Chapter 26 Home Work Solutions
______________________________________________________________________________
1.
SSM REASONING The substance can be identified from Table 26.1 if its index of refraction is known.
The index of refraction n is defined as the speed of light c in a vacuum divided by the speed of light v in the
substance (Equation 26.1), both of which are known.
SOLUTION Using Equation 26.1, we find that
n
c 2.998 108 m/s

 1.362
v 2.201108 m/s
Referring to Table 26.1, we see that the substance is ethyl alcohol .
__________________________________________________________________________
3. REASONING The refractive index n is defined by Equation 26.1 as n = c/v, where c is the speed of light in
a vacuum and v is the speed of light in a material medium. We will apply this definition to both materials A
and B, and then form the ratio of the refractive indices. This will allow us to determine the unknown speed.
SOLUTION Applying Equation 26.1 to both materials, we have
nA 
c
vA
nB 
and
c
vB
Dividing the equation for material A by that for material B gives
nA c / vA vB


nB c / vB vA
Solving for vB, we find that
n 
vB  vA  A   1.25 108 m/s 1.33  1.66 108 m/s
 nB 

5.

SSM WWW REASONING Since the light will travel in glass at a constant speed v, the time it takes
to pass perpendicularly through the glass is given by t  d/v , where d is the thickness of the glass. The
speed v is related to the vacuum value c by Equation 26.1: n  c/v.
SOLUTION Substituting for v from Equation 26.1 and substituting values, we obtain
d nd 1.5  4.0 10 –3 m 


 2.0 10 –11 s
8
v
c
3.00 10 m/s
______________________________________________________________________________
t
9. SSM REASONING AND SOLUTION
a. We know from the law of reflection (Section 25.2), that the angle of reflection is equal to the angle of
incidence, so the reflected ray is reflected at 43 .
b. Snell’s law of refraction (Equation 26.2: n1 sin 1  n2 sin  2 can be used to find the angle of refraction.
Table 26.1 indicates that the index of refraction of water is 1.333. Solving for 2 and substituting values,
we find that
n1 sin 1 (1.000) (sin 43)

 0.51
or
 2  sin –1 0.51  31
n2
1.333
______________________________________________________________________________
13. SSM REASONING We will use the geometry of the situation to determine the angle of incidence.
Once the angle of incidence is known, we can use Snell's law to find the index of refraction of the
unknown liquid. The speed of light v in the liquid can then be determined.
sin  2 
SOLUTION From the drawing in the text, we see that the angle of incidence at the liquid-air interface is
 5.00 cm 
1  tan –1 
  39.8
 6.00 cm 
The drawing also shows that the angle of refraction is 90.0°. Thus, according to Snell's law (Equation
26.2: n1 sin 1  n2 sin  2 ), the index of refraction of the unknown liquid is
n1 
n2 sin 2 (1.000) (sin 90.0)

 1.56
sin1
sin 39.8
From Equation 26.1 ( n  c/ v ), we find that the speed of light in the unknown liquid is
c 3.00 108 m/s

 1.92 108 m/s
n1
1.56
____________________________________________________________________________
v
27.
SSM REASONING AND SOLUTION According to Equation 26.4, the critical angle is related to the
refractive indices n1 and n2 by sin  c  n2 / n1 , where n1 > n2. Solving for n1, we find
n2
1.000

 1.54
sin  c sin 40.5
______________________________________________________________________________
43. REASONING The angle of each refracted ray in the crown glass can be obtained from Snell’s law
(Equation 26.2) as ndiamond sin 1 = ncrown glass sin 2, where 1 is the angle of incidence and 2 is the angle
of refraction.
n1 
SOLUTION The angles of refraction for the red and blue rays are:
Blue ray
n
sin 1 
  2.444  sin 35.00 
  sin 1 
  66.29
 ncrown glass 
1.531




 2  sin 1 
diamond
n
sin 1 
 2.410 sin 35.00 
  sin 1 
  65.43
 ncrown glass 
1.520



 2  sin 1 
Red ray
diamond
The angle between the blue and red rays is
 blue   red  66.29  65.43  0.86
______________________________________________________________________________
49. SSM REASONING The ray diagram is constructed by drawing the paths of two rays from a point on
the object. For convenience, we will choose the top of the object. The ray that is parallel to the principal
axis will be refracted by the lens so that it passes through the focal point on the right of the lens. The ray
that passes through the center of the lens passes through undeflected. The image is formed at the
intersection of these two rays. In this case, the rays do not intersect on the right of the lens. However, if
they are extended backwards they intersect on the left of the lens, locating a virtual, upright, and enlarged
image.
SOLUTION
a. The ray-diagram, drawn to scale, is shown below.
Scale:
20 cm
20 cm
F
Image
F
Object
From the diagram, we see that the image distance is di = –75 cm and the magnification is +2.5 . The
negative image distance indicates that the image is virtual. The positive magnification indicates that the
image is larger than the object.
b. From the thin-lens equation [Equation 26.6: (1/ do )  (1/ di )  (1/ f ) ], we obtain
1 1 1
1
1
 –

–
di
f d o 50.0 m 30.0 cm
or
d i = –75.0 cm
The magnification equation (Equation 26.7) gives the magnification to be
d
–75.0 cm
m– i –
 +2.50
do
30.0 cm
______________________________________________________________________________
53.
REASONING The distance from the lens to the screen, the image distance, can be obtained directly from
the thin-lens equation, Equation 26.6, since the object distance and focal length are known. The width and
height of the image on the screen can be determined by using Equation 26.7, the magnification equation.
SOLUTION
a. The distance di to the screen is
1 1 1
1
1
 


 2.646  104 mm1
di f do 105.00 mm 108.00 mm
so that di  3.78  103 mm = 3.78 m .
b. According to the magnification equation, the width and height of the image on the screen are
Width
 d 
 3.78  103 mm 
2
hi  ho   i    24.0 mm   
   8.40  10 mm
 d 
108
mm


 o
The width is 8.40  102 mm .
Height
 d 
 3.78  103 mm 
3
hi  ho   i    36.0 mm   
   1.26  10 mm
 d 
108
mm


 o
The height is 1.26  103 mm .
67. REASONING The thin-lens equation can be used to find the image distance of the first image (the image
produced by the first lens). This image, in turn, acts as the object for the second lens. The thin-lens
equation can be used again to determine the image distance for the final image (the image produced by the
second lens).
SOLUTION For the first lens, the object and image distances, do,1 and di,1, are related to the focal length f
of the lens by the thin-lens equation
1
1
1


do1 di1 f
(26.6)
Solving this expression for the image distance produced by the first lens, we find that
1
1
1
1
1
 


di1 f do1 12.00 cm 36.00 cm
or
di1  18.0 cm
This image distance indicates that the first image lies between the lenses. Since the lenses are separated by
24.00 cm, the distance between the image produced by the first lens and the second lens is 24.00 cm  18.0
cm = 6.0 cm. Since the image produced by the first lens acts as the object for the second lens, we have that
do2 = 6.0 cm. Applying the thin-lens equation to the second lens gives
1
1
1
1
1
 


di2 f do2 12.00 cm 6.0 cm
or
di 2  12 cm
The fact that this image distance is negative means that the final image is virtual and lies to the left of the
second lens.
______________________________________________________________________________
75. REASONING We will apply the thin-lens equation to solve this problem. In doing so, we must be careful
to take into account the fact that the lenses of the glasses are worn at a distance of 2.2 or 3.3 cm from her
eyes.
SOLUTION
a. The object distance is 25.0 cm – 2.2 cm, since it is measured relative to the lenses, which are worn
2.2 cm from the eyes. As discussed in the text, the lenses form a virtual image located at the near point.
The image distance must be negative for a virtual image, but the value is not –67.0 cm, because the glasses
are worn 2.2 cm from the eyes. Instead, the image distance is –67.0 cm + 2.2 cm. Using the thin-lens
equation, we can find the focal length as follows:
1 1 1
1
1

 

f do di 25.0 cm  2.2 cm 67.0 cm  2.2 cm
or
f  35.2 cm
b. Similarly, we find
1 1 1
1
1

 

or
f  32.9 cm
f do di 25.0 cm  3.3 cm 67.0 cm  3.3 cm
______________________________________________________________________________
77. SSM REASONING The closest she can read the magazine is when the image formed by the contact
lens is at the near point of her eye, or di = 138 cm. The image distance is negative because the image is a
virtual image (see Section 26.10). Since the focal length is also known, the object distance can be found
from the thin-lens equation.
SOLUTION The object distance do is related to the focal length f and the image distance di by the thinlens equation:
1
1 1
1
1
  

do f di 35.1 cm 138 cm
or
do  28.0 cm
(26.6)