Chapter 35 Solutons
The Moon's radius is 1.74 traveled by the light is:
35.1
d = 2(3.84
× 10 6
m and the Earth's radius is 6.37
× 10 8 m – 1.74 × 10 6 m – 6.37 × 10 6 m) = 7.52 × 10 8
× 10 6
m. The total distance
m
7.52 × 10 8 m This takes 2.51 s, so v = = 2.995 × 10 8 m/ s = 299.5 Mm/ s 2.51 s
∆ x
2(1.50 × 10 8 km)(10 km)(1000 00 m/ km) (22.0 (22.0 min)(60.0 min)(60.0 s/ min)
35.2
∆ x = ct ;
35.3
The experiment is most convincing if the wheel turns fast enough to pass outgoing light t = 2l c through one notch notch and returning light light through the next:
c=
=
t
2l θ = ω t = ω c
ω =
so
cθ
2l
=
= 2.27 × 10 8 m / s = 227 M m / s
(2.998 × 10 8 )[2π / / (720)] 2(11.45 × 10 3 )
=
114 rad/ s
The returning light would be blocked by a tooth at one-half the angular speed, giving another data p oint. oint.
35.4
(a) (a)
For For the light light beam beam to make it through both slots, slots, the time time for for the light light to travel the d istanc istancee d must equal the time for the disk to rotate through the angle θ , if c is the speed of light, d c
(b)
=
θ , so ω
ω d ω θ
W e a re re g iv iv en en th th a t d = 2.50 2.50 m ,
c=
35.5
c=
ω d ω θ
θ =
1.00° π ra d 60.0
180°
(2.50 m )(3.49 × 10 4 r a d = − 2.91 × 10 4 ra d
U s in g Sn e ll' s la w , sin θ 2
θ 2 = 25.5°
λ 2
=
λ 1 n1
= =
n1 n2
= 2 .91 91 × 10 − 4 s
ω = 5555
rad,
) = 3.00 × 108
m s
=
rev re v 2 π ra d s
300 Mm/ Mm/ s
sin θ 1
442 nm
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= 3.49 × 10 4 1.00 rev
r ad s
Chapter 35 Solutions
35.6
35.7
c
(a)
f =
(b)
λ glass
=
(c)
v glass
=
λ
=
n1 sin θ1
3. 00 00 × 10 8 m / s − 6.328 × 10 7 m
λ air
=
n cair n
=
632.8 n m 1.50
=
323
4.74 × 10 14 H z
=
422 nm
3. 00 00 × 10 8 m / s 1.50
8 = 2. 00 00 × 10
m / s = 200 M m / s
= n2 sin θ 2
sin θ 1
= 1.333 .333 sin sin 45.0 45.0°
sin θ 1
= (1.33)(0.707) = 0.943 Figure for Goal Solutio n
θ 1 = 70.5° → 19.5° above the horizon
Goal Solution An u nd erwater scuba d iver sees sees the Sun Sun at an ap paren t angle of 45.0 45.0°° from from the vertical. vertical. What is the actual direction of the Sun? G:
The sunlight refracts as it enters the water from the air. Because the water has a higher index of refraction, refraction, the light light slows slows dow n and bend s toward th e vertical vertical line line th at is nor ma l to the interface. Therefore, the elevation angle of the Sun above the water will be less than 45 ° as shown in th e diagram to the right, even though it appears to the diver that the sun is 45 ° above the horizon.
O:
We can use Snell’s law of refraction to find the precise angle of incidence.
A:
Snell’s law is:
n 1 s i n θ1
which gives
sin θ 1 sin θ 1
The sunlight is at θ 1
L:
= 70.5°
= n2 sin θ 2
= 1.333
sin 45.0 45.0 °
= (1.333)(0.707) = 0.943
to the vertical, so the Sun is 19.5 ° above the horizon.
The calc calculated ulated resu lt agrees with ou r pred iction. iction. When ap plying Snell’ Snell’ss law, it is is easy to mix up the index values and to confuse angles-with-the-normal angles-with-the-normal and angles-with-the-surfac angles-with-the-surface. e. Making a sketch sketch and a pred iction iction as we d id here helps avoid careless careless mistakes. mistakes.
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Chapter 35 Solutions
324
*35.8
(a)
n1 sin θ1
= n2 sin θ 2
1.00 sin 30.0° = n sin 19.24° n = 1.52
(c)
35.9
35.10
f =
=
(d)
v
(b)
λ =
(a )
c
λ c n v f
= = =
3. 00 00 × 10 8 m / s − 6.328 × 10 7 m 3. 00 00 × 10 8 m / s 1.52 1. 98 98 × 10 8 m / s 4. 74 74 × 10 14 / s
Flin t G la la ss ss :
= =
(b)
W a ter :
(c)
C u bi bic Zi Zir co co ni n ia :
n1 sin θ1
v
n1 sin θ1
θ 2
n c n
= = v
= n2 sin θ 2 ;
n 2 = 1.90 =
35.11
v
c
c ; v
v=
= n2 sin θ 2 ;
= sin −1
=
4.74 × 10 14 Hz
8 = 1.98 98 × 10
=
8
m s
1.66 3. 00 00 × 108 m s 1.333 c n
m / s = 198 M m/ m/ s
417 nm
3. 00 00 × 10
=
in air air and in in syrup. syrup.
=
= 1.81 81 × 10 8
m s
=
181 M m / s
= 2 .25 25 × 10 8
m s
=
225 M m / s
3. 00 00 × 108 m s 2.20
8 = 1.36 36 × 10
m s
=
136 M m / s
1.333 sin 37.0° = n 2 sin 25.0° c = 1.58 × 10 8 m / s = 158 M m / s 1.90
θ 2 = sin –1
(1.00)(sin 30°) 1.50
=
n 1 s in n 2
θ 1
19.5°
θ 2 an d θ 3 are alternate interior angles formed by the ray cutting pa rallel norm als. So, θ 3 = θ 2 = 19.5 19.5°° . 1.50 sin θ 3 = (1.00) sin θ 4
θ 4 = 30.0°
Chapter 35 Solutions
35.12
(a )
W a t e r λ =
λ 0
(b)
G la s s
λ =
λ 0
*35.13
=
436 n m
= n w sin θ 2
sin θ 2
=
1 1.333
1.333
1.52
sin θ 1
=
=
327 nm
=
287 nm
1 1.333
sin (90.0° − 28.0°) = 0.662
= sin −1 0.66 662 = 41.5°
h=
(a )
n
436 n m
s in θ 1
θ 2
35.14
n
=
d
ta n θ 2
=
3.00 m tan 41.5 1.5°
=
3.39 m
Fr o om m g e om om e tr tr y y,, 1. 1.25 m = d sin 40.0° so d = 1.94 m
(b)
50 .0 .0° a b ov ov e h o r iz iz o n t al al
, o r p a r al alle l t o t h e
incident ray
*35.15
The incident light reaches the left-hand mirror at distance (1.00 m) tan 5.00° = 0.0875 m above its bottom edg e. The reflected reflected light first reaches the right-hand mirror at height 2(0.0875 m) = 0.175 m It bounces between the mirrors with this distance between points of contact with either. Since
1.00 m 0.175 m
= 5.72 5.72,, the light reflects
five five times from the right-hand mirror an d six times from from th e left. left.
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325
326
*35.16
Chapter 35 Solutions
At entry,
n1 s i n θ 1
= n2 sin θ 2
or
1.00 s in in 30 30.0° = 1. 1.50 s in in θ 2
θ 2 = 19.5° The distance h the light travels in the medium is given by (2.00 (2.00 cm ) h
cos θ 2 =
or
h=
( 2.00 cm) cm ) cos 19.5°
α
The angle of deviation upon entry is The offset offset distan ce comes from from
*35.17
sin α =
d : h
The d istance, istance, h, traveled traveled by the light light is h =
*35.18
t =
h v
=
2. 12 12 × 10 − 2 m
2.00 cm cos 19.5° v
− = 1.06 06 × 10 10
8
2. 00 00 × 10 m / s
= θ1 − θ 2 = 30.0° − 19.5° = 10.5°
d = (2.21 cm) sin 10.5° = 0.388 cm
The speed of light in the material is
Therefore,
= 2.12 cm
= 2.12 cm
c
3. 00 00 × 10 8 m / s
n
1.50
= =
= 2.00 00 × 10 8
m/ s
s = 106 106 ps
Applying Snell's law at the air-oil interface, n a ir sin θ = n o il s in 20. 0°
y ie ld s
θ = 30.4°
App lying lying Snell's nell's law at th e oil-water oil-water interface interface n w sin θ ′ = n oil sin 20. 0°
*35.19
y ie ld s
θ ′ = 22.3°
time difference = (time for light to travel 6.20 m in ice) – (time to travel 6.20 m in air)
∆t = 6.20 m − 6.20 m v ice
v
bu t
c
=
c n
(6.20 m ) 1.309 − 1 = (6.20 − ∆t = (6.20 (6.20 m ) (0.309) = 6.39 × 10 9 c
c
c
s = 6.39 6.39 ns
Chapter 35 Solutions
Consid er glass w ith an ind ex of refraction of 1.5, 1.5, w hich is 3 m m thick The sp eed of light i n the glass is
*35.20
3 × 108 m / s 1.5 3 × 10
The extra travel time is
−3 m
2 × 10 8 m / s
= 2 × 108 −
For light of wavelength 600 nm in vacuum and wavelength 3 × 10 − 3 m −7 4 × 10 m
the extra op tical tical path , in w avelengths, is
*35.21
−
m/ s
3 × 10
−3 m
3 × 108 m / s 600 n m 1.5
~ 10–11 s
= 400 n m
in glass,
3 × 10 − 3 m ~ 103 wavelengths 7 − × 6 10 m
(1.00) sin θ1 = (1.66) sin θ 2
Refraction proceeds according to (a) (a)
(1)
= v 2 cos θ 2
For the normal comp comp onent of veloc velocity ity to be be constant, constant,
v 1 cos θ 1
or
(c ) co s θ1 = (c 1.66) co s θ 2
We multiply Equations (1) and (2), obtaining:
sin θ1 co s θ1
= sin θ2 co s θ2
sin 2θ1
or
sin θ1
= sin 2θ 2
= 1.66 co s θ 1 ,
The p hysical
o r t a n θ 1
= 1.66
θ 1 = 58.9°
wh ich ich yields yields (b)
(2)
The solution θ1 = θ 2 = 0 does not satisfy satisfy Equat ion (2) (2) and m ust be rejected rejected . s o lu t io n is 2θ1 = 180° − 2θ 2 or θ 2 = 90.0° − θ 1 . Th en en Eq u at at io n (1) b e co m es es :
Light Light enteri entering ng the glass glass slows slows down and makes a smaller smaller angle with with the norm al. Both effe effect ctss reduce the velocity component parallel to the surface of the glass, so that component cannot remain constant, or will remain constant only in the trivial case
35.22
327
See the sketch sketch show ing the path of the light α an d γ are angles of incidence at ray. mirrors 1 and 2. For triangle abca, 2α + 2 γ + β = 180° or
β
= 180° − 2(α + γ )
(1)
Now for triangle triangle bcdb,
(90.0° − α ) + (90.0° − γ ) + θ = 180° or
θ = α + γ
(2)
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θ 1
= θ 2 = 0
Chapter 35 Solutions
328
Substituting Equation (2) into Equation (1) gives
β = 180° − 2θ
N ote: From Equat ion (2), (2), γ = θ − α . Thus, the ray will foll follow ow a path like that shown only if α < θ . For α > θ , γ is negative and multiple reflections from each mirror will occur before the incident an d reflect reflected ed rays intersect. intersect.
35.23
Let n ( x ) be the index of refraction at distance x below the top of the atmosphere and its value at the planet surface. surface. Then, n ( x = h) = n be its n ( x ) = 1.000 +
(a) (a)
The total time required required to traverse traverse the atmosphere atmosphere is is t =
t =
(b)
n − 1.000 x h
h dx
∫ 0
v
h n x n − 1.000 h (n − 1.000) h 2 h n + 1.000 1 h x dx = + = ∫ 0 ( ) dx = ∫ 0 1.000 + = 2 h c c c ch c 2 3
20.0 × 10 m 3.00 × 10
8
m
1.005 + 1.000 = 2 s
66.8 µ s
The travel travel time time in the absenc absencee of an atmosphere atmosphere w ould ould be h / c . presence of an atmosphere is
n + 1.000 = 1.0025 2 .0025
Thus, the time in th e
times larger or 0.250% 0.250% long er .
Let n ( x ) be the index of refraction at distance x below the top of the atmosphere and n ( x = h) = n be its its value at the planet surface. surface. Then,
35.24
n ( x ) = 1.000 +
(a) (a)
The total time required required to traverse traverse the atmosphere atmosphere is is t =
(b)
n − 1.000 x h
h dx
∫ 0
v
hn x h n − 1.000) h 2 = ∫ 0 ( ) dx = 1 ∫ 0 1.000 + n − 1.000 x dx = h + ( = 2 h c c c ch
The travel travel time time in in the abs absenc encee of of an atmosphere would be h / c . presence of an atmosphere is
n + 1.000 times larger 2
h n + 1.000 c
2
Thus, the time in th e
Chapter 35 Solutions
35.25
From Fig. 35.20
n v = 1.470 1.470 at 400 nm
Then
(1.00) sin θ
δ r − δ v
n r = 1.458 at 700 nm
and
(1.00) s in θ
sin θ
sin sin 30.0°
n1 sin θ 1
sin sin 30.0 30.0°
− sin −1
1.458
sin θ
= n2 sin θ 2
1.470
θ 2
so
(1.00)( (1.00)(sin sin 30.0 30.0 ° )
=
0.171°
n sin θ 1 = sin −1 1 n 2
θ 2
= sin −1
θ 3
= ([(90.0° − 19.5° ) + 60.0°] − 180°) + 90.0° =
n 3 sin θ 3
=
1.50
= n 4 sin θ 4
19.5°
θ 4
so
40.5°
n sin θ 3 (1.50)(si (1.50)(sin n 40.5 40.5 ° ) = sin −1 3 = sin −1 = 1.00 n 4
Taking Φ to be the apex angle and δ m in to be the angle angle of of minim um Equation 35.9, the index of refraction of the prism material is
35.27
sin n
Solving for δ mi n ,
35.28
= 1.458 sin θ r
− sin −1 = θ r − θ v = sin −1 1.458 1.470
∆δ = sin −1
35.26
= 1.470 sin θ v
an d
329
=
δ mi n
Φ = 2 sin sin −1 n sin − Φ = 2 sin sin −1[( 2.20) sin (25. 0°)] − 50.0° = 2
(a)
(1.0 (1.00) 0) sin 75.0 75.0°° = 1.45 1.458 8 sin sin θ 2;
(b)
Le t
θ 3
So
60.0° − θ 2
θ 2
deviation, deviation, from
Φ + δ m in 2 sin ( Φ 2)
n (700 nm) = 1.458
+ β = 90.0° ,
77.1°
θ 2 = 41.5°
+ α = 90.0° ;
then α + β + 60.0° = 180°
− θ 3 = 0 ⇒ 60.0° − 41.5° = θ 3 =
(c) (c)
1.45 1.458 8 sin 18. 18.5° 5° = 1.0 1.00 0 sin sin θ 4
(d)
γ = (θ 1 − θ 2 ) + [β − (90.0° − θ 4 )]
18.5°
θ 4 = 27.6°
γ = 75.0° − 41.5° + (90.0° − 18.5° ) − (90.0° − 27.6° ) = 42.6°
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86.8°
330
35.29
Chapter 35 Solutions
=
For the incoming ray,
sin θ 2
Using the figure to the right,
(θ 2 )violet
(θ 2 )re d
sin θ 1 n
sin 50.0 50.0° = sin −1 sin = 27.48° 1.66 sin 50.0 50.0°
= sin −1 θ 3′
For the outgoing ray,
= 60.0° – θ 2
(θ 4 )violet (θ 4 )re d
35.30
n
=
Φ,
35.31
sin θ 4
= n sin θ 3
= sin −1[1.66 sin 32.52° ] = 63.1 63.17 7°
= sin −1[1.62 sin 31.78° ]= 58.56 .56°
Φ + δ m in ≈ Φ so is also a small 2 ( sin θ ≈ θ when θ << 1 rad ), we have:
δ mi n
approximation
≈
and
4.61°
Φ + δ mi n 2 sin (Φ 2)
For small
n
= 28.22°
∆ θ 4 = (θ 4 )violet − (θ 4 )re d = 63.17° – 58.56° =
The d ispersion ispersion is the d iffe ifference rence
sin
1.62
(Φ + δ m in ) Φ2
2
=
Φ + δ mi n Φ
At the first refraction,
or
δ m in
≈ (n − 1)Φ
angle.
where
Then, using the small ang le
Φ
(1 .00) sin θ1 = n sin θ 2
The critical angle at the second surface is given by n sin θ 3
= 1.00,
or
θ 3
= sin −1
1.00
1.50
= 41.8° .
But, θ 2 = 60.0° −θ 3 . Thus, to avoid avoid total internal reflection reflection at th e second surface (i.e., have θ 3 < 41.8° ), it is necessary that θ 2 > 18.2° . Since sin θ 1 = n sin θ 2 , this requirement becomes sin θ 1 > (1.50) s in in (18. 2°) = 0.468, or θ 1 > 27.9°
is in in rad ians.
Chapter 35 Solutions
= n sin θ 2 .
At the first refraction, (1.00 ) sin θ 1 the second surface is given by
35.32
n sin θ 3
= 1.00,
θ 3
or
The critical critical angle at
= sin −1(1.00 n )
But (90.0° − θ 2 ) + ( 90.0° − θ 3 ) + Φ = 180° , wh ich ich g ives θ 2
< sin −1(1.00 n ) and θ 2 > Φ − sin −1 (1.00 n ) .
Thus, to have θ 3 necessary necessary th at
− sin θ 1 > n sin Φ − sin 1
n
=
Si S in ce s in θ1
1.00 n
tan ta n
*35.34
sin (δ + φ ) sin ( φ / / 2) sin 5° ( φ ) = 1.544sin − cos5° 1 2
= n s in θ 2 ,
2 θ 1 > sin −1 n
1
1
1
2
2
2
an d
φ = 18.1°
so
= Φ − θ 2 α = θ 1 − θ 2 β = θ 4 − θ 3 θ 3
The total deviation is therefore δ = α + β = θ 1 + θ 4
At exit:
− 1 s in in Φ − cos Φ
2
Φ + (90.0° − θ 2 ) + (90.0° − θ 3 ) = 180°
Thus, θ 3
1.00 n
1
1.544sin
Note for use in every part:
A t e n tr tr y :
sin 5° + sin ( φ ) cos5° ( φ ) = sin (5° + φ ) = cos ( φ ) sin
so
At exit, the deviation is
n1 sin θ 1
= n 2 sin θ 2
or
= 60.0° − 30.0° = 30.0° 1.50 sin 30.0 30.0 ° = 1.00 sin θ 4
− θ 2 − θ 3 = θ 1 + θ 4 − Φ sin 48.6 48.6° sin θ 2 = sin −1 = 30.0° 1.50
or
so the path through the prism is symmetric symmetric when θ 1 (b)
δ = 48.6° + 48.6° − 60.0° = 37.2°
(c)
A t e n tr tr y :
sin θ 2
=
A t e xit :
sin θ 4
= 1.50 sin (31.6°) ⇒
A t e n tr tr y: y:
sin θ 2
=
(d )
this requirement becomes
θ 1 > sin −1 n sin Φ − sin −1
or
At the first surface is
(a )
= Φ − θ 3 .
avoid total internal reflection at the second surface, it is
Through the application of trigonometric identities,
35.33
331
A t exit : sin θ 4
sin 45.6° 1.50
s in 51.6°
⇒
θ 2
= 28.4° θ 4
= 51.7°
θ 4
= sin −1[1.50 .50 sin sin (30.0°)] = 48.6°
= 48.6° .
θ 3
= 60.0° − 28.4° = 31.6°
δ = 45.6° + 51.7° − 60.0° = 37.3°
⇒
θ 2
= 31.5°
θ 3
= 1.50 sin (28.5°) ⇒
θ 4
= 45.7°
δ = 51.6° + 45.7° − 60.0° = 37.3°
1.50
= 60.0° − 31.5° = 28.5°
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Chapter 35 Solutions
332
n sin θ = 1.
35.35
(a)
θ = sin −1
1 = 2.419
(b)
θ = sin −1
1 = 1.66
(c)
θ = sin −1
1 = 1.309
s in θ c
35.36
35.37
From Table 35.1, 35.1,
=
n2 n1
θ c
;
24.4°
37.0°
49.8°
n = sin −1 2 n1
(a )
D i a m on on d: d:
θ c
= = sin −1 2.419
33.4°
(b)
Flin t g la la ss ss :
θ c
= = sin −1 1.66
53.4°
(c)
Ice : Sin ce ce n 2 > n 1, there is is no critic critical al angle .
sin θ c
=
n2 n1
1.333
1.333
(Equation 35.10)
n 2 = n 1 sin 88.8° = (1.0003)(0.9998) = 1.000 08
*35.38
sin θ c
=
n air n pipe
=
1.00 1.36
= 0.735
θ c = 47.3°
Geometry show s that the angle of refracti refraction on at th e end is
θ r = 90.0° – θ c = 90.0° – 47.3° = 42.7° Th en en , Sn Sn el ell' s la w a t th th e en en d , gives
35.39
For total internal reflection,
1.00 s in in θ = 1.36 sin 42.7°
θ = 67.2°
n1 sin θ 1
= n2 sin 90.0°
(1.50) sin θ 1 = (1.33)(1.00)
or
θ 1 = 62.4°
Chapter 35 Solutions
35.40
333
To avoid internal reflection and come out through the vertical vertical face, face, light light inside the cube mu st have
θ 3
< sin −1 (1/ n) > 90.0° − sin −1 (1/
So
θ 2
Bu t
θ1 < 90.0°
n)
n sin θ 2
an d
<1
In t h e cr it ica l ca se, si s in − (1/ n) = 90. 0° − sin − ( 1/ n) 1
1/ n = sin 45.0°
35.41
1
n = 1.41
n1 sin θ1
From Snell's law,
= n2 sin θ 2
At the extreme an gle of viewing, viewing, θ 2 = 90.0° (1.59)(sin θ 1) = (1.00) · sin 90.0° So
θ 1 = 39.0°
Therefore, Therefore, the dep th of the air bu bble is r d
ta n θ 1
*35.42
(a)
tan ta n θ 1
1.08 cm < d < 1.17 cm
sin θ 2
=
sin θ 1
sin θ c
(c)
r p
or
sin 90. 0°
(b)
< d <
v2 v1
=
an d θ 2
= 90.0°
1850 m s 343 m s
a t th th e cr cr iitt ic ica l a n gl gle
so θ c
= s in −1 0.185 =
10.7°
Soun d can be totally totally reflec reflected ted if it is traveling traveling in the medium w here it it travels travels slower:
air
Sound in air air fall falling ing on the wall wall from from most directi directions ons is is 100 100% % refl reflec ected ted , so the w all all is a good mirror.
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334
*35.43
Chapter 35 Solutions
For plastic with index of refraction n ≥ 1.42 surrou nd ed by air, the critic critical al angle for for total internal reflection is given by 1 ≤ sin −1 1 = 44.8° θ c = sin −1 n 1.42 In the gasoline gauge, skylight skylight from a bove travels dow n the p lastic lastic.. The rays close close to to t h e vertical are totally reflected from both the sides of the slab and from facets at the lower end of the plastic, plastic, wh ere it is not immersed in gasoline. This light return s u p inside th e plastic and makes it look bright. bright. Wher e the plastic plastic is immersed in gasoline, with ind ex of refraction refraction about 1.50, 1.50, total internal reflec reflection tion should n ot h app en. The light passes out of the lower end of the p lastic lastic with little little reflec reflected, ted, making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, the index of refraction of the plastic should be
n < 2.12
,
since
θ c
*35.44
1.50 = sin −1( 2.12 ) = 45.0° .
Assume the lifeguard’s path makes angle θ 1 with the northsouth normal to the shoreline, and angle θ 2 with this normal in the wa ter. By Fermat’s Fermat’s principle, his his path sh ould follow the law of refraction: sin θ 1 sin θ 2
=
v1 v2
=
7. 00 00 m s 1. 40 40 m s
= 5. 00
or
θ 2
= sin −1
sin θ 1 5
The lifeguard on land travels eastward a distance x = (16.0 m ) t a n θ 1 . Then in the water, h e t r a v e ls 26.0 m − x = (20 .0 m ) t a n θ 2 fu r t h er e a s t . Th u s, 26.0 m = (16.0 m ) t a n θ1 + (20.0 m ) t a n θ 2 or
26.0 26.0 m
− sin θ 1 = (16.0 16.0 m ) ta n θ 1 + ( 20.0 20.0 m ) t a n sin 1 5
We home in on the solution as follows:
θ 1 (d eg ) r ig h t -h a n d si si d e
50.0 22.2 m
60.0 3 1. 2 m
54.0 25.3 m
54.8 25.99 m
54.81 26.003 m
The life lifeguard guard should start run ning at 54.8 54.8°° east of north .
*35.45
= n1 sin θ 1
Let the air and glass be mediu m 1 and 2, respectively. By Snell's Snell's law law ,
n 2 sin θ 2
or
1.56 sin θ 2
= sin θ 1
1.56 1.56 sin θ 2
= sin 2θ 2
But the conditions of the problem are such that θ 1
= 2θ 2 .
We now use the double-angle trig identity suggested.
1.56 sin θ 2
or
cos θ 2
Thus, θ 2
= 38.7°
and
θ 1 = 2θ 2
=
77.5°
= 2 sin sin θ 2 cos θ 2
= 1.56 = 0.780 2
Chapter 35 Solutions
*35.46
(a)
θ1′ = θ 1
=
n1 sin θ1
335
30.0°
= n2 sin θ 2
(1.00) sin 30.0° = 1.55 sin θ 2
θ 2
(b)
=
18.8°
θ 1′ = θ 1 = 30.0° θ 2
n sin θ 1 1.55 sin 30.0° = sin −1 1 = sin −1 = 1 n 2
(c) (c) and (d)
The other entries entries are comp comp uted similarly, similarly, and are shown in the table below.
(c) air air into into glass glass,, angles angles in degree degreess in ci d en en c e 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0
50.8°
r e f l e ct i o n 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0
r e fr a ct i o n 0 6.43 12.7 18.8 24.5 29.6 34.0 37.3 39.4 40.2
(d) glas glasss into air, air, angles angles in degree degreess i n cid en en ce 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0
r e fle ct i o n 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0
r e fr a ct i o n 0 15.6 32.0 50.8 85.1 n on e* n on e* n on e* n on e* n on e*
*total internal reflection
35.47
=
1
=
3
For water,
sin θ c
Thus
θ c
an d
d = 2 [(1.00 m)tan θ c ]
4/ 3
4
= sin −1 (0.750) = 48.6°
d = (2.00 (2.00 m)tan 48.6 48.6° = 2.27 m
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Figure for Goal So lution
Chapter 35 Solutions
336
Goal Solution A small und erwater p ool light light is 1.00 1.00 m below the su rface. rface. The light light emerging from the water forms a circle circle on the wat er's surface. What is the diam eter of this circle circle??
G:
Only the light light that is directed directed u pw ards and hits the water’s water’s surface surface at less than the critic critical al angle w ill be transm itted to the air so that someon e outside can see see it. The light light that hits th e su rface rface farth er from the center at an angle greater than θ c will be totally reflected within the water, unable to be seen from from th e outside. From the d iagram above, the diameter of this this circle circle of light light app ears to be about 2 m.
O:
We can apply Snell’ Snell’ss law to find find the critical critical angle, and th e d iameter can th en be found from th e geometry.
A:
The critical angle is found when the refracted ray just grazes the surface ( θ 2 = 90°). The ind ex of refraction of water is n 2 = 1.33, an d n 1 = 1.00 1.00 for air, so n1 sin θ c
L:
= n2 sin
90°
g iv iv es es
θ c
= sin −1
= sin −1(0.750) = 48.6° 1.333
=
1
r
The rad ius then satisfies satisfies
tan θ c
So the diameter is
48.6° = 2.27 m d = 2 r = 2(1.00 m ) tan 48.6
(1.00 (1.00 m )
Only the light rays within a 97.2 ° cone above the lamp lamp escape escape the w ater and can be seen by an outside observer (Note: (Note: this angle angle does not depend on th e depth of the light source). The p ath of a light light ray is always reversible, so if a person were located beneath the water, they could see the whole hemisphere above the water surface within this cone; this is a good experiment to try the next time you go swimming!
*35.48
Call θ 1 the angle of incidence and of reflection on the left face and θ 2 those angles on the right face. Let α represent the complement of θ 1 an d β be the complement of θ 2 . Now α = γ an d β = δ because they are pairs of alternate interior angles. We have A
an d
= γ + δ = α + β
B = α + A
+ β = α + β + A =
2 A
Chapter 35 Solutions
*35.49
(a) (a)
337
We see see the Sun Sun swinging around a circl circlee in in the extended extended p lane of our parallel parallel of latitude. latitude. Its angular speed is
ω =
∆θ 2 π ra d = = 7.27 × 10−5 ∆t 86 400 400 s
rad s
The d irection irection of sunlight crossing crossing the cell cell from from the w indow changes at this rate, mov ing o n the opp osite osite wall at speed v
(b)
= r ω ω = ( 2.37 m )(7.27 × 10 − 5
rad s
) = 1.72 × 10− 4
m s
=
0.17 .172 mm s
The mirror mirror folds folds into the cel celll the motion that would occur occur in a room tw ice ice as as wide: v
= r ω ω = 2 (0. 174
mm m m s ) = 0. 34 345 m m s
(c) (c) and (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the window as fix fixed, ed, and both patches patches of light light move
*35.50
n1 sin θ 1
By Snell's law, With v =
northward and d ownw ard at 50.0 50.0°° .
c , n
c v1
This is also true for sound . Here,
sin θ 1
sin 12.0° 340 m / s
= n2 sin θ 2 = =
c v2
sin θ 2
or
s in θ1 v1
=
sin θ 2 v2
sin θ 2 1510 m / s
θ 2 = arcsin (4.44 (4.44 sin 12.0°) 12.0°) = 67.4°
*35.51
=c=
2.998 × 10 8 m s
(a)
n
(b)
n1 sin θ 1
v
61.15
km 1.00 h 1.00 × 10 3 m 1.00 km h r 3600 s
= n2 sin θ 2
so
=
1.76 × 10
7
(1.76 × 107 ) sin θ 1 = (1.00) sin 90.0°
− θ 1 = 3. 25 25 × 10 6 degree This problem is misleading. The sp eed of energy transport is slow, slow, but the speed of the wa vefront ad vance is norm ally fast. fast. The cond cond ensate's index of refraction refraction is not far from unity.
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338
*35.52
Chapter 35 Solutions
Violet light: sin 25.0° = 1.689 s in θ 2 ⇒ θ 2 = 14.490° (1.00) si y v
= (5.00 cm ) t a n θ 2 = (5.00 cm ) ta n 14 .490° = 1.2622 cm
Red Light: sin 25.0° = 1.642 sin θ 2 ⇒ θ 2 = 14.915° (1.00) si y R
= (5.00 cm ) t a n 14.915° = 1.3318 cm
The emergent beams are both at 25. 25.0° 0° from the normal. Thus, w
= ∆y co s 25.0°
∆ y = 1.3318 cm − 1.2622 cm = 0.0396 cm
w h er e
w
35.53
= (0.396 m m ) co s 25 .0° =
0.359 mm
Horizontal light rays from the setting Sun pass above the the hiker. The light light rays are twice refracted and once reflected, as in Figure (b) below, by just the certain special raindrops at 40.0° to 42.0° from the hiker's shadow, and reach the hiker as the rainbow. The hiker sees a greater percentage of the violet inner edge, so we consider the red outer edge. The radius R of the circle of drop lets is is
Figure (a)
R = (8.00 km)(sin 42.0°) = 5.35 km
Then the angle φ , between the vertical and the radius where the bow touches the ground, is given by cos φ =
2.00 k m R
=
2.00 k m 5.35 km
= 0.374
or
The ang le filled filled by th e visible bow is 360° 360° – (2 (2 so the visible bow is 224° = 62.2% of a circle circle 360°
φ = 68.1°
× 68.1°) = 224°, Figure (b)
Chapter 35 Solutions
35.54
(1.00) s in θ1 =
From Snell’s nell’s law,
4 3
sin θ 2
z
r
so
=
R
sin θ 2
3
=
sin θ 1
actual actual dep th sin 2 θ 2 z
So
d
=
=
3 4
z
= =
r cos θ 1
d R cos θ 2
=
cos θ 1
3
4 1 − sin 2 θ 2
2
= 9 1 − cos 2 θ 1) 16 (
cos θ 1 1−
Fish Fish at dep th d Image at depth z
R
sin θ 1
3 4
θ 2
Fish
apparent depth
Bu t
r
d
4
eye
θ 1
x
= R sin θ 2 = r sin θ 1
x
339
9 16
+
9 16
=
cos 2 θ 1
cos θ 1
3 4
7 + 9 cos 2 θ 1
or
z
=
3d cos θ 1 7 + 9 cos 2 θ 1
16
A s t h e b e a m e n t e r s t h e sla b , (1.00 0 0 ) sin 50.0 ° = (1.48 4 8 ) sin θ 2 giving θ 2 = 31.2° . Th e b e a m t h en en strikes the top of the slab at x1 = 1.55 m m ta n (31.2°) fr om om t h e le ft ft e nd nd . Th er er ea ea ft ft eerr , t he he be be am am strikes a face face each time it has traveled a d istance of 2 x1 along the length of the slab. slab. Since Since th e slab is is 420 420 mm long, the beam has an add itional itional 420 420 m m − x1 to travel after the first reflection. The number of additional reflections is
35.55
420 420 m m − x1 2 x 1
=
420 mm
− 1. 55 55 m m
t an an (31. 2°)
3. 10 10 m m t an an (31. 2°)
= 81.5
or 81 reflections since the answer must be an integer. The total total number of reflect reflections ions mad e in the slab slab is then 82 .
*35.56
(a)
S 1′
(b)
S 1′
S 1
If medium medium 1 is is glass glass and medium 2 is is air, air,
S 1
2
n 2 − n1 1.52 − 1.00 2 = = = n 2 + n1 1.52 + 1.00
0.0426
2
n 2 − n1 1.00 − 1.52 2 = = 0.0426; = n 2 + n1 1.00 + 1.52
There is no difference difference
(c)
S 1′ S 1 S 1′
2
1.76 × 107 − 1.00 1.76 × 107 + 1.00 − 2.00 = = 7 7 1.76 × 10 + 1.00 1.76 × 10 + 1.00
2
2
2 .00 2 .00 ≈ 1.00 − 2 = 1.00 − 2.27 × 10−7 = 1.00 − 7 7 1.76 × 10 + 1.00 S 1 1.76 × 10 + 1.00 This suggests he app earance w ould be
or
100%
very shiny, reflecting reflecting pr actically actically all incident light
See, however, the note concluding the solution to problem 35.51.
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.
Chapter 35 Solutions
340
*35.57
(a )
= 1 and
W i t h n1
n2
= n , the reflected fractional intensity is
2 n = − 1 . S 1 n + 1
S 1′
The remaining intensity must be transmitted: S 2 S 1
(b)
*35.58
2
= 1 − n − 1 = ( n + 1
A t e n tr tr y ,
S 2
At exit,
S 3
Overall,
S 3
Define T =
S 1
S 2
S 1
4n
(n + 1)
− (n − 1)2 n 2 + 2n + 1 − n 2 + 2n − 1 4n = = 2 2 2 (n + 1) (n + 1) (n + 1)
n + 1)
2
2
4( 2. 41 419) n − 1 = 1 − = n + 1 (2.419 + 1)2 = 0.828
= 0.828 S S = 3 2 = (0.828)2 = 0.685 S 2 S 1
or
68.5%
as the transmission coefficient for one
2
encounter with an interface. For d iamon d and a ir, it is 0.828 0.828,, as in p ro blem 57. As shown in the figure, the total amount transmitted is
+ T 2 (1 − T )2 + T 2 (1 − T ) 4 + T 2 (1 − T )6 + T 2 (1 − T )2n + . . . 2
T
+ ...
We have 1 − T = 1 − 0.828 = 0.172 transmission is
so
[
the
2 2 4 6 (0.828) 1 + (0.172 ) + (0.172) + (0.172) + . . .
total
]
+ (0.172) 4 + (0.172)6 + . . . . 2 2 4 6 N ote tha t (0.172 0.172) F = (0.172) + (0.172) + (0.172) + . . . , a n d 2 2 4 6 1 + (0.172) F = 1 + (0.172) + (0.172) + (0.172) + . . . = F . To sum this series, define F = 1 + (0.172)
Then,
2
1 = F − (0.172) F or
F =
2
1 1 − (0.172)
2
2
The overall transmission is then
(0.828) = 0.706 o r 70.6% 2 1 − (0.172)
Chapter 35 Solutions
n sin 42.0° = sin 90.0°
35.59
sin θ 1
= n sin 18.0°
=
1
so
n
an d
sin θ 1
sin 42.0°
=
341
= 1.49
sin 18.0° sin 42.0°
θ 1 = 27.5° Figure for Goal Solu tion
Goal Solution The light beam sh ow n in Figu re P35.59 P35.59 strikes sur sur face 2 at the critical critical angle. Deter min e th e angle o f incidence θ 1 .
G:
From the d iagram it app ears that the an gle of incidence incidence is is about 40 °.
O:
We can find θ 1 by app lying Snell’ Snell’ss law at the first interface wh ere th e light is refracted. At su rface 2, 2, know ing that the 42.0 42.0° angle of reflection is the critical angle, we can work backwards to find θ 1 .
A:
Define n1 to be the index of refraction of the surrounding medium and n 2 to be that for the prism ma terial. We can use the critical critical angle of 42.0 42.0 ° to find find the ratio n 2 n1 : 42.0° = n1 sin 90.0 90.0° n 2 sin 42.0 So ,
n2 n1
=
1 sin sin 42.0 42.0°
= 1.49
Call the angle of refraction θ 2 at the surface surface 1. The ray inside inside th e p rism forms a triangle triangle w ith sur faces faces 1 and 2, so the sum of the interior angles of this triangle mu st be 180 180°. °. Thu s,
(90.0° − θ 2 ) + 60.0° +(90.0° −42.0°) = 180°
= 18.0°
Therefore,
θ 2
Applying Snell’s law at surface 1,
n1 sin θ 1
sin θ 1
= n 2 sin sin
= (n 2
18.0 18.0°
n1 ) sin θ 2
= (1.49) sin 18.0°
θ 1 = 27.5°
L:
The result is a bit less than the 40.0 ° we expected, but this is probably because the figure is not drawn to scale scale.. This pr oblem w as a bit tricky becau becau se it it requ ired fou r k ey concepts concepts (refraction, reflection , critic critical al angle, and g eometry) in in order to find find the solution. solution. One practical practical extension extension of this problem is to consider what would happen to the exiting light if the angle of incidence were varied slightly. Would all the light light still still be reflec reflected ted off surface surface 2, or or w ould some light light be refracted refracted and pass thr ou gh this second surface? surface?
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Chapter 35 Solutions
342
Light passing the top of the pole makes an angle of incidence φ 1 = 90.0° − θ . It falls falls on th e water surface at distance
35.60
s1
L d =( − )
ta n θ
from the pole,
and has an angle of refracti refraction on φ 2 from (1.00)sin φ 1 Then s2 = d ta n φ 2 and the whole shadow length length is s1 + s2
=
s1 + s2 =
35.61
(a )
L − d
sin φ 1 + d ta n sin −1 n ta n θ
40.0° − cos θ = 2.00 m + 2.00 m t a n s in −1 cos 40.0 + d t a n s in 1 = ( ) n tan 40.0° 1.33 tan ta n θ
L − d
= sin −1 1.00 = 42.2° 1.49
Then from the geometry,
θ 2 = 90.0° – θ 3 = 47.8°
From Snell's law,
sin θ 1
This has no solution.
= (1.49) sin sin 47.8 7.8° = 1.10
Therefore, Therefore, total internal reflecti reflection on
= sin −1
Fo r p ol o ly st st yr yr en en e surrounded by water ,
θ 3
an d
θ 2 = 26.8°
always hap pens .
1.33 1.49
= 63.2°
θ 1 = 30.3°
From Snell's law,
(c)
3.79 m
Fo r p ol oly st st y re re ne ne surrounded by air , internal reflecti reflection on requires
θ 3
(b)
= n sin φ 2 .
No internal internal refract refraction ion is possible possible since since the beam beam is is initi initial ally ly traveli traveling ng in a medium of lower index of refraction.
*35.62
δ
= θ1 − θ 2 = 10.0°
Thus,
an d
n1 sin θ 1
θ 1
= n2 sin θ 2
with
n1
= 1,
n2
=
4 3
= sin −1(n 2 sin θ 2 ) = sin −1[n 2 sin(θ 1 − 10.0° )]
Chapter 35 Solutions
343
(You can use a calculator to home in on an approximate solution to this equation, testing different values of θ 1 until you find that θ 1 = 36.5 36.5°° . Alternatively, you can can solve solve for θ 1 exactly, as shown below.) sin θ 1
We are given that
3
This is the sine of a difference, so
4
= 4 sin(θ 1 − 10.0° ) 3
sin θ 1
= sin θ 1 cos 10.0 10.0° − cos θ 1 sin sin 10.0°
s in 10.0° co s θ1
Rearranging,
3 = co s 10.0° − s in θ 1 4
sin 10.0° co s 10. 0° − 0. 75 750
35.63
tan ta n θ 1 2
=
t a n θ1
4.00 cm
tan ta n θ 2
an d
h
=
= ta n θ 1
an d
θ 1 = t a n −1 0.740 = 36.5°
2.00 cm h
= (2 .00 t a n θ2 )2 = 4.00 t a n 2 θ 2
sin 2 θ 1 2
(1 − sin θ 1 )
= 4.00
sin 2 θ 2
(1)
2
(1 − sin θ 2 )
Snell's law in this case is: n1 s in θ1
= n2 sin θ 2
s in θ 1
sin 2 θ 1
Squaring both sides,
=
sin θ 2
= 1.777 sin 2 θ 2
1.777 sin 2 θ 2
Substituting (2) into (1),
Defining x
= 1.333
1 − 1.777 sin 2 θ 2 0.444
sin 2 θ ,
(1 − 1.777 x )
=
From x we can solve for θ 2 :
θ 2
h=
(2.00 cm ) tan ta n θ 2
=
1 − sin 2 θ 2
(1 − x )
0.444 − 0.444 x
Thus, the height is
= 4.00
sin 2 θ 2
1
Solving for x ,
= sin −1
(2)
= 1 − 1.777 x
0.417
= 40.2°
(2.00 cm ) t a n ( 40.2°)
=
2.37 cm
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an d
x
= 0.417
344
35.64
Chapter 35 Solutions
Observe in the sketch that th e angle of incidence incidence at point P is γ , and using triangle OPQ: sin γ = L / R . Also,
cos γ = 1 − sin 2 γ =
Applying Snell’s law at point P,
(1.00) sin γ = n sin φ
Thus,
sin φ =
and
cos φ = 1 − sin 2 φ =
sin γ n
=
R
2
− L2
R
L nR 2
n R
2
− L2
nR
From triangle OPS, φ + (α + 90.0°) + (90.0° − γ ) = 180° or the angle of incidence at point S is Snell’s law at poin t S gives (1.00) sin θ = n sin α = n sin ( γ − φ ) , o r α = γ − φ . Then, app lying Snell’
L R
sin θ = n [sin γ cos φ − cos γ sin φ ] = n
sin θ =
35.65
L R
2
n 2 R 2 − L2
−
R 2 − L2
n 2 R 2 − L2
an d
nR
−
R 2 − L2 R
L R 2
θ = sin −1
L nR n 2 R 2 − L2
−
R 2 − L2
To derive the law of reflection , locate point O so that the time of travel from point A to point B will be minimum. Th e total light light p ath is L = a se c θ1 + b se c θ 2 The time of travel is
t =
1 ( a sec θ + b sec θ ) 1 2 v
If point O is displaced by dx , then dt =
1 ( a sec θ ta n θ d θ θ 2 ) = 0 1 1 θ 1 + b sec θ 2 ta n θ 2 d θ v
(1)
(since for minimum time dt = 0). Also,
c + d = a ta n θ 1 + b ta n θ 2
so, so ,
θ 1 + b sec 2 θ 2 d θ θ 2 a sec 2 θ 1 d θ
Divide equations (1) and (2) to find
= constant =0
θ 1 = θ 2
(2)
Chapter 35 Solutions
35.66
345
As shown in the sketch, the angle of incidence at point A is:
(d 2) = sin −1 1.00 m = 30.0° 2.00 m R
θ = sin −1
If the emerging ray is to be parallel to the incident ray, the path must be symmetric about the center line CB o f the cylind cylind er . In the isosceles isosceles triang le A BC , γ = α a n d β = 180° − θ . Th er efo r e, α + β + γ = 180° becomes
2α
+ 180° − θ = 180°
or α =
θ 2
= 15.0°
Then, applying Snell’s law at point A , n sin α or
35.67
(a) (a)
n
=
sin θ sin α
s in 30.0° sin 15.0°
=
1.93
At the boundary of the air air and glass, glass, the crit critic ical al angle angle is given given by 1 sin θ c = n
=
Consider the critical ray PB B′ :
tan ta n θ c
Squaring the last equation gives:
sin 2 θ c
Since sin θ c
(b)
=
= (1.00) sin θ
=
1 n
, this becomes
(c)
= 1.52 and t = 0.600 0.600 cm cm ,
=
cos θ c
d
=
4t 2
sin 2 θ c 1 − sin 2 θ c
d = 4t
2
d = 2 n − 1 4t
d =
sin θ c
or
t
1
d =
So lv lv in in g fo fo r d ,
Thus, if n
cos 2 θ c
d 4
or
n
=
1 + ( 4t d )
2
4t n2
−1
4(0.600 .600 cm ) 2 (1.52) − 1
=
2.10 cm
Since ince violet violet light has a larger larger index of refract refractio ion, n, it will will lead lead to a small smaller er criti critical cal angle
and the inner inner edge of of the white halo will be tinge tinged d w ith ith violet
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light. light.
Chapter 35 Solutions
346
From the sketch, observe that the angle of incidence at point A is the same as the
35.68
prism angle θ at point O . Given tha t θ = 60.0° , application of Snell’s law at point A gives 1.50 sin β = 1.00 sin 60.0°
β = 35.3°
or
From triangle A OB , we calculate the angle of incidence (and reflection) at point B .
θ
35.69
(a )
+ (90.0° − β ) + (90.0° − γ ) = 180°
= θ − β = 60.0° − 35.3° = 24.7°
γ
so
Now, using triangle BCQ :
(90.0° − γ ) + (90.0° − δ ) + (90.0° − θ ) = 180°
Thus the angle of incidence at point C is
δ
Finally, Snell’s law applied at point C gives
1.00 sin φ = 1.50 sin 5.30°
or
φ = sin − 1(1.50 sin 5.30°) = 7.96°
G iv en en th t h at at θ1 = 45.0° and θ 2 the first surface gives
= 76.0° ,
= (90.0° − θ ) − γ = 30.0° − 24.7° = 5.30°
Snell’s Snell’s law at
n sin α = (1.00) s in 45.0°
(1)
Observe that the angle of incidence at the second surface is β = 90.0° − α . Th u s, s, Sn e ll’s la w a t t h e second surface yields n sin β
= n sin (90.0° − α ) = (1.00) sin 76.0° ,
or
n co s α = sin 76.0°
(b)
(2)
Dividing Equation (1) (1) by Equation (2), (2),
tan α =
Then, from Equation (1),
n
=
sin 45.0° sin 76.0°
sin 45.0° sin α
=
= 0.729 729
sin 45.0° sin 36.1°
or
=
α = 36.1°
1.20
From From the sketch, sketch, observe observe that the distance distance the light light travels travels in the plastic plastic is d = L sin α . Also, the speed of light in the plastic is v = c n , so the time required to travel through the plastic is t =
d v
=
nL c sin α
=
(1.20)(0.500 m )
(3.00 × 10
8
)
m s sin sin 36.1°
= 3.40 × 10− 9 s =
3.40 ns
Chapter 35 Solutions
35.70
sin θ 1 0.174 0.342 0.500 0.643 0.766 0.866 0.940 0.985
sin θ 2 0.131 0.261 0.379 0.480 0.576 0.647 0.711 0.740
sin θ 1 / sin θ 2 1.3304 1.3129 1.3177 1.3385 1.3289 1.3390 1.3220 1.3315
The straightness of the graph line demonstrates Snell's proportionality. The slope of the line is n = 1.32 3 276 ± 0.01 01 an d
n = 1.328 ± 0.8%
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Harcourt Harcourt Inc. nc. All All ri hts reserv reserved. ed.
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