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PHYS 132, WEEK #5, CHAPTER 23
23.7.
Model:
Light rays travel in straight lines and follow the law of reflection.
Visualize:
Solve:
We are asked to obtain the distance
h
=
x
1
+
5.0 cm. From the geometry of the diagram,
tan
q
i
=
x
1
10 cm
tan
r
=
x
2
15 cm
x
1
+
x
2
=
10 cm
Because
r
=
i
, we have
x
1
10 cm
=
x
2
15 cm
=
10 cm

x
1
15 cm
15 cm
( )
x
1
=
100 cm
2

10
x
1
x
1
=
4.0 cm
Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror.
23.11.
Model:
Use the ray model of light and the law of reflection.
Visualize:
We only need one ray of light that leaves your toes and reflects in your eye.
Solve:
From the geometry of the diagram, the distance from your eye to the toe’s image is
2
d
=
(400 cm)
2
+
(165 cm)
2
=
433 cm
Assess:
The light appears to come from your toes’ image.
23.12.
Model:
Use the ray model of light and Snell’s law.
Visualize:
Solve:
According to Snell’s law for the airwater and waterglass boundaries,
n
air
sin
air
=
n
water
sin
water
n
water
sin
water
=
n
glass
sin
glass
From these two equations, we have
n
air
sin
air
=
n
glass
sin
glass
sin
glass
=
n
air
n
glass
sin
air
=
1.0
1.50
ˆ
sin 60
°
glass
=
sin

1
sin 60
°
1.5
ˆ
˜ =
35.3
°
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View Full Document23.16.
Model:
Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total
internal reflection.
Visualize:
Solve:
The critical angle of incidence is given by Equation 23.9:
q
c
=
sin

1
n
cladding
n
core
ˆ
˜
=
sin

1
1.48
1.60
ˆ
=
67.7
°
Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90
°

67.7
°
=
22.3
°
.
Assess:
We can have total internal reflection because
n
core
>
n
cladding
.
23.22.
Model:
Use the ray model of light.
Visualize:
Solve:
Using Snell’s law,
n
air
sin 30
° =
n
red
sin
red
red
=
sin

1
sin 30
°
1.52
ˆ
=
19.2
°
n
air
sin 30
° =
n
violet
sin
violet
violet
=
sin

1
sin 30
°
1.55
ˆ
=
18.8
°
Thus the angular spread is
D
=
red

violet
=
19.2
° 
18.8
° =
0.4
°
23.26.
Model:
Use ray tracing to locate the image.
Solve:
The figure shows the raytracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane
where the three special rays converge. The image is inverted and is located at
s
¢
=
20.0 cm to the right of the converging lens.
23.28.
Model:
Use ray tracing to locate the image.
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This note was uploaded on 09/27/2008 for the course PHYS 131133 taught by Professor All during the Spring '08 term at Cal Poly.
 Spring '08
 ALL
 Light

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