286
P35.7
Chapter 35
Let d represent the perpendicular distance
from the person to the mirror. The distance
between lamp and person measured parallel
to the mirror can be written in two ways:
2d tan θ + d tan θ = d tan φ . The condition on the
distance traveled by the light
is 2 d = 2 d + d . We have the two
cos φ cos θ cos θ
q
f
d tan q
2d
d
equations 3 tan θ = tan φ and 2 cos θ = 3 cos φ .
To eliminate φ we write
9 sin 2 θ sin 2 φ
=
cos 2 θ
cos 2 φ
2d tan q
q
d tan f
d
FIG. P35.7
4 cos 2 θ = 9 cos 2 φ
9 cos 2 φ sin 2 θ = cos 2 θ (1 − cos 2 φ )
4
4 cos 2 θ sin 2 θ = cos 2 θ ⎛ 1 − cos 2 θ ⎞
⎝
⎠
9
4
4 sin 2 θ = 1 − (1 − sin 2 θ ) 36 sin 2 θ = 9 − 4 + 4 sin 2 θ
9
5
sin 2 θ =
θ = 23.3°
32
*P35.8
The excess time the second pulse spends in the ice is 6.20 m/[(3.00 × 108 m/s)/1.309] = 27.1 ns
P35.9
Using Snell’s law,
sin θ 2 =
n1
sin θ1
n2
θ 2 = 25.5°
λ2 =
λ1
= 442 nm
n1
FIG. P35.9
*P35.10 The law of refraction n1 sin θ1 = n2 sin θ 2 can be put into the more general form
c
c
sin θ1 = sin θ 2
v1
v2
sin θ1 sin θ 2
=
v1
v2
In this form it applies to all kinds of waves that move through space.
sin 3.5°
sin θ 2
=
343 m s 1 493 m s
sin θ 2 = 0.266
θ 2 = 15.4°
The wave keeps constant frequency in
f =
v1 v2
=
λ1 λ2
λ2 =
v2 λ1 1 493 m s ( 0.589 m )
=
= 2.56 m
v1
343 m s
The light wave slows down as it moves from air into water but the sound speeds up by a large
factor. The light wave bends toward the normal and its wavelength shortens, but the sound wave
bends away from the normal and its wavelength increases.
The Nature of Light and the Laws of Geometric Optics
P35.11
287
We find the angle of incidence:
n1 sin θ1 = n2 sin θ 2
1.333 sin θ1 = 1.52 sin 19.6°
θ1 = 22.5°
The angle of reflection of the beam in water is then also 22.5° .
P35.12
P35.13
c 3.00 × 10 8 m s
= 4.74 × 1014 Hz
=
λ 6.328 × 10 −7 m
(a)
f =
(b)
λglass =
λair 632.8 nm
=
= 422 nm
n
1.50
(c)
vglass =
cair 3.00 × 108 m s
=
= 2.00 × 108 m s = 200 Mm s
n
1.50
n1 sin θ1 = n2 sin θ 2
sin θ1 = 1.333 sin 45°
sin θ1 = (1.33)( 0.707 ) = 0.943
θ1 = 70.5° → 19.5° above the horizon
FIG. P35.13
⎛n
⎞
*P35.14 From Snell’s law, sin θ = ⎜ medium ⎟ sin 50.0°
⎝ n
⎠
12 cm
liver
But
nmedium c vmedium
v
=
= liver = 0.900
nliver
c vliver
vmedium
so
θ = sin −1 ⎡⎣( 0.900 ) sin 50.0° ⎤⎦ = 43.6°
50°
nmedium
d
n liver
h
q
From the law of reflection,
d=
h=
P35.15
12.0 cm
= 6.00 cm, and
2
Tumor
FIG. P35.14
d
6.00 cm
=
= 6.30 cm
tan θ tan ( 43.6° )
n1 sin θ1 = n2 sin θ 2 :
⎛ n sin θ1 ⎞
θ 2 = sin −1 ⎜ 1
⎝ n2 ⎟⎠
⎧1.00 sin 30° ⎫
θ 2 = sin −1 ⎨
⎬ = 19.5°
⎩ 1.50
⎭
θ 2 and θ 3 are alternate interior angles formed by the ray cutting
parallel normals.
So,
θ 3 = θ 2 = 19.5°
1.50 sin θ 3 = 1.00 sin θ 4
θ 4 = 30.0°
FIG. P35.15
288
P35.16
Chapter 35
sin θ1 = nw sin θ 2
q 1 = 62°
1
1
sin θ 2 =
sin θ1 =
sin ( 90.0° − 28.0° ) = 0.662
1.333
1.333
θ 2 = sin −1 ( 0.662 ) = 41.5°
h=
d
3.00 m
=
= 3.39 m
tan θ 2 tan441.5°
air
n = 1.00
28°
q2
water
n = 1.333
h
3.0 m
FIG. P35.16
P35.17
At entry,
n1 sin θ1 = n2 sin θ 2
or
1.00 sin 30.0° = 1.50 sin θ 2
θ 2 = 19.5°
The distance h the light travels in the medium is given by
cos θ 2 =
or
h=
2.00 cm
h
2.00 cm
= 2.12 cm
cos19.5°
FIG. P35.17
α = θ1 − θ 2 = 30.0° − 19.5° = 10.5°
The angle of deviation upon entry is
The offset distance comes from sin α =
P35.18
d
:
h
d = ( 2.21 cm ) sin 10.5° = 0.388 cm
The distance h traveled by the light is
h=
2.00 cm
= 2.12 cm
cos19.5°
The speed of light in the material is
v=
c 3.00 × 108 m s
=
= 2.00 × 108 m s
n
1.50
Therefore,
t=
h
2.12 × 10 −2 m
= 1.06 × 10 −10 s = 106 ps
=
8
v 2.00 × 10 m s
Image Formation
(c)
313
1 1 1
1
1
= − =
−
=0
q f p 10.0 cm 10.0 cm
Thus,
q = infinity
No image is formed . The rays are reflected parallel to each
other.
P36.9
(a)
(b)
1 1 2
+ =
p q R
becomes
1
2
1
=
−
q 60.0 cm 90.0 cm
q = 45.0 cm
and
M=
1 1 2
+ =
p q R
becomes
1
2
1
=
−
q 60.0 cm 20.0 cm
−q
( −60.0 cm )
M=
=−
= 3.00
p
( 20.0 cm )
q = −60.0 cm and
(c)
P36.10
−q
45.0 cm
=−
= −0.500
p
90.0 cm
The image (a) is real, inverted, and diminished. That of (b) is
virtual, upright, and enlarged. The ray diagrams are similar to
Figures 36.13(a) and 36.13(b) in the text, respectively.
FIG. P36.9
With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length
f =
R
= 1.25 m
2
In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is concentrated in a sound image at distance q from the back of the niche, where
1 1 1
+ =
p q f
so
1
1
1
+ =
2.00 m q 1.25 m
q = 3.33 m
P36.11
(a)
(b)
M = −4 = −
q
p
q = 4p
q − p = 0.60 m = 4 p − p
p = 0.2 m
1 1 1
1
1
= + =
+
f p q 0.2 m 0.8 m
f = 160 mm
M =+
1
q
=−
2
p
q = 0.8 m
p = −2q
q + p = 0.20 m = − q + p = − q − 2q
q = −66.7 mm
p = 133 mm
1 1 2
1
1
+ = =
+
p q R 0.133 m − 0.066 7 m
R = −267 mm
Image Formation
Section 36.4
P36.24
Thin Lenses
Let R1 = outer radius and R2 = inner radius
⎡1
1
1 ⎤
1
1
⎤ = 0.050 0 cm −1
−
= ( n − 1) ⎢ − ⎥ = (1.50 − 1) ⎡⎢
⎥⎦
.
m
.
f
R
R
2
00
2
50
cm
⎣
⎣ 1
2 ⎦
so
P36.25
f = 20.0 cm
For a converging lens, f is positive. We use
(a)
1 1 1
+ = .
p q f
1 1 1
1
1
1
= − =
−
=
q f p 20.0 cm 40.0 cm 40.0 cm
M =−
q = 40.0 cm
q
40.0
=−
= −1.00
p
40.0
The image is real, inverted , and located 40.0 cm past the lens.
(b)
1 1 1
1
1
= − =
−
=0
q f p 20.0 cm 20.0 cm
q = infinity
No image is formed. The rays emerging from the lens are parallel to each other.
(c)
1 1 1
1
1
1
= − =
−
=−
q f p 20.0 cm 10.0 cm
20.0 cm
M =−
q = −20.0 cm
q
( −20.0 )
=−
= 2.00
p
10.0
The image is upright, virtual and 20.0 cm in front of the lens.
P36.26
(a)
1 1 1
+ = :
p q f
f = 6.40 cm
so
P36.27
1
1
1
+
=
32.0 cm 8.00 cm f
q
8.00 cm
=−
= −0.250
p
32.0 cm
(b)
M =−
(c)
Since f > 0, the lens is converging .
We are looking at an enlarged, upright, virtual image:
M=
h′
q
=2=−
h
p
1 1 1
+ =
p q f
q
( −2.84 cm )
=−
= +1.42 cm
2
2
so
p=−
gives
1
1
1
+
=
1.42 cm ( −2.84 cm ) f
f = 2.84 cm
FIG. P36.27
319
320
P36.28
Chapter 36
1 1 1
+ = :
p q f
p −1 + q −1 = constant
We may differentiate through with respect to p:
−1 p −2 − 1q −2
dq
=0
dp
dq
q2
= − 2 = −M 2
dp
p
1
1
1
1 1 1
=
we see q = −3.5 p and f = 7.50 cm for a
+ = with +
p −3.5 p 7.5 cm
p q f
converging lens.
*P36.29 Comparing
(a)
To solve, we add the fractions:
−3.5 + 1
1
=
−3.5 p
7.5 cm
3.5 p
= 7.5 cm
2.5
p = 5.36 cm
(b)
q = −3.5 ( 5.36 cm ) = −18.8 cm
M =−
q
−18.8 cm
=−
= +3.50
p
5.36 cm
(c)
I
F
O
L
F
FIG. P36.29(c)
The image is enlarged, upright, and virtual.
(d)
P36.30
The lens is being used as a magnifying glass. Statement: A magnifying glass with focal
length 7.50 cm is used to form an image of a stamp, enlarged 3.50 times. Find the object
distance. Locate and describe the image.
The image is inverted:
M=
h′
−1.80 m
−q
=
= −75.0 =
h 0.024 0 m
p
(a)
q + p = 3.00 m = 75.0 p + p
p = 39.5 mm
(b)
q = 2.96 m
1 1 1
1
1
= + =
+
f p q 0.039 5 m 2.96 m
q = 75.0 p
f = 39.0 mm
Image Formation
P36.31
(a)
1
1
1
+ =
20.0 cm q ( −32.0 cm )
1 1 1
+ =
p q f
so
321
1
1 ⎞
q = −⎛
+
⎝ 20.0 32.0 ⎠
−1
= −12.3 cm
The image is 12.3 cm to the left of the lens.
q
( −12.3 cm )
= 0.615
=−
p
20.0 cm
(b)
M =−
(c)
See the ray diagram to the right.
*P36.32 (a)
1
1 1
+
=
pa qa
f
becomes
FIG. P36.31
1
1
1
+ =
30.0 cm q1 14.0 cm
or
qa = 26.2 cm
⎛ −q ⎞
hb′ = hM a = h ⎜ a ⎟ = (10.0 cm ) ( −0.875) = −8.75 cm
⎝ pa ⎠
1
1
1
or
qd = 46.7 cm
+
=
20.0 cm qd 14.0 cm
hc′ = hM d = (10.0 cm ) ( −2.33) = −23.3 cm
b
c
The square is imaged as a trapezoid.
FIG. P36.32(b)
(b)
1 1
1
1 1 1
+ = becomes + =
p q 14 cm
p q f
or 1 / p = 1 / 14 cm − 1 / q
⎛ −q ⎞
⎛ 1
1⎞
h ′ = hM = h ⎜ ⎟ = (10.0 cm ) q ⎜
− ⎟
⎝ p⎠
⎝ 14 cm q ⎠
(c)
The quantity ∫
qd
qa
h ′ dq adds up the geometrical (positive) areas of thin vertical ribbons
comprising the whole area of the image. We have
46.7 cm
2
qd
qd
⎞
⎛ q − 1⎞ dq = (10.0 cm ) ⎛ q
)
q
h
dq
10
.
0
cm
−
=
(
′
⎜⎝ 28 cm
⎟⎠
∫qa
∫qa
⎝ 14 cm ⎠
26.2 cm
⎛ 46.72 − 26.22
⎞
Area = (10.0 cm ) ⎜
− 46.7 + 26.2⎟ cm = 328 cm 2
⎝
⎠
28