Chapter 35 solution P35.5
(a)
(b)
From geometry,
1.25 m = dsin 40.0°
so
d = 1.94 m .
Mirror 2
50.0° above the horizontal
i2= 50°
40°
P
1.25 m
or parallel to the incident ray.
50°
Light
beam
40.0°
Mirror 2
d
40° i1= 40°
50°
Mirror 1
1.25 m
50°
Mirror 1
FIG. P35.5 P35.7
Let d represent the perpendicular distance from
the person to the mirror. The distance
between lamp and peson measured
parallel to the mirror can be written in two
ways: 2dtan θ + dtan θ = dtan φ . The
condition on the distance traveled by the
2d
2d
d
=
+
light is cos φ cos θ cos θ . We have the
two equations 3 tanθ = tan φ and
2 cosθ = 3 cos φ . To eliminate φ we write
9 sin2 θ sin2 φ
=
cos2 θ
cos 2 φ
P35.11 n1 sin θ 1 = n 2 sin θ 2 sin θ1 = 1.333 sin 45°
d tanφ
θ
d tanθ
2d
d
d
FIG. P35.7 4cos2 θ = 9cos2 φ
sin θ1 = ( 1.33)( 0.707 ) = 0.943
θ1 = 70.5° → 19.5° above the horizon
P35.13
2d tanθ
θ
φ
We find the angle of incidence:
n1 sin θ 1 = n 2 sin θ 2
1.333 sinθ 1 = 1.52 sin 19.6° θ1 = 22.5°
The angle of reflection of the beam in water is then also 22.5° . FIG. P35.11 P35.19 by At entry, or 1.00 sin 30.0° = 1.50sin θ 2 θ2 = 19.5° . The distance h the light travels in the medium is given 2.00 cm
cos θ 2 =
h
or h=
n1 sin θ 1 = n 2 sin θ 2 The angle of deviation upon entry is
α = θ1 − θ 2 = 30.0° − 19.5° = 10.5° . The offset distance comes from sin α =
P35.21 Applying Snell’s law at the air-­‐oil interface, n air sin θ = n oil sin 20.0° yields θ = 30.4° . €
d
:
h
d = (2.21 cm) sin 10.5° = 0.388 cm . Applying Snell’s law at the oil-­‐water interface n w sinθ ʹ′ = n oil sin 20.0° yields θ ʹ′ = 22.3° . FIG. P35.21 P35.25 Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is sin (Φ + δ min ) 2
n=
sin(Φ 2)
Solving for δ min ,
⎛
Φ ⎞
δ min = 2sin −1 ⎜ n sin ⎟ − Φ = 2sin −1 [(2.20) sin(25.0°) ] − 50.0° = 86.8° . ⎝ € 2 ⎠
[
€
2.00 cm
= 2.12 cm . cos19.5°
FIG. P35.19 ]
P35.31 For the incoming ray,
sin θ1
sin θ2 =
. n
Using the figure to the right,
(θ 2 ) violet = sin −1 ⎛⎜ sin 50.0° ⎞⎟ = 27.48° ⎝ 1.66 ⎠
FIG. P35.31 (θ 2 ) red = sin −1 ⎛⎜ sin 50.0° ⎞⎟ = 28.22° . ⎝
1.62
⎠
For the outgoing ray,
θ3 = 60.0° − θ 2 and sin θ4 = n sin θ 3 :
(θ 4 ) violet = sin −1 [1.66sin 32.52° ] = 63.17° (θ 4 ) red = sin −1 [1.62 sin 31.78° ] = 58.56° . The angular dispersion is the difference
Δθ4 = (θ 4 ) violet − (θ4 )red = 63.17° − 58.56° = 4.61° . P35.33 n sinθ = 1 . From Table 35.1, €
€
(a)
⎛ 1 ⎞
θ = sin −1 ⎜
⎟ = 24.4°
⎝ 2.419 ⎠
(b)
⎛ 1 ⎞
θ = sin −1 ⎜
⎟ = 37.0°
⎝ 1.66 ⎠
(c)
⎛ 1 ⎞
θ = sin −1 ⎜
⎟ = 49.8°
⎝ 1.309 ⎠
n air
1.00
=
= 0.735 θ c = 47.3° n pipe 1.36
P35.36 sin θ c =
refraction at the end is Geometry shows that the angle of information. The 2-­‐µ m diameter is unnecessary φ = 90.0° − θ c = 90.0° − 47.3° = 42.7° . Then, Snell’s law at the end,
1.00 sinθ = 1.36 sin 42.7° gives θ = 67.2° . FIG. P35.36