# Physics 41 Chapter 37 Sample Problems ```Physics 41 Chapter 35 HW Key
1. Suppose a woman stands in front of a mirror as shown. Her eyes are 1.65 m above the floor, and the top of her
head is 0.13m higher. Find the height above the floor of the top and bottom of the
shortest mirror in which she can see both the top of his head and his feet. How
long is the mirror? How is the distance related to the woman’s height?
24.1
a
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. And the
L
bottom must go down to halfway between his eyes and the floor. This
1.65 m
result independent of how far he stands from the wall. So,
0.13 m
1.65 m
b
=
a = 0.065=
m , b = 0.825 m and
2
2
0.13 m 1.65 m
=
L 1.65 m + 0.13 m − a=
− b 1.78 m −
−
= 0.89 m . So, the bottom is b = 0.825 m from floor and the
2
2
top is b=
+ L 0.825 m + 0.89
=
m 1.715 m from floor.
0.13 m
2. When the light illustrated in the diagram passes through the glass block, it is
shifted laterally by the distance d. Taking n = 1.50, find the value of d.
P35.21
At entry, n1 sin θ1 = n2 sin θ 2
1.00sin 30.0° =1.50sin θ 2
or
=
θ 2 19.5° .
The distance h the light travels in the medium is given by
cosθ 2 =
or =
h
The angle of deviation upon entry is
2.00 cm
h
2.00 cm
= 2.12 cm .
cos19.5°
FIG. P35.21
α= θ1 − θ 2= 30.0° − 19.5°= 10.5° .
The offset distance comes from sin α =
=
d
=
cm ) sin 10.5°
( 2.21
0.388 cm .
d
:
h
3. The light beam shown in the figure makes an angle of 20.0° with the normal line NN’ in the linseed oil. Determine
the angles θ and θ’. (The index of refraction of linseed oil is 1.48.)
P35.23 Applying Snell’s law at the air-oil interface,
=
nair sin θ noil sin 20.0°
yields
=
θ 30.4° .
Applying Snell’s law at the oil-water interface
=
nw sin θ ′ noil sin 20.0°
yields
=
θ ′ 22.3° .
FIG. P35.23
4. Light of wavelength 700 nm is incident on the face of a fused quartz prism at an angle of 75.0° (with respect to the
normal to the surface). The apex angle of the prism is 60.0°. Use the value of
n ( 700 nm ) = 1.458 and calculate the angle (a) of refraction at this first surface,
(b) of incidence at the second surface, (c) of refraction at the second surface,
and (d) between the incident and emerging rays.
P35.26 From the figure on page 992: n ( 700 nm ) = 1.458
60.0°
(a)
(b)
; θ2
( 1.00) sin 75.0° =1.458sin θ 2 =
41.5°
θ1
Let
° 180° .
θ 3 + β= 90.0° , θ 2 + α= 90.0° then α + β + 60.0=
So
60.0° − θ 2 − θ 3 = 0 ⇒ 60.0° − 41.5°= θ 3 =
(c)
1.458sin 18.5° =1.00sin θ 4
(d)
=
γ
=
θ4
18.5° .
27.6°
(θ1 − θ 2 ) + β − ( 90.0° − θ 4 ) 
=
γ 75.0° − 41.5° + ( 90.0° − 18.5°) − ( 90.0° − 27.6=
°)
42.6°
α
β
θ2
θ3
FIG. P35.30
θ4
γ
5. Determine the maximum angle θ for which the light rays incident on the end
of the pipe in Figure P35.38 are subject to total internal reflection along the walls
of the pipe. Assume that the pipe has an index of refraction of 1.36 and the
outside medium is air.
sin
=
θc
P35.36
nair
1.00
θ c 47.3°
= = 0.735 =
npipe 1.36
Geometry shows that the angle of refraction
at the end is
=
φ 90.0° − =
θ c 90.0° − 47.3
=
° 42.7° .
Then, Snell’s law at the end,
=
1.00sin θ 1.36sin 42.7°
gives
FIG. P35.38
=
θ 67.2° .
The 2-µm diameter is unnecessary information.
6
A hiker stands on an isolated mountain peak near sunset and
observes a rainbow caused by water droplets in the air 8.00 km away.
The valley is 2.00 km below the mountain peak and entirely flat.
What fraction of the complete circular arc of the rainbow is visible to
the hiker? (See Fig. 35.24.)
P35.47 Horizontal light rays from the setting Sun pass above the
hiker. The light rays are twice refracted and once reflected, as in
Figure (b). The most intense light reaching the hiker, that which
represents the visible rainbow, is located between angles of 40°
and 42° from the hiker’s shadow.
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 droplets is
=
R
=
km ) sin 42.0°
( 8.00
Figure (a)
5.35 km .
Then the angle φ, between the vertical and the radius
where the bow touches the ground, is given by
cosφ
=
2.00 km 2.00 km
= = 0.374
R
5.35 km
or
=
φ 68.1° .
The angle filled by the visible bow is
360° − ( 2 × 68.1=
°) 224°
so the visible bow is
224°
= 62.2% of a circle .
360°
Figure (b)
FIG. P35.53
7. At what angle phi should the laser be aimed at the mirrored ceiling in
to hit the midpoint of the far wall on the right?
order
Light rays travel in straight lines and follow the law of reflection.
Visualize:
To determine the angle φ , we must know the point P on the mirror where the ray is incident. P is a distance x2 from the far wall and a
horizontal distance x1 from the laser source. The ray from the source must strike P so that the angle of incidence θi is equal to the angle
of reflection θr.
Solve: From the geometry of the diagram,
tan φ =
1.5 m 3 m
=
x1
x2
⇒ tan φ =
x1 + x2 = 5 m ⇒
(
)
( )
3m
1.5 m
10
⇒ 1.5 m x1 = 15 m 2 − 3 m x1 ⇒ x1 =
=
m
x1
5 m − x1
3
3m 9
= 0.90 ⇒ φ = 42°
=
10
x1
8. What is the smallest incident angle θ1 for which a laser beam will undergo total
internal reflection on the hypotenuse of the glass prism shown?
23.57. Model: Use the ray model of light and the phenomenon of refraction.
Visualize:
Solve: (a) The critical angle θc for the glass-air boundary is
 1.0 
θ c sin −1  =
nglass
=
sin θ c nair sin 90° ⇒=
 41.81°
 1.50 
For the triangle ABC,
θ glass 1 + 120° + ( 90° − θ c=
) 180° ⇒ θ glass=1 180° − 120° − ( 90° − 41.81°=) 11.81°
Having determined θglass 1, we can now find θair 1 by using Snell’s law:
 1.50 × sin11.81° 
=
sin −1 
nair sin θ air 1 = nglass sin θ glass 1 ⇒ θ=
air 1
 17.88°
1.0


Thus, the smallest angle θ1 for which a laser beam will undergo TIR on the hypotenuse of this glass prism is 17.9°.
(b) After reflecting from the hypotenuse (face 3) the ray of light strikes the base (face 2) and refracts into the air. From the triangle
BDE,
( 90° − θ
glass 2
) + 60° + ( 90° − θ =)
c
180° ⇒ θ glass 2= 90° + 60° + 90° − 41.81° − 180°= 18.19°
Snell’s law at the glass-air boundary of face 2 is
 n sin θ glass 2 
−1  1.50sin18.19° 
=
=
nglass sin θ glass 2 = nair sin θ air 2 ⇒ θ air 2 = sin −1  glass
 sin 
 27.9°
nair
1.0




Thus the ray exits 27.9° left of the normal.
9. A narrow beam of light containing red (660nm) and blue (470nm) wavelengths travels from air through a 1.00 cm
thick flat piece of crown glass and back to air again. The beam strikes at a 30 degree incident angle. (a) At what
angles do the two colors emerge? (b) By what distance are the red and blue separated when they emerge?
n1
n 2 , n'2
n3
θ2
θ1
θ3
θ 3'
θ 2'
Let unprimed represent red light and primed represent blue.
.
n2 = 1512
.
n2′ = 1524
y' y
x
(a) n1 sin θ 1 = n2 sin θ 2 = n3 sin θ 3 since n3 = n1 , θ 3 = θ 1 and similarly for θ ′3 . θ 3 = θ ′3 = θ 1 = 30.0° .
FG sinθ IJ = sin F sin 30.0°I = 19.311° ; θ ′ = sin F sin 30.0°I = 19153
H 1512
K
H 1524
K . °
H n K
.
.
y
= ⇒ y = x tan θ = c100
. × 10 mh tan 19.311° = 3504
.
× 10 m
x
(b) θ 2 = sin −1
1
−1
−1
2
2
tan θ 2
−2
−3
2
. × 10−5 m
× 10−3 m ; ∆y = y − y ′ = 309
y ′ = (0100
.
m) tan 19153
. ° = 3473
.
```