Φ21 Fall 2006
1 Problem K23.13
HW24 Solutions
◦
A diamond is underwater. A light ray enters one face of the diamond, then travels at an angle of 30.0 with
respect to the normal. What was the ray's angle of incidence on the diamond?
Solution: Snell's Law gives the answer. The index of refraction of water is
nwater = 1.33,
while
ndiamond =
2.42.
n1 sin θ1
θ1
= n2 sin θ2
(
)
n2
= sin−1
sin θ2
n1
◦
= 65
2 Problem K23.22
A sheet of glass has nred = 1.52 and nviolet = 1.55. A narrow beam of white light is incident on the glass at
◦
30.0 . What is the angular spread of the light inside the glass?
Solution: The white light separates into a rainbow with the dierent angles determined by the dierent
indices of refraction.
θ2,red
θ2,violet
∆θ
(
)
nair sin 30◦
= 19.205◦
nred
)
(
nair sin 30◦
= sin−1
= 18.819◦
nviolet
= 0.386◦
=
sin−1
3 Problem K23.26
An object is 20 cm in front of a converging lens with a focal length of 10 cm.
Use the lens equation to
determine the location of the image. Draw a ray diagram and bring it with you to recitation.
Solution: In this case, the object is a real object, so
f = +0.2 m.
do = +0.20 m.
The converging lens has a focal length
The location of the image is
1
1
1
+
=
do
di
f
The positive value of
di
(
⇒ di =
1
1
−
f
do
)−1
(
=
1
1
−
0.1 m 0.2 m
)−1
= 0.2 m
means the image is a real image on the opposite side of the lens from the real object.
Figure 1: Ray diagram for the given problem. Ray 1 (the parallel ray) is the top one, Ray 2 (the Focal
ray) is on the bottom, and Ray 3 (the center ray) is in the middle.
1
Part B. Is the image upright or inverted?
Solution: From the ray diagram, the image is inverted. Also, note that
m=
hi
ho
= − ddoi = −1.
4 Problem YF34.24 (modied)
An object 37.5 cm in front of a certain lens is imaged 8.2 cm in front of the lens (on the same side as the
object). Part A. Use the lens equation to answer this question, but then draw a ray diagram and bring it
to recitation. What is the focal length of the lens?
Solution: The object distance is
equation:
(
f=
do = +37.5 cm
1
1
+
di
d1
)−1
(
=
while the image distance is
1
1
+
37.5 cm −8.2 cm
di = −8.2 cm.
From the lens
)−1
= −10.5 cm
Part B: Is the lens converging or diverging? Solution: The lens is diverging because it has a negative focal
length.
Figure 2: Ray diagram for this problem. The object is the blue arrow, and the image is the green arrow.
Ray 1 is on top, Ray 2 is in the middle, and Ray 3 is on the bottom.
Part C. If the object is 6.0 cm tall, how tall is the image? Part D. Is the image upright or inverted?
Solution: The image height is
hi = −ho
di
(−8.2 cm)
= − (6.0 cm)
= 1.31 cm
do
37.5 cm
The image is upright.
5 Problem YF 34.8
An object is a distance of 24.9 cm from the CENTER of a silvered spherical glass Christmas tree ornament
which has a diameter of 5.50 cm.
Part A. What is the position of its image?
Use the mirror equation
to answer this question, but draw a ray diagram and bring it with you to recitation. Hint: be careful to
determine
Solution:
do
(the distance to the SURFACE of the mirror) correctly.
The object distance is
do = 24.9 cm − (5.50 cm) /2 = 22.15 cm. The focal distance negative
f = − (5.50 cm/4) = −1.375 cm. The
because the mirror is convex and its magnitude is half the radius,
image distance is
(
di =
1
1
−
f
do
)−1
(
=
1
1
−
(−1.375 cm) (22.15 cm)
Part B. What is the magnication? Solution: The magnication is
2
)−1
= −1.29 cm
cm
m = − dd1i = − −1.29
22.15 cm = 0.058.
6 Problem YF 34.14
A spherical, concave, shaving mirror has a radius of curvature of 32.1 cm. Part A. What is the magnication
of a person's face when it is a distance
do
= 13.0 cm from the center of the mirror? Use the mirror equation
to answer this question, but draw a ray diagram and bring it with you to recitation. Part B. What is the
distance
di
to the image?
do = 13.0 cm. The focal length
f = 32.1 cm/2 = 16.05 cm. The image distance is
Solution: The object distance is
half the radius,
(
di =
1
1
−
f
do
(
)−1
is positive for a concave mirror and it is
)−1
=
1
1
−
16.05 cm 13.0 cm
m=−
di
68.4 cm
=
= 5.26
do
13.0 cm
= −68.4 cm
The magnication is
Part C. Is the image real or virtual?
Solution: The image is virtual.
The image distance is negative and placing a screen behind the mirror
would not project the image onto the screen.
3