1. Does a light ray traveling from one medium into another
always bend toward the normal?
CHAPTER 15 REVIEW
Pages 587-592
1,6,14,21,24,27,37,45, 46,
49,53,61,63
6. Two colors of light (X and Y ) are sent through a glass
prism, and X is bent more than Y. Which color travels
more slowly in the prism?
NO. If the light is traveling from a high index
of refraction (n) to low n, the light is bent
away from the normal.
AIR (n = 1.00)
WATER (n = 1.33)
normal
14. The light ray shown below makes an angle of 20.0° with
the normal line at the boundary of linseed oil and water.
Determine the angles !1 and !2. Note that n = 1.48 for
linseed oil.
1.00 sin !1 = 1.48 sin 20°
n = 1.00
Color X travels more
slowly in the prism.
n = 1.48
"1 = sin !1 (1.48 sin 20°) =
30.4˚
1.33 sin ! 2 = 1.48 sin 20°
n = 1.33
20°
)=
" 2 = sin !1 (1.481sin
.33
22.4˚
21. In order to get an upright image, slides must be placed
upside down in a slide projector. What type of lens must
the slide projector have? Is the slide inside or outside
the focal point of the lens?
•  The projector needs a converging lens,
as this is the only type of lens that will
form a real image, which is inverted.
•  The slide must be outside the focal
point, as anything inside the focal point
would form a virtual image.
24. An object is placed in front of a diverging lens with a
focal length of 20.0 cm. For each object distance, find
the image distance and the magnification. Describe
each image.
USING THE LENS EQUATION:
a. p = 40.0 cm
b. p = 20.0 cm
c. p = 10.0 cm
1 1 1
+ =
p q f
1
1
1
+ =
20.0cm q ! 20.0cm
q = -10.0 cm
MAGNIFICATION
M =!
q
! 10.0cm
=!
p
20.0cm
24. An object is placed in front of a diverging lens with a
focal length of 20.0 cm. For each object distance, find
the image distance and the magnification. Describe
each image.
USING THE LENS EQUATION:
a. p = 40.0 cm
b. p = 20.0 cm
c. p = 10.0 cm
1 1 1
+ =
p q f
1
1
1
+ =
40.0cm q ! 20.0cm
q = -13.3 cm
MAGNIFICATION
M =!
q
! 13.3cm
=!
p
40.0cm
= 0.333
24. An object is placed in front of a diverging lens with a
focal length of 20.0 cm. For each object distance, find
the image distance and the magnification. Describe
each image.
USING THE LENS EQUATION:
a. p = 40.0 cm
b. p = 20.0 cm
c. p = 10.0 cm
1 1 1
+ =
p q f
1
1
1
+ =
10.0cm q ! 20.0cm
q = -6.67 cm
MAGNIFICATION
= 0.500
M =!
q
! 6.67cm
=!
p
10.0cm
= 0.667
27. Is it possible to have total internal reflection for light
incident from air on water? Explain.
37. Assuming that "= 589 nm, calculate the critical angles
for the following materials when they are surrounded by
air:
n = 1.923
Using Snell s Law:
a. zircon
To experience total internal
reflection, light must be moving
from a substance of higher n to
lower n, which causes light to be
refracted away from the normal.
Since air has the lowest n, this is
not possible.
b. fluorite
ni sin!i = nr sin!r
c. ice
!ci =
(1.923
)
n
sin-1
(1)(1)
i
!c = 31.3˚
37. Assuming that "= 589 nm, calculate the critical angles
for the following materials when they are surrounded by
air:
n = 1.434
Using Snell s Law:
a. zircon
b. fluorite
ni sin!i = nr sin!r
c. ice
!ci =
(1.434
)
n
sin-1
(1)(1)
i
!c = 44.2˚
37. Assuming that "= 589 nm, calculate the critical angles
for the following materials when they are surrounded by
air:
n = 1.309
Using Snell s Law:
a. zircon
b. fluorite
ni sin!i = nr sin!r
c. ice
!ci =
(1.309
)
n
sin-1
(1)(1)
i
!c = 49.8˚
45. Where must and object be placed to have a
magnification of 2.00 in each of the following cases?
a. A converging lens of focal length 12.0 cm.
P = 6.00 cm
46. A diverging lens is used to form a virtual image of an
object. The object is 80.0 cm in front of the lens, and
the image is 40.0 cm in front of the lens. Determine the
focal length of the lens.
p = 80.0 cm
q = -40.0 cm
1
1
1
+
=
80cm !40cm f
b. A diverging lens of focal length 12.0 cm.
f = -80.0cm
NOT POSSIBLE
49. The index of refraction for red light in water is 1.331,
and that for blue light is 1.340. If a ray of white light
traveling in air enters the water at an angle of incidence
of 83.0°, what are the angles of refraction for
the red and blue components of the light?
ni sin!i = nr sin!r
!r = sin-1
(
1sin 83.0˚
nr
)
48.2˚
!r = 47.8˚
!r =
53. A jewel thief decides to hide a stolen diamond by placing it at
the bottom of a crystal-clear fountain. He places a circular piece
of wood on the surface of the water and anchors it directly
above the diamond at the bottom of the fountain, as shown in
Figure 15-21. If the fountain is 2.00 m deep, find the minimum
diameter of the piece of wood that would prevent the diamond
from being seen from outside the water.
n = 1.00
n = 1.33
2.00 m
r
Critical angle
!c = sin-1
1
1.33
r
tan!c =
2.00 m
r = 2.27m
DIAMETER = 4.54m
61. A submarine is 325 m horizontally out from the shore and 115 m
beneath the surface of the water. A laser beam is sent from the
submarine so that it strikes the surface of the water at a point 205 m
from the shore. If the beam strikes the top of a building standing
directly at the water s edge, find the height of the building. (Hint: To
determine the angle of incidence, consider the right triangle formed
by the light beam, the horizontal line drawn at the depth of the
submarine, and the imaginary line straight down from where the
beam strikes the surface of the water.)
1.  First, find angle of incidence in the water using trig
!i = tan-1
120m
115m
= 46.2˚
1.33 sin 46.2˚ = 1.00 sin !r
!r = 73.8˚
h = 205m tan 16.2˚ = 59.6m
2.  Find angle of refraction in air using Snell s Law
3.  Find height of building using trig
115 m
115 m
205 m
205 m
325 m
325 m
63. A nature photographer is using a camera that has a lens with a
focal length of 4.80 cm. The photographer is taking pictures of
ancient trees in a forest and wants the lens to be focused on a very
old tree that is 10.0 m away.
a.  How far must the lens be from the film in order for the resulting
picture to be clearly focused?
1
1
1
+ =
10.0m q 0.0480m
q = 4.82 cm
b. How much would the lens have to be moved to take a picture of
another tree that is only 1.75 m away?
1
1
1
+ =
1.75m q 0.0480m
q = 4.93 cm
The lens must be
moved 0.11
cm