are traveling. Consider your spaceship as reference
frame S, and the planet as reference frame S. The velocity of the planet relative to you is
v 0.90c. The velocity of the probe relative to the planet is ux 0.95c. Solve for the velocity of
the probe relative to your spaceship, ux , using Eq. 36-7a.
ux ux v 0.95c 0.90c 0.34c
1 0.90 0.95
1 vux
c2
27. We set frame S as the frame at rest with the spaceship. In this frame the module has speed
u uy 0.82c. Frame S is the frame that is stationary with respect to the Earth. The spaceship, and
therefore frame S moves in the x-direction with speed 0.76c in this frame, or v 0.76c. We use
Eq. 36-7a and 36-7b to determine the components of the module velocity in frame S. Then using
trigonometry we combine the components to determine the speed and direction of travel.
ux ux v 0 0.76c 0.76c ; uy uy 1 v2 c2 0.82c 1 0.762 0.533c
1 vux / c2 1 0 1 vux / c2 1 0
u ux2 u 2 0.76c2 0.533c2 0.93c ; tan 1 uy tan1 0.533c 35
y ux 0.76c
28. The velocity components of the particle in the S frame are ux u cos and uy u sin. We find the
components of the particle in the S frame from the velocity transformations given in Eqs. 36-7a and
36-7b. Those transformations are for the S frame moving with speed v relative to the S frame. We
can find the transformations from the S frame to the S frame by simply changing v to –v and primed
to unprimed variables.
ux v ux ux v ; uy 1 v2 c2 uy uy 1 v2 c2
1 vux c2 1 vux c2
1 vux c2 1 vux c2
ux uy
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Chapter 36 The Special Theory of Relativity
uy 1 v2 c2
tan
uy 1 vux c2 uy 1 v2 c2 u sin 1 v2 c2 sin 1 v2 c2
ux
ux v ux v u cos v cos v u
1 vux c2
29. (a) In frame S the horizontal component of the stick length will be contracted, while the vertical
component remains the same. We use the trigonometric relations to determine the x- and y-
components of the length of the stick. Then using Eq. 36-3a we determine the contracted length
of the x-component. Finally, we use the Pythagorean theorem to determine stick length in
frame S.
lx l0 cos ; l y l0 sin ly ; lx lx 1 v2 c2 l0 cos 1 v2 c2
l lx2 ly2 l02 cos2 1 v2 c2 l02 sin2 l0 1 v cos c2
(b) We calculate the angle from the length components in the moving frame.
tan1 ly l0 sin tan tan1 tan
tan1 tan1 1 v2 c2
lx l0 cos 1 v2 c2
30. (a) We choose the train as frame S and the Earth as frame S. Since the guns fire simultaneously in
S , we set these times equal to zero, that is tA tB 0. To simplify the problem we also set the
location of gunman A equal to zero in frame S when the guns were fired, xA 0. This places
gunman B at xB 55.0 m. Use Eq. 36-6 to determine the time that each gunman fired his
weapon in frame S.
tA tA vxA 0 v0 0
c2 c2
tB vxB 1 0 35 m/s 55.0 m 2.14 1014 s
c2 35.0 m/s 3.00 108 m/s
3.00 108 m/s 2
tB 2
1
Therefore, in Frame S, A fired first.
(b) As found in part (a), the difference in time is 2.14 1014 s .
(c) In the Earth frame of reference, since A fired first, B was struck first. In the train frame, A is
moving away from the bullet fired toward him, and B is moving toward the bullet fired toward
him. Thus B will be struck first in this frame as well.
31. We set frame S as the frame moving with the observer. Frame S is the frame in which the two light
bulbs are at rest. Frame S is moving with velocity v with respect to frame S. We solve Eq. 36-6 for
the time t in terms of t, x, and v. Using the resulting equation we determine the time in frame S that
each bulb is turned on, given that in frame S the bulbs are turned on simultaneously at tA tB 0.
Taking the difference in these times gives the time interval as measured by the observing moving
with velocity v.
x x vt x x vt
t t vx = t v x vt t1 v2 vx t vx t t vx
c2 c2 c2 c2 c2 c2
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tA t A vxA 0 v0 0 ; tB tB vxB 0 vl vl
c2 c2 c2 c2 c2
t tB tA vl
c2
According to the observer, bulb B turned on first.
32. We set up the two frames such that in frame S, the first object is located at the origin and the second
object is located 220 meters from the origin, so xA 0 and xB 220 m. We set the time when event
A occurred equal to zero, so tA 0 and tB 0.80 s. We then set the location of the two events in
frame S equal, and using Eq. 36-6 we solve for the velocity.
xA xB xA vtA xB vtB ; v xA xB 0 220 m 2.5 108 m/s
tA tB 0 0.88 s
33. From the boy’s frame of reference, the pole remains at rest with respect to him. As such, the pole
will always remain 12.0 m long. As the boy runs toward the barn, relativity requires that the
(relatively moving) barn contract in size, making the barn even shorter than its rest length of 10.0 m.
Thus it is impossible, in the boy’s frame of reference, for the barn to be longer than the pole. So
according to the boy, the pole will never completely fit within the barn.
In the frame of reference at rest with respect to the barn, it is possible for the pole to be shorter than
the barn. We use Eq. 36-3a to calculate the speed that the boy would have to run for the contracted
length of the pole, l, to equal the length of the barn.
l l0 1 v2 c2 v c 1 l2 l02 c 1 10.0 m2 12.0 m2 0.5528c
If persons standing at the front and back door of the barn were to close both doors exactly when the
pole was completely inside the barn, we would have two simultaneous events in the barn’s rest frame
S with the pole completely inside the barn. Let us set the time for these two events as tA tB 0. In
frame S these two events occur at the front and far side of the barn, or at xA 0 and xB 10.0m.
Using Eq. 36-6, we calculate the times at which the barn doors close in the boy’s frame of reference.
tA tA vxA 0 v0 0
c2 c2
tB tB vxB 1 0 0.552810.0 m 2.211 108 s
c2 1 0.55282
3.00 108 m/s
Therefore, in the boy’s frame of reference the far door of the barn closed 22.1 ns before the front
door. If we multiply the speed of the boy by this time difference, we calculate the distance the boy
traveled between the closing of the two doors.
x vt 0.5528 3.00 108 m/s 2.211108 s 3.67 m.
We use Eq. 36-3a to determine the length of the barn in the boy’s frame of reference.
l l0 1 v2 c2 10.0 m 1 0.55282 8.33 m
Subtracting the distance traveled between closing the doors from the length of the pole, we find the
length of the barn in the boy’s frame of reference.
l0,pole x 12.0 m 3.67 m 8.33 m lbarn
Therefore, in the boy’s frame of reference, when the front of the pole reached the far door it was
closed. Then 22.1 ns later, when the back of the pole reached the front door, that door was closed.
In the boy’s frame of reference these two events are not simultaneous.
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436
Chapter 36 The Special Theory of Relativity
34. The momentum of the proton is given by Eq. 36-8.
p mv mv 1.67 1027 kg 0.75 3.00 108 m s 5.7 1019 kgm s
1 v2 c2 1 0.752
35. (a) We compare the classical momentum to the relativistic momentum.
pclassical mv
mv
p relativistic
1 v2 c2 1 0.10 2 0.995
1 v2 c2
The classical momentum is about 0.5% in error.
(b) We again compare the two momenta.
p classical mv 1 v2 c2 1 0.602 0.8
p relativistic mv
1 v2 c2
The classical momentum is 20% in error.
36. The momentum at the higher speed is to be twice the initial momentum. We designate the initial
state with a subscript “0”, and the final state with a subscript “f”.
mvf vf2
pf 1 vf2 c2 2 1 vf2 c2 4 vf2 4 0.26c2 0.29c2
p0 mv0 v02 vf2
1 c2 1 0.262
1 v02 c2 1 v02 c2
vf2 0.29 c 2 vf 0.47 c
1.29
37. The two momenta, as measured in the frame in which the particle was initially at rest, will be equal
to each other in magnitude. The lighter particle is designated with a subscript “1”, and the heavier
particle with a subscript “2”.
p1 p2 m1v1 m2v2
1 v12 c2 1 v22 c2
v12 m2 2 v22 c2 6.68 1027 kg 2 0.60c2 9.0c2
m1 1 v22 1.67 1027 kg 1 0.602
1 v12 c2
v1 0.90 c 0.95c
38. We find the proton’s momenta using Eq. 36-8.
p0.45 mpv1 mp 0.45c 0.5039mpc ; p0.80 mpv2 mp 0.80c 1.3333mpc
1 0.452 1 0.802
1 v12 1 v22
c2 c2
p0.98 mpv2 mp 0.98c 4.9247mpc
v22 1 0.982
1 c2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(a) p2 p1 100 1.3333mpc 0.5039mp c 100 164.6 160%
p1 0.5039mpc
(b) p2 p1 100 4.9247mpc 1.3333mpc 100 269.4 270%
p1 1.3333mpc
39. The rest energy of the electron is given by Eq. 36-12.
E mc2 9.11 1031 kg 3.00 108 m s 2 8.20 1014 J
8.20 1014 J
1.60 1013 J MeV 0.511MeV
40. We find the loss in mass from Eq. 36-12.
m
E 200 MeV 1.60 1013 J MeV 3.56 1028 kg 4 1028 kg
c2
3.00 108 m s 2
41. We find the mass conversion from Eq. 36-12.
m
E 8 1019 J
c2 3.00 108 m s 2 900 kg
42. We calculate the mass from Eq. 36-12.
m 2
E 1 1 1.6726 1027 kg 2.9979 108 m s
c2 c2 mc2 c2 1.6022 1013 J MeV 938.2 MeV c2
43. Each photon has momentum 0.50 MeV/c. Thus each photon has mass 0.50 MeV. Assuming the
photons have opposite initial directions, then the total momentum is 0, and so the product mass will
not be moving. Thus all of the photon energy can be converted into the mass of the particle.
Accordingly, the heaviest particle would have a mass of 1.00 MeV c2 , which is 1.78 1030 kg. .
44. (a) The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.
W K Kfinal 1 mc2 1 1938.3MeV 1.39 104 MeV
1 0.9982
13.9GeV
(b) The momentum of the proton is given by Eq. 36-8.
p mv 1
938.3MeV c2 0.998c 1.48 104 MeV c 14.8GeV c
1 0.9982
45. We find the energy equivalent of the mass from Eq. 36-12.
E mc2 1.0 103 kg 3.00 108 m s 2 9.0 1013J
We assume that this energy is used to increase the gravitational potential energy.
m E 9.0 1013J 9.2 109 kg
hg 1.0 103 m 9.80 m s2
E mgh
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438
Chapter 36 The Special Theory of Relativity
46. The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.
W1 0.90 1 mc2 ; W2 K0.99c K0.90c 0.99 1 mc2 0.90 1 mc2
1 mc2 0.90 1 mc2 11
0.90 1 mc2
W2 0.99 0.99 0.90 1 0.992 1 0.902 3.7
W1 0.90 1 1 1
1 0.902
47. The kinetic energy is given by Eq. 36-10.
K 1 mc2 mc2 2 1 v 43c 0.866c
1 v2 c2
48. The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 36-13 to find the
momentum.
E K mc2 ;
2 2 2 K2 2K
pc2 E2
mc2 K mc2 mc2 mc2
pc mc 2 938.3 MeV
K 2 2K mc2 K 950 MeV
K 1 2 950 MeV 1 2 1638 MeV
p 1638 MeV c 1.6GeV c
49. We find the speed in terms of c. The kinetic energy is given by Eq. 36-10 and the momentum by Eq.
36-8.
2.80 108 m s
v 3.00 108 m s 0.9333c
K 1 mc2 1 1938.3MeV 1674.6 MeV 1.67 GeV
1 0.93332
p mv 1
938.3MeV c2 0.9333c 2439 MeV c 2.44GeV c
1 0.93332
50. We use Eq. 36-10 to find the speed from the kinetic energy.
1 c2
1 v2 1 mc2
K 1mc2
vc 1 1 c 1 1 0.957 c
K 2 1.25 MeV 2
mc2 0.511MeV
1 1
51. Since the proton was accelerated by a potential difference of 125 MV, its potential energy decreased
by 125 MeV, and so its kinetic energy increased from 0 to 125 MeV. Use Eq. 36-10 to find the
speed from the kinetic energy.
1 c2
1 v2 1 mc2
K 1mc2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
vc 1 1 c 1 1 0.470c
K 2 125 MeV 2
mc2 938.3 MeV
1 1
52. We let M represent the rest mass of the new particle. The initial energy is due to both incoming
particles, and the final energy is the rest energy of the new particle. Use Eq. 36-11 for the initial
energies.
E 2 mc2 Mc2 M 2 m 2m
1 v2 c2
We assumed that energy is conserved, and so there was no loss of energy in the collision.
The final kinetic energy is 0, so all of the kinetic energy was lost.
1 c2
1 v2 1 2mc2
Klost Kinitial 2 1 mc2
53. Since the electron was accelerated by a potential difference of 28 kV, its potential energy decreased
by 28 keV, and so its kinetic energy increased from 0 to 28 MeV. Use Eq. 36-10 to find the speed
from the kinetic energy.
1 c2
1 v2 1 mc2
K 1mc2
vc 1 1 c 1 1 0.32 c
K 12 0.028 MeV 12
mc2 0.511MeV
54. We use Eqs. 36-11 and 36-13 in order to find the mass.
E 2 p2c2 m2c4 K mc2 2 K 2 2Kmc2 m2c4
m p2c2 K 2 121MeV c2 c2 45MeV2 140 MeV c2 2.5 1028 kg
2Kc2 245MeVc2
The particle is most likely a probably a 0 meson.
55. (a) Since the kinetic energy is half the total energy, and the total energy is the kinetic energy plus
the rest energy, the kinetic energy must be equal to the rest energy. We also use Eq. 36-10.
K K mc2
1 E 1 K mc2
2 2
K 1 mc2 mc2 2 1 v 3 c 0.866 c
4
1 v2 c2
(b) In this case, the kinetic energy is half the rest energy.
K 1 mc2 1 mc2 3 1 v 5 c 0.745c
2 2 1 v2 c2 9
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440
Chapter 36 The Special Theory of Relativity
56. We use Eq. 36-10 for the kinetic energy and Eq. 36-8 for the momentum.
1 938.3 MeV
1 1 mc2 1
K 1mc2 1 v2
c2 1 8.15 107 m s 2
3.00 108 m
s
36.7 MeV
938.3 MeV 8.15 107 m s
3.00 108 m
p mv mv 1 mc2 v c 1 s 265 MeV c
1 v2 c c
c2 1 v2 c2 1 8.15 107 m s 2
3.00 108 m s
Evaluate with the classical expressions.
Kc 1 mv 2 1 mc2 v 2 1 938.3 MeV 8.15 107 m s 2 34.6 MeV
2 2 c 2 3.00 108 m s
pc mv 1 mc 2 v 938.3 MeV 8.15 107 m s 255 MeV c
c c 3.00 108 m
s
Calculate the percent error.
errorK Kc K 100 34.6 36.7 100 5.7%
K 36.7
errorp pc p 100 255 265 100 3.8%
p 265
57. (a) The kinetic energy is found from Eq. 36-10.
1
1 v2 1 mc2
K 1mc
c2 1 1 1.7 104 kg 3.00 108 m s 2
1 0.182
2.5411019 J 2.5 1019 J
(b) Use the classical expression and compare the two results.
K 2 2.479 1019 J
1 mv 1 1.7 104 kg 0.18 3.00 108 m s
2 2
2.479 1019 J 2.5411019 J
%error 100 2.4%
2.5411019 J
The classical value is 2.4% too low.
58. The kinetic energy of 998 GeV is used to find the speed of the protons. Since the energy is 1000
times the rest mass, we expect the speed to be very close to c. Use Eq. 36-10.
1 c2
1 v2 1 mc2
K 1mc2
vc 1 1 c 1 1 c to 7 sig. fig.
K 2 998GeV 2
mc2 0.938GeV
1 1
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441
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
K 1 mc 998GeV 1
B mc2
mv2 mv 0.938 GeV 1.673 1027 kg 3.00 108 m s
rqv rq
3.3T
rq 1.0 103 m 1.60 1019 C
59. By conservation of energy, the rest energy of the americium nucleus is equal to the rest energies of
the other particles plus the kinetic energy of the alpha particle.
mAmc2 mNp m c2 K
mNp mAm m K 241.05682 u 4.00260 u 5.5 MeV 1u c2 237.04832 u
c2 c2 931.49 MeV
60. (a) For a particle of non-zero mass, we derive the following relationship between kinetic energy
and momentum.
pc2 E2 2 2 2 K2 2K
E K mc2
; mc2 K mc2 mc2 mc2
4 mc2 2 4 pc2
2
K 2 2K mc2 pc2 0
2mc2
K
For the kinetic energy to be positive, we take the positive root.
2mc2 4 mc2 2 4 pc2 mc2 2 pc2
K 2 mc2
If the momentum is large, we have the following relationship.
K mc2 mc2 2 pc2 pc mc2
Thus there should be a linear relationship between kinetic energy and momentum for large
values of momentum.
If the momentum is small, we use the binomial expansion to derive the classical relationship.
mc2 2 pc2 mc2 mc2 pc 2
K mc2 1 mc2
mc2 mc2 1 1 pc 2 p2
2 mc2 2m
Thus we expect a quadratic relationship for m0
small values of momentum. The adjacent
graph verifies these approximations. K m0
(b) For a particle of zero mass, the relationship is
simply K pc. See the included graph. The
spreadsheet used for this problem can be p
found on the Media Manager, with filename
“PSE4_ISM_CH36.XLS,” on tab “Problem
36.60.”
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442
Chapter 36 The Special Theory of Relativity
61. All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use Eq.
36-11. Both masses are the same.
Etotal E1 E2 1mc2 2mc2 1 2 mc2 1 1 105.7 MeV
1 0.552
1 0.432
243.6 MeV 240 MeV
62. We use Eqs. 36-11 and 36-13.
pc2 E2 2 2 2 K2 2K
E K mc2
; mc2 K mc2 mc2 mc2
K 2 2K mc2
p c
63. (a) We assume the mass of the particle is m, and we are given that the velocity only has an x-
component, ux. We write the momentum in each frame using Eq. 36-8, and we use the velocity
transformation given in Eq. 36-7. Note that there are three relevant velocities: ux , the velocity
in reference frame S; ux , the velocity in reference frame S; and v, the velocity of one frame
relative to the other frame. There is no velocity in the y or z directions, in either frame. We
reserve the symbol for 1 , and also use Eq. 36-11 for energy.
1 v2 c2
px mux ; py 0 ; pz 0
1 ux2 c2
ux ux v ux ux v ; uy uy 1 vux c2 0 ; uz uz 1 vux c2 0
1 vux c2 1 vux c2 1 v2 c2 1 v2 c2
mux
1 ux2 c2
px ; pz 0since uz 0
; py 0 since uy 0
Substitute the expression for ux into the expression for px.
m ux v
1 vux c2
px mux m ux v 1
1 ux2 c2
1 1 ux v2 1 vux c2 c2 2 ux v2
c2 c2
1 vux c2 2 1 vux 2 c2
1 vux
1 vux c2 2
ux v 1 mux v
1 vux c2
m ux 2 ux v2
1 1 vux c2 2 v 1 vux c2 2
c2 c2
1 vux c2
mux v mux v
2 ux2 v2 2 ux2 v2
1 2 vux vux c2 2uxv c2 1 vux c2 c2
c2 c2 c2 c2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
mux mv
2 c2
mux v 1 ux2 c2 1 u x
1 v2 c2 1 ux2 c2 1 v2 c2
mux mc2 v px mc2 v
1 ux2 c2 1 ux2 c2 c2 1 ux2 c2 c2
px vE c2
1 v2 c2 1 v2 c2 1 v2 c2
It is obvious from the first few equations of the problem that py py 0 and pz pz 0.
E mc2 mc2 mc2
1 ux2 c2 1 ux v2 2 ux v2
c2
1 vux c2 2 c2
1
1 vux c2
2 1 vux c2
1 vux c2
2
mc2 mvux
mc2 1 vux c2 2
mc2 mvux 1 ux2 c2 1 u x c2
1 vux c2 2 ux v 2 1 v2 c2 1 ux2 c2 1 v2 c2
c2
E pxv
1 v2 c2
(b) We summarize these results, and write the Lorentz transformation from Eq. 36-6, but solved in
terms of the primed variables. That can be easily done by interchanged primed and unprimed
quantities, and changing v to v.
px vE c2 ; py py ; py py ; E E pxv
1 v2 c2 1 v2 c2
px
x x vt ; y y ; z z ; t t vx c2
1 v2 c2 1 v2 c2
These transformations are identical if we exchange px with x, py with y, pz with z, and E c2
with t (or E c with ct).
64. The galaxy is moving away from the Earth, and so we use Eq. 36-15b.
f0 f 0.0987 f0 f 0.9013 f0
f f0 cv v 1 f f0 2 c 1 0.90132 c 0.1035c
cv 1+ f f0 2 1 0.90132
65. For source and observer moving towards each other, use Eq. 36-14b.
f f0 cv f0 1 v c 95.0 MHz 1 0.70 226 MHz 230 MHz
cv 1 v c 1 0.70
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444
Chapter 36 The Special Theory of Relativity
66. We use Eq. 36-15a, and assume that v c.
cv 0 1 v c 0 1 v c1 v c 0 1 v c 1
cv 1 v c 1 v c1 v c 1 v2 c2
0
0 1 v c 1 v2 c2 1/ 2 0 1 v c 0 0 v c 0 v
0 0 c
67. (a) We apply Eq. 36-14b to determine the received/reflected frequency f. Then we apply this same
equation a second time using the frequency f as the source frequency to determine the Doppler-
shifted frequency f . We subtract the initial frequency from this Doppler-shifted frequency to
obtain the beat frequency. The beat frequency will be much smaller than the emitted frequency
when the speed is much smaller than the speed of light. We then set c v c and solve for v.
f f0 cv f f cv f0 cv cv f0 c v
cv cv cv cv c v
f beat f f0 f0 c v f0 c v f0 2v f0 2v v cf beat
c v c v cv c 2 f0
v
3.00 108 m/s 6670 Hz 27.8 m/s
2 36.0 109 Hz
(b) We find the change in velocity and solve for the resulting change in beat frequency. Setting
the change in the velocity equal to 1 km/h we solve for the change in beat frequency.
v cf beat v cf beat f beat 2 f0v
2 f0 2 f0 c
fbeat
2 36.0 109 Hz 1km/h 1m/s 70 Hz
3.00 108 3.600 km/h
m/s
68. We consider the difference between Doppler-shifted frequencies for atoms moving directly towards
the observer and atoms moving directly away. Use Eqs. 36-14b and 36-15b.
f f0 c v f0 cv f 0 c v cv f 0 2v 2v c
c v cv c v cv c2 v2 f0 1 v2 c2
We take the speed to be the rms speed of thermal motion, given by Eq. 18-5. We also assume that
the thermal energy is much less than the rest energy, and so 3kT mc2.
v vrms 3kT v 3kT f 2 3kT 1 3kT 1/ 2 2 3kT
m c mc2 f0 mc2 mc2 mc2
We evaluate for a gas of H atoms (not H2 molecules) at 550 K. Use Appendix F to find the mass.
f
f0
2 3kT 2 3 1.38 1023 J K 550 K s 2 2.5 105
mc2
1.008 u 1.66 1027 kg u 3.00 108 m
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
69. At the North Pole the clock is at rest, while the clock on the equator travels the circumference of the
Earth each day. We divide the circumference of the Earth by the length of the day to determine the
speed of the equatorial clock. We set the dilated time equal to 2.0 years and solve for the change in
rest times for the two clocks.
v
2 R 2 6.38 106 m 464 m/s
T
24 hr 3600 s/hr
t t0,eq t0,eq t 1 v2 / c2 t 1 v2
1 v2 / c2 t0,pole t 2c2
t t0,pole
1 0
t0,eq t0,pole t 1 v2 t
2c2
2.0 yr 464 m/s2 3.156 107s/yr
2 3.00 108 m/s 2
tv2 75s
2c2
70. We take the positive direction in the direction of the motion of the second pod. Consider the first
pod as reference frame S, and the spacecraft as reference frame S. The velocity of the spacecraft
relative to the first pod is v 0.60c. The velocity of the first pod relative to the spacecraft is
ux 0.50c. Solve for the velocity of the second pod relative to the first pod, ux , using Eq. 36-7a.
ux ux v 0.50c 0.60c 0.846 c
1 0.60 0.50
1 vux
c2
71. We treat the Earth as the stationary frame, and the airplane as the moving frame. The elapsed time in
the airplane will be dilated to the observers on the Earth. Use Eq. 36-1a.
tEarth 2 rEarth ; tplane tEarth 1 v2 c2 2 rEarth 1 v2 c2
v v
2 rEarth 2 rEarth 1 1 v2 rEarthv
v v c2 c2
t
tEarth tplane 1 1 v2 c2 1
2
1300 km 1m s
3.6 km
6.38 106 m h h
8.0 108s
3.00 108 m s 2
72. (a) To travelers on the spacecraft, the distance to the star is contracted, according to Eq. 36-3a.
This contracted distance is to be traveled in 4.6 years. Use that time with the contracted
distance to find the speed of the spacecraft.
v xspacecraft xEarth 1 v2 c2
tspacecraft tspacecraft
vc 1 c 1 0.6829 c 0.68c
2 2
1 ctspacecraft 1 4.6 ly
xEarth 4.3 ly
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446
Chapter 36 The Special Theory of Relativity
(b) Find the elapsed time according to observers on Earth, using Eq. 36-1a.
tEarth tspaceship 4.6 y 6.3y
1 v2 c2 1 0.68292
Note that this agrees with the time found from distance and speed.
tEarth xEarth 4.3ly 6.3yr
v 0.6829 c
73. (a) We use Eq. 36-15a. To get a longer wavelength than usual means that the object is moving
away from the Earth.
0
cv 1.0700 1.0702 1
cv 1.0702 1 c v 0.067c
(b) We assume that the quasar is moving and the Earth is stationary. Then we use Eq. 16-9b.
f f0 c c 1 0 1 v c 1.0700 v 0.070 c
v 0
1 c 1 v c
74. We assume that some kind of a light signal is being transmitted from the astronaut to Earth, with a
frequency of the heartbeat. That frequency will then be Doppler shifted, according to Eq. 36-15b.
We express the frequencies in beats per minute.
f f0
cv f 2 f 2 602 302
cv 0 c 602 302
vc 0.60c
f 2 f02
75. (a) The velocity components of the light in the S frame are ux 0 and uy c. We transform those
velocities to the S frame according to Eq. 36-7.
ux ux v 0v v ; uy uy 1 v2 c2 c 1 v2 c2 c 1 v2 c2
vux c2 1 0 1 vux c2 10
1
tan1 uy tan1 c 1 v2 c2 tan1 c2 1
ux v v2
(b) u ux2 u2y v2 c2 1 v2 c2 v2 c2 v2 c
(c) In a Galilean transformation, we would have the following.
ux ux v v ; uy uy c ; u v2 c2 c ; tan 1 c
v
76. We take the positive direction as the direction of motion of rocket A. Consider rocket A as reference
frame S, and the Earth as reference frame S. The velocity of the Earth relative to rocket A is
v 0.65c. The velocity of rocket B relative to the Earth is ux 0.85c. Solve for the velocity of
rocket B relative to rocket A, ux , using Eq. 36-7a.
ux ux v 0.85c 0.65c 0.45c
1 0.65 0.85
1 vux
c2
Note that a Galilean analysis would have resulted in ux 0.20c.
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447
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
77. (a) We find the speed from Eq. 36-10.
1 c2 14,000mc2
1 v2 1 mc2
K 1mc2
vc 1 1 2 c c 1 2
14, 001 2 14,001
c 1 2
v c 1 2 3.00 108 m s 14,001 0.77 m s
2 14,001 2
(b) The tube will be contracted in the rest frame of the electron, according to Eq. 36-3a.
l0 l 1 v2 c2 3.0 103 m 1 1 1 2 0.21m
14, 001
78. The electrostatic force provides the radial acceleration. We solve that relationship for the speed of
the electron.
F Felectrostaticcentripetal 1 e2 m v2
4 0 r2 electron
r
1.60 1019 C 2
0.53 1010 m
v
1 e2 8.99 109 N m2 C2 2.18 106 m s 0.0073c
40 m relectron 9.11 1031 kg
Because this is much less than 0.1c, the electron is not relativistic.
79. The minimum energy required would be the energy to produce the pair with no kinetic energy, so the
total energy is their rest energy. They both have the same mass. Use Eq. 36-12.
E 2mc2 2 0.511MeV 1.022 MeV 1.64 1013 J
80. The wattage times the time is the energy required. We use Eq. 36-12 to calculate the mass.
E Pt mc2
m Pt 75W 3.16 107s 1000 g 2.6 105 g
c2 1kg
3.00 108 m s 2
81. Use Eqs. 36-13, 36-8, and 36-11.
E 2 p2c2 m2c4 E p2c2 m2c4 1/ 2
dE 1/ 2 2 pc2 pc2 pc2 mvc2
E E mc2
dp
1 p2c2 m2c4 v
2
82. The kinetic energy available comes from the decrease in rest energy.
K mnc2 mpc2 mec2 mvc2 939.57 MeV 938.27 MeV 0.511MeV 0 0.79 MeV
83. (a) We find the rate of mass loss from Eq. 36-12.
E mc2 E m c2
E 4 1026 J s
t 3.00 108 m s
m 1 2 4.44 109 kg s 4 109 kg s
c2
t
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448
Chapter 36 The Special Theory of Relativity
(b) Find the time from the mass of the Sun and the rate determined in part (a).
t
mEarth 5.98 1024 kg 4.27 107 y 4 107 y
m t 4.44 109 kg s 3.156 107 s y
(c) We find the time for the Sun to lose all of its mass at this same rate.
t
mSun 1.99 1030 kg 1.42 1013 y 11013 y
m t 4.44 109 kg s 3.156 107 s y
84. Use Eq. 36-8 for the momentum to find the mass.
p mv mv
1 v2 c2
m p 2.24 108 m s 2
3.07 1022 kgm s 1 3.00 108 m
1 v2 c2 s 9.12 1031 kg
v
2.24 108 m s
This particle has the mass of an electron, and a negative charge, so it must be an electron.
85. The total binding energy is the energy required to provide the increase in rest energy.
E 2mp+e 2mn mHe c2
2 1.00783 u 21.00867 u 4.00260 u c2 931.5 MeV c2 28.32 MeV
u
86. The momentum is given by Eq. 36-8, and the energy is given by Eq. 36-11 and Eq. 36-13.
P mv mc2v Ev v pc2 pc2 pc
c2 c2 E m2c2 p2
m2c4 p2c2
87. (a) The magnitudes of the momenta are equal. We use Eq. 36-8.
p mv mv c2 1 mc2 v c 1 938.3 MeV 0.985 5356 MeV c
1 v2 c c
1 v2 c2 1 0.9852
5.36 GeV c 5.36GeV c 1c 1.602 1010 J GeV
108
3.00 m s 1GeV
2.86 1018 kgm s
(b) Because the protons are moving in opposite directions, the vector sum of the momenta is 0.
(c) In the reference frame of one proton, the laboratory is moving at 0.985c. The other
proton is moving at 0.985c relative to the laboratory. We find the speed of one proton
relative to the other, and then find the momentum of the moving proton in the rest frame of the
other proton by using that relative velocity.
ux v ux 0.985c 0.985c 0.9999 c
1 0.9850.985
1 vux
c2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
449
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
mux 1 mc2 ux c 1 938.3 MeV 2 0.985
1 ux2 c c
p mux c2 1 0.9852 62081MeV c
c2 1 u 2 1 2 0.985 2
x
1 0.9852
62.1GeV c 62.1GeV c 1c 1.602 1010 J GeV
108
3.00 m s 1GeV
3.311017 kgm s
88. We find the loss in mass from Eq. 36-12.
E 484 103 J 5.38 1012 kg
c2 3.00 108 m s
m 2
Two moles of water has a mass of 36 103 kg. Find the percentage of mass lost.
5.38 1012 kg 1.49 1010 1.5 108 %
36 103 kg
89. Use Eq. 36-10 for kinetic energy, and Eq. 36-12 for rest energy.
K m c2 m c2
1 Enterprise converted
1 mEnterprise
mconverted
1 c2 1 1 6 109 kg 3107 kg
1 v2 1 0.102
90. We set the kinetic energy of the spacecraft equal to the rest energy of an unknown mass. Use Eqs.
36-10 and 36-12.
K 1 mshipc2 mc2
m 1 mship
1 c2 1 1 1.8 105 kg 7.2 104 kg
1 v2 1 mship 1 0.702
From the Earth’s point of view, the distance is 35 ly and the speed is 0.70c. That data is used to
calculate the time from the Earth frame, and then Eq. 36-1a is used to calculate the time in the
spaceship frame.
t d 35 y c 50 y ; t0 t 1 v2 c2 50 y 1 0.702 36 y
v
0.70c
91. We assume one particle is moving in the negative direction in the laboratory frame, and the other
particle is moving in the positive direction. We consider the particle moving in the negative
direction as reference frame S, and the laboratory as reference frame S. The velocity of the
laboratory relative to the negative-moving particle is v 0.85c, and the velocity of the positive-
moving particle relative to the laboratory frame is ux 0.85c. Solve for the velocity of the positive-
moving particle relative to the negative-moving particle, ux.
ux ux v 0.85c 0.85c 0.987 c
1 0.85 0.85
1 vux
c2
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450
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