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Chapter 30: Atomic Physics 25.1 THE RAY ASPECT OF LIGHT 1. 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 height? Solutio
n 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 stands from the wall. Therefore, , and The bottom is from the floor and the top is from th e floor. 25.2 THE LAW OF REFLECTION 2. 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 following figure. Solutio
n The incident ray reflects at an angle with the first mirror. The first reflection ray forms a right triangle with the two mirrors. Hence . The outgoing C/kg 1076.111=
emq
, C/kg 1057.9 7=
pmq
. 1084.1 1839C/kg 1057.9C/kg 1076.13
711
====
pe
ep
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==
= = == = =
−−
ddV mVm
3g/cm0.81 2. 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 following figure. ray makes an angle with a line parallel to the first mirror and so the outgoing 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 when the mirror is rotated by an angle . Solutio
n As shown in the figure, a ray of light that strikes a mirror at an angle is deflected from its original path by an angle of . Therefore, rotating the mirror by an additional angle will change the ray's direction by . 4. A flat mirror is neither converging nor diverging. To prove this, co nsider two rays originating from the same point and diverging at an angle . Show that after striking a plane mirror, the angle between their directions remains . Solutio
n The two incident rays, along with segment make a triangle. The projection of the reflected rays backward forms another triangle. Using the law of reflection and geometry, we can see that the two triangles have two interior angles and which are the same. Therefore the third angle must also be the same: 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
n From Table 25.1, we find the indices of refraction: ( )
3 20
462533
27--27 -15
kg/m1054.61051027.35.0234 and kg 10 1.66 197 with kg 10 1.66 amu 1 amu, 197 mass Gold m, 10 diameter Nucleus
==
= = == = =
−−
ddV mVm
m 100.46−
()()( )
( )()( )3363
3 14
14 2 2
18810 kg/m 4.10 10 m 4810 kg/m 2.71 10 kg3 8 6
2.71 10 kg 9.80 m/s 2.00 10 m
2.6 10 C2033 qVF qE mgd
mVV
d
mgdqVV
−
−
−−
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=
= = =
= = =
μm 10.0m 1000.1)m 00.1(m 10m 105
1015
A
an
N
AN
an==
===−
−−
dddddd
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μm 10.0
, 1 , 2 ;) (1 1 1 1
f i 2
f2
i2
fi
2
i2
f==
−=
−= n nnnnn
R n nR nm 122m 1022.114)1.2(
10 097.1m72
7==
−
=−
, 1 , 4 ;) (1 1 1 1
f i 2
f2
i2
fi
2
i2
f==
−=
−= n nnnnn
R n nR nm 97.2m 1072.9116)1.4(
10 097.1m82
7==
−
=−
2 22
B4e ekqmha=
Ba
m 10 529.010−
m 1029.5)C 10 602.1)(1)(C/mN 10 988.8)(kg 10 109.9(4)sJ 10 626.6(
4
112 19 2 2 9 31 22 34
2 22
B
−− −−
= = =e ekZqmπha
0E
.2
22 42
0hkmqEe e= 5. What is the speed of light in water? In glycerine? 6. What is the speed of light in air? In crown glass? Solutio
n (5 sig. fig. used to show difference with value in vacuum). 7. Calculate the index of refraction for a medium in which the speed of light is , and identify the most likely substance based o n Table 25.1. Solutio
n Use the equation . From Table 25.1, the substance is polystyrene . 8. In what substance in Table 25.1 is the speed of light ? Solutio
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 away, would the light first arrive on Earth? Solutio
n 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 be tween the ray in the water and the perpendicular to the water is eV 6.13J 10 1.60eV 1)J 10 1716.2()sJ 1063.6()C/mN 1000.9)(kg 1011.9()C 1060.1(2 2
19-18
02 3422 2 9 31 4 19 2
22 42
0
=
= = =
−−− −
EhkmqEe e
4=n
eV. 850.016eV 6.13 eV 6.13
2−=−=−=nEn
eV 850.0
n
2eV 6.13
nEn−=
4 0.4eV 85.0eV 6.13 eV 6.132/1
==
−−=−=
nEn
n
2=n
()( )m 10 2.12 1m 1029.521011 2
B2
−−
===Zanrn
() R hc == m 10 097.1 /eV6.137 5. What is the speed of light in water? In glycerine? Solutio
n 11. Components of some computers communicate with each other through optical fibers having an index of refraction . What time in nanoseconds is required 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. Solutio
n (a) ()( )
( )( )m/10 1.097 m/s 1000.3sJ 10 626.6J/eV 10 602.1eV 6.137
8 3419
=
−− 1 1 1
2
i2
f
−=
n nR
)(=in
( ) nm 365 m 1065.3m/10 097.14 17
72
f
2
f=====−
Rn
nR
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