EXTRA PROBLEMS
EXAM II
Chapter 22
19.
The length of the pulse is d  ct, so the number of wavelengths in this length is found
by dividing this length by the wavelength.
N
(ct )


(3.00 108 m/s)(34 1012 s)
(1062 109 m)
 9600 wavelengths
The time for the length of the pulse to be one wavelength is the wavelength divided by the speed
of light.
t  

c

(1062 109 m)
(3.00 10 m/s)
8
 3.54 1015 s
Chapter 23
18.
The mirror must be convex. Only convex mirrors produce images that are upright
and smaller than the object. The object distance of 16.0 m and the magnification of  0.33 are
used to find the image distance. The focal length and radius of curvature can then be found.
m
 di
do
 di   md o
1
1 1
 
d o di
f

f 
d o di
d (mdo ) mdo 033(160 m)
 o


 7881 m
d o  di
d o  md o
m 1
033  1
r  2 f  2(7881 m)  1576 m  16 m
The negative focal length confirms that the mirror must be convex.
37.
We find the critical angle for light leaving the
water, so that the refracted angle is 90.
sin C 
nair
nliquid
 1.00 
 C  sin 1 
  48.75
 1.33 
If the light is incident at a greater angle than this, then it
will totally reflect. Find R from the diagram.
R  H tan 1  (82.0 cm) tan 48.75  93.5 cm
air
R
n
H
1
1
54.
(a)
Use Eqs. 23–8 and 23–9.
1
1 1
 
d o di
f
m
hi
d
 i
ho
do
 di 
do f
(1.30 m)(0.135 m)

 0.1506 m  151 mm
do  f
1.30 m  0.135 m
di
0.1506 m
ho  
(2.40 cm)  0.278 cm
do
1.30 m
 hi  
The image is behind the lens a distance of 151 mm and is real, inverted, and reduced.
(b)
Again use Eqs. 23–8 and 23–9.
1 1 1
 
d o di f
m
hi
d
 i
ho
do
 di 
do f
(1.30 m)( 0.135 m)

 0.1223 m  122 mm
do  f 1.30 m  (0.135 m)
di
(0.1223 m)
ho  
(2.40 cm)  0.226 cm
do
1.30 m
 hi  
The image is in front of the lens a distance of 122 mm and is virtual, upright, and reduced.
Chapter 24
13.
For constructive interference, the path difference is a multiple of the wavelength,
as given by Eq. 24–2a. The location on the screen is given by x  tan  , as seen in Fig. 24–7(c).
For small angles, we have sin   tan   x / . For adjacent fringes, m  1.
d sin   m
x  m

d
 d
 (1)
x
 m
 x
(544  109 m)(4.0 m)
(1.0  10
3
m
d

 2.2  103 m
m)
24.
We find the angle to the first minimum using Eq. 24–3a. The distance on the screen from
the central maximum is found using the distance to the screen and the tangent of the angle. The
width of the central maximum is twice the distance from the central maximum to the first
minimum.
sin 1 

D
 450  109 m 

 1  sin 1    sin 1 
 0.02578
 1.0  103 m 
D


x1  tan 1  (6.0 m) tan 0.02578  2.70  103 m
x  2 x1  2(2.70  103 m)  5.4  103 m  0.54 cm
41.
The maximum angle is 90°. The slit separation is the reciprocal of the line spacing.
d sin   m  m 
d sin 

 1

cm  sin 90

6500


 2.43
633 107 cm
Thus the second order is the highest order that can be seen.
96.
The reduction being investigated is that which occurs when the polarized light passes
through the second Polaroid. Let I1 be the intensity of the light that emerges from the first
Polaroid and I 2 be the intensity of the light after it emerges from the second Polaroid. Use Eq.
24–5.
(a)
I 2  I1 cos2   025I1    cos1 0.25  60
(b)
I 2  I1 cos2   0.10I1    cos1 0.10  72
(c)
I 2  I1 cos2   0010 I1    cos 1 0.010  84