Official© Solutions Manual for College Phyisics, OpenStax

Chapter 30: Atomic Physics

25.1 THE RAY ASPECT OF LIGHT
Suppose a man stands in front of a mirror as shown in Figure 25.50. His eyes are
1.65 m above the floor, and the top of his head is 0.13 m higher. Find the height
above the floor of the top and bottom of the smallest mirror in which he can see
both the top of his head and his feet. How is this distance related to the man’s
1. height?
Solutio
n q
= 1.76 1011 C/kg
me
From ray-tracing and the law of reflection, we know that the angle of incidence is
equal to the angle of reflection, so the top of the mirror must extend to at least
halfway between his eyes and the top of his head. The bottom must go down to
halfway between his eyes and the floor. This result is independent of how far he
q m q/m 1.76 1011 C/kg
= 9.57 10 7 C/kg , p = e = = 1839 = 1.84 103 .
mp me q/m p 9.57 107 C/kg
stands from the wall. Therefore, , and
−27
mp 1.6726  10 kg
= = 1836 = 1.84  10 3 ,
me 9.1094  10 −31 kg

q q 1.60 10−19 C
= 9.57 107 C/kg  m = = = 1.67 10−27 kg .
The bottom is m 9.57 107 C/kg 9.57 107 C/kg from the floor and the top is
rn r rr (0.500 m)(10 −10 m)
= N  ra = n A = = 5.00 10 4 m = 50 km
ra rA rN 10 -15 m
from the floor.



25.2 THE LAW OF REFLECTION
Show that when light reflects from two mirrors that meet each other at a right
angle, the outgoing ray is parallel to the incoming ray, as illustrated in the
2. following figure.
Solutio
n
−15
10 m Nucleus diamet r =10-15 m, Gold mas =197amu,1amu =1.6 10-27kg
4 d 
3

( )

The incident ray reflects at an angle with the first mirror. The first reflection
m
 = with m =197 1.6 10-27kg andV =    0.5d3
V 3 2
3.2710−25
 = −46 = 6.541020kg/m3
510



3
ray forms a right triangle with the two mirrors. Hence 0.81 g/cm . The outgoing

, Show that when light reflects from two mirrors that meet each other at a right
angle, the outgoing ray is parallel to the incoming ray, as illustrated in the
2. following figure.
Nucleus diamet r =10-15 m, Gold mas =197amu,1amu =1.6 10-27kg
4 d 
3

( )

ray makes an angle with a line parallel to the first mirror and so the outgoing
m
 = with m =197 1.6 10-27kg andV =    0.5d3
V 3 2
3.2710−25
 = −46 = 6.541020kg/m3
510




ray is parallel to the incident ray.

3. Light shows staged with lasers use moving mirrors to swing beams and create
colorful effects. Show that a light ray reflected from a mirror changes direction by qV




4.010−6 m when the mirror is rotated by an angle .
F = qE = = mg
d
m
 =  V
V
 4  d  (810 kg/m ) (4.1010 m) −14
3 3 −6 3

= (810 kg/m3 )   = = 2.7110 kg
 3  8  6
mgd (2.7110 kg)(9.80 m/s )(2.0010 m) −18
−14 2 −2

q= = = 2.610 C
V 2033 V




Solutio
n dn dN dn 10−15 m 
=  dN = dA =  −10 ( .1 00 m) = .1 0010−5 m =10.0 μm
da dA da 10 m 
As shown in the figure, a ray of light that strikes a mirror at an angle 10.0 μm is
1  1 1  1 ( inn f )2 
= R 2 − 2  = 2 2 ;ni = 2,nf =1,
deflected from its original path by an angle of  n n  Rn −n  . Therefore, rotating the mirror f i i f


 m (2.1) 
2
1  1 1  1 (ninf )2 
by an additional angle will change the ray's direction by . = 7  =1.2 10−7 m=12 nm
1.09710  −14 
= R −    =  ; ni = 4, nf =1,
  nf2 ni2  R ni2 −nf2 




4. A flat mirror is neither converging nor diverging. To prove this, consider two rays
 m (4.1) 
2



originating from the same point and diverging at an angle . Show that after = 7 =9.7210−8m=97.2nm
1.09710 16−1

h2
striking a plane mirror, the angle between their directions remains . aB =
4 2mekqe2



Solutio
n

aB
−10
The two incident rays, along with segment 0.529 10 m make a triangle. The projection of
the reflected rays backward forms another triangle. Using the law of reflection h2 (6. 2610−34 Js)2


and geometry, we can see that the two triangles have two interior angles and
aB = 2 2 = 2 −31 9 2 2 −19 2
4 mπ ekZqe 4 (9.10910 kg)(8.98 10 Nm /C )(1)(1.60210 C)
=5.2910−1 m


2 2 qe4 me k 2
E0
which are the same. Therefore the third angle must also be the same: E = 0
h2
.


and the angle between the outgoing rays is the same as that between the
incoming rays.



25.3 THE LAW OF REFRACTION
5. What is the speed of light in water? In glycerine?
Solutio From Table 25.1, we find the indices of refraction:
n

,5. What is the speed of light in water? In glycerine?
2 2 qe4 me k 2 2 2 (1.60  10 −19 C) 4 (9.11  10 −31 kg)(9.00  109 N  m 2 / C 2 ) 2
E0 = =
h2 (6.63  10 −34 J  s) 2
 1 eV 
E0 = (2.1716  10 −18 J) -19 
= 13.6 eV
 1.60  10 J  n=4
6. What is the speed of light in air? In crown glass?
Solutio − 13.6 eV − 13.6 eV
n En = = = −0.850 eV.
n2 16
(5 sig. fig. used to show difference with value in vacuum).

0.850 eV
7. Calculate the index of refraction for a medium in which the speed of light is
n , and identify the most likely substance based on Table 25.1.
Solutio − 13.6 eV
En =
n n2
Use the equation . From Table 25.1, the substance
is polystyrene.
8. − 13.6 eV  − 13.6 eV 
1/ 2

n= = = 4.0 = 4
In what substance in Table 25.1 is the speed of light ?

En  − 0.85 eV 



Solutio
n n
9. There was a major collision of an asteroid with the Moon in medieval times. It was
described by monks at Canterbury Cathedral in England as a red glow on and
around the Moon. How long after the asteroid hit the Moon, which is n = 2
away, would the light first arrive on Earth?
Solutio
rn = =
2
(
n 2 aB (2) 5.29  10 −11 m )
= 2.12  10 −10 m
n Z 1
10. A scuba diver training in a pool looks at his instructor as shown in Figure 25.53.
What angle does the ray from the instructor’s face make with the perpendicular to
the water at the point where the ray enters? The angle between the ray in the
water and the perpendicular to the water is (13.6 eV)/hc = .1 09710 m = R
7

, (13.6eV)( .1 60210 J/eV) = 1.097107 /m
5. What is the speed of light in water? In glycerine?
Solutio

−19
n




( .6 62610 Js)( .3 0010 m/s)
−34 8



11. Components of some computers communicate with each other through optical
 1 1 

fibers having an index of refraction . What time in nanoseconds is required
1
= R 2 − 2  
  nf ni 



for a signal to travel 0.200 m through such a fiber?
Solutio
n
12. (a) Given that the angle between the ray in the water and the perpendicular to the
water is 25.0°, and using information in Figure 25.53, find the height of the
instructor’s head above the water, noting that you will first have to calculate the
angle of refraction. (b) Find the apparent depth of the diver’s head below water as
seen by the instructor. Assume the diver and the diver's image are the same
horizontal distance from the normal.




ni ( = )
Solutio
n




(a)

2
1 R nf 4 − 7
= 2  = = 7 =3.6510 m = 365nm
( )
 nf R 1.09710 /m