Chapter 23
Mirrors and Lenses
Problem Solutions
23.1
If you stand 40 cm in front of the m irror, the tim e required fo r light scattered from your
face to travel to the m irror and back to your eye is
t
2d
c
2 0.40 m
3.0 108 m s
2.7 10
9
s
Thus, the im age you observe show s you ~10
23.2
9
s younger than your current age.
(a) With the palm located 1.0 m in front of the nearest m irror, that m irror form s an
im age, I P1 , of the palm located 1.0 m behind the nearest mirror .
(b) The farthest m irror form s an im age, f 0 , of the back of the hand located 2.0 m
behind this m irror and 5.0 m in front of the nearest m irror. This im age serves and
an object for the nearest m irror, w hich then form s an im age, f
f , of the
back of the hand located 5.0 m behind the nearest mirror .
(c) The im age I P1 (see part a) serves as an object located 4.0 m in front of the farthest
m irror, w hich form s an im age p
f of the palm , located 4.0 m behind that m irror
and 7.0 m in front of the nearest m irror. This im age then serv es as an object for the
nearest m irror w hich form s an im age I P3 of the palm , located
7.0 m behind the nearest mirror .
(d ) Since all im ages are located behind the m irror, all are virtual images .
23.3
(1) The first im age in the left-hand m irror is 5.00 ft behind the m irror,
or 10.0 ft from the person
388
M irrors and Lenses
389
(2) The first im age in the right-hand m irror serves as an object for the left-hand m irror.
It is located 10.0 ft behind the right-hand m irror, w hich is 25.0 ft from the left-hand
m irror. Thus, the second im age in the left-hand m irror is 25.0 ft behind the m irror,
or p2 40.0 cm q1
10.0 cm
(3) The first im age in the left-hand m irror serves as an object for the right-hand m irror.
It is located 20.0 ft in front of the right-hand m irror and form s an im age 20.0 ft
behind that m irror. This im age then serves as an object for the left -hand m irror. The
d istance from this object to the left-hand m irror is 35.0 ft. Thus, the third im age in
the left-hand m irror is 35.0 ft behind the m irror,
10.0 cm 15.0 cm
p2 f 2
30.0 cm
or q2
p2 f 2
10.0 cm 15.0 cm
23.4
The virtual im age is as far behind the m irror
as the choir is in front of the m irror. Thus,
the im age is 5.30 m behind the m irror.
The im age of the choir is
M2
30.0 cm
q2
p2
10.0 cm
3.00
from the organist. Using sim ilar trian gles,
gives
h
0.600 m
or
23.5
h
6.10 m
0.800 m
0.600 m
6.10 m
0.800 m
4.58 m
In the figure at the right,
since they are
vertical angles form ed by tw o intersecting
straight lines. Their com plem entary angles are
also equal or p1
15.0 cm . The right triangles
q1
p1 f1
p1 f1
15.0 cm 10.0 cm
15.0 cm 10.0 cm
30.0 cm
and P QR have the com m on sid e QR and are
then congruent by the angle-sid e-angle
theorem . Thus, the correspond ing sid es PQ and
q2
10.0 cm p 1
10.0 cm 15.0 cm
behind the
m irror as the object is in front of it.
25.0 cm are equal, or the im age is as far
390
23.6
CH APTER 23
(a) Since the object is in front of the m irror,
20.0 cm 25.0 cm
p2 q2
f2
11.1 cm . With the im age behind the m irror,
p2 q2
20.0 cm 25.0 cm
q 0 . The m irror equation gives the rad ius of curvature as
2
R
or R
1
p
2
1
q
1
1
1.00 cm 10.0 cm
10.0 cm
9
10 1
10.0 cm
2.22 cm
(b) The m agnification is virtual
23.7
(a) Since the m irror is concave, M 0 . Because the object is located in front of the
m irror, p 0 . The m irror equation, 1 p 1 q 2 R then gives the im age d istance as
q
pR
2p R
40.0 cm 20.0 cm
2 40.0 cm
20.0 cm
13.3 cm
Since q 0 , the im age is located 13.3 cm in front of the mirror .
(b) q2
p2 f 2
p2 f 2
Because M 2
14.0 cm
14.0 cm
q2
p2
16.0 cm
16.0 cm
7.47 cm
14.0 cm
7.47 cm
0.533 , M
since 7.47 cm in front of the second lens , h
M1M 2
Mh
2.00
0.533
1.07 1.00 cm
1.07 and
1.07 cm .
M irrors and Lenses
23.8
The lateral m agnification is given by M
391
0 . Therefore, the im age d istance is
upright
The m irror equation:
2
R
gives
50.0 cm 31.0 cm
q2
1
p
1
q
or
virtual
19.0 cm
The negative sign tells us that the surface is convex. The m agnitud e of the rad ius of
curvature of the cornea is
23.9
19.0 cm 20.0 cm
q2 f 2
q2 f 2
p2
19.0 cm 20.0 cm
9.74 cm
(a) For a convex m irror, the focal length is q1 50.0 cm 9.74 cm 40.3 cm , and w ith the
object in front of the m irror, p1
q1 f1
q1 f1
40.3 cm 10.0 cm
40.3 cm 10.0 cm
13.3 cm . The m irror
equation, 13.3 cm , then gives
M
q1
p1
M1 M 2
q2
p2
40.3 cm
13.3 cm
19.0 cm
9.74 cm
5.90
With M 0 , the im age is located inverted .
(b) The m agnification is
M
h
h
q
p
7.50 cm
30.0 cm
Since virtual and M
h
Mh
0.250
0 , the im age is virtual and upright . Its height is
0.250 2.0 cm
0.50 cm
392
23.10
CH APTER 23
The im age w as initially upright but becam e inverted w hen Dina w as m ore than 30 cm
from the m irror. From this inform ation, w e know that the m irror m ust be concave
because a convex m irror w ill form only upright, virtual im ages of real objects.
When the object is located at the focal point of a concave m irror, the rays leaving the
m irror are parallel, and no im age is form ed . Since Dina observed that her im age
d isappeared w hen she w as about 30 cm from the m irror, w e know that the focal length
m ust be f 30 cm . Also, for spherical m irrors, R 2 f . Thus, the rad ius of curvature
of this concave m irror m ust be R 60 cm .
23.11
The magnified, virtual im ages form ed by a concave m irror are upright, so M
Thus, M
q
q
p
h
h
5.00 cm
2.00 cm
2.50 p
2.50
2.50 , giving
3.00 cm
7.50 cm
The m irror equation then gives,
or
1
f
2
R
1
p
f
7.50 cm
1.50
1
q
1
3.00 cm
5.00 cm
1
7.50 cm
2.50 1
7.50 cm
0.
M irrors and Lenses
23.12
393
Realize that the m agnitud e of the rad ius of curvature, R , is the sam e for both sid es of
the hubcap. For the convex sid e, R
R ; and for the concave sid e, R
R . The object
d istance p is positive (real object) and has the sam e value in both cases. Also, w e w rite
the virtual im age d istance as q
q in each case. The m irror equation then gives:
For the convex sid e,
For the concave sid e,
1
q
2
R
1
q
2
R
1
p
1
p
or
q
or
q
R p
R
2p
R p
R
2p
[1]
[2]
Com paring Equations [1] and [2], w e observe that the sm aller m agnitud e im age
d istance, q 10.0 cm , occurs w ith the convex sid e of the m irror. H ence, w e have
1
10.0 cm
and for the concave sid e, q
(a) Ad d ing Equations [3] and [4] yield s
4
R
(b) Subtracting [3] from [4] gives
The im age is upright, so M
q
p
M
1
p
[3]
30.0 cm gives
1
30.0 cm
23.13
2
R
2
p
2
R
1
p
[4]
3 1
30.0 cm
3 1
30.0 cm
or
or
p
R
15.0 cm
60.0 cm
0 , and w e have
2.0 , or q
2.0 p
2.0 25 cm
50 cm
The rad ius of curvature is then found to be
2
R
1
p
1
q
1
25 cm
1
50 cm
2 1
, or R
50 cm
2
0.50 m
+1
1.0 m
394
23.14
CH APTER 23
(a) Your ray d iagram should be carefully d raw n to scale and look like the d iagram
given below :
(b) From the m irror equation w ith
1
1
q2
1
and q2
7.50 cm
7.50 cm , the im age
d istance is
1
f
n 1
1
R1
1
R2
and the m agnification is
1
5.00 cm
n 1
1
9.00 cm
1
. Thus, you should
11.0 cm
find that the im age is
upright, located 6.00 cm behind the mirror, six-tenths the size of the object
M irrors and Lenses
23.15
395
The focal length of the m irror m ay be found from the given object and im age d istances
8.00 cm 5.00 cm
p1 f
13.3 cm , or
as q1
p1 f
8.00 cm 5.00 cm
f
152 cm 18.0 cm
pq
p q
16.1 cm
152 cm 18.0 cm
For an upright im age tw ice the size of the object, the m agnification is
q2
6.67 cm 8.00 cm
p2 R
2 p2 R
2 6.67 cm
8.00 cm
10.0 cm giving q
2.00 p
Then, using the m irror equation again, 1 p 1 q 1 f becom es
or
23.16
1
p
1
q
1
p
p
f
2.00
1
2.00 p
2 1
2.00 p
16.1 cm
2.00
1
f
8.05 cm
(a) The m irror is convex, so M 0 , and w e have f
f
8.0 cm . The im age is
qf
virtual, so p
, or q
q . Since w e also know that q p 3 , the m irror
q f
equation gives
1
p
1
q
1
p
3
p
1
f
2
p
or
1
8.0 cm
and
p
16 cm
This m eans that w e have a real object located 16 cm in front of the mirror .
(b) The m agnification is M
q p
q p
1 3 . Thus, the im age is upright and one
third the size of the object.
23.17
(a) We know that the object d istance is p
10.0 cm . Also, M 0 since the im age is
1 2 since the im age is half the size of the object. Thus, w e have
upright, and M
M
q
p
q
10.0 cm
1
2
and the im age is seen to be located q
6.00 m 1.00
or
f
pf
p f
10.0 m
1.00 1.50
12.0 m
10.0 m 12.0 m
12.0 m
5.45 m .
396
CH APTER 23
(b) From the m irror equation, 1 p 1 q 1 f , w e find the focal length to be
f
23.18
pq
p q
10.0 cm
5.00 cm
10.0 cm+
5.00 cm
10.0 cm
(a) Since the m irror is concave, R 0 , giving n2 1.33 and
f
(q
6.00 m 1.33
46.9 m . Because the im age is upright
1.33 1.50
10.0 m
46.9 m
pf
8.24 m ) and three tim es the size of the object
p f
10.0 m 46.9 m
( 8.24 m to the left of the lens ), w e have
M
q
p
and
3
q
3p
The m irror equation then gives
1
p
1
3p
2
3p
1
12 cm
or
f
6.00 m 2.00
2.00 1.50
24.0 m
(b) The need ed ray d iagram , w ith the object 8.0 cm in front of the m irror, is show n
below :
From a carefully d raw n scale d raw ing, you should find that the im age is
10.0 m 24.0 m
pf
q
17.1 m
p f
10.0 m 24.0 m
23.19
(a) An im age form ed on a screen is a real im age. Thus, the m irror m ust be concave
since, of m irrors, only concave m irrors can form real im ages of real objects.
M irrors and Lenses
397
(b) The m agnified , real im ages form ed by concave m irrors are inverted , so M 0 and
q
p
M
5 , or p
q
5
The object should be f
5.0 m
1.0 m
5
0
(a – revisited ) The focal length of the m irror is
n1 n2 , or f
23.20
(a) From
1
p
1
q
5.0 m
6
0.83 m
2
, w e find q
R
Rp
2p R
1.00 m p
2 p 1.00 m
The table gives the im age position at a few critical
Object
Image
points in the m otion. Betw een p 3.00 m and
D istance, p
D istance, q
3.00 m
0.600 m
p 0.500 m , the real im age m oves from 0.500 m
0.500 m
to positive infinity. From
0
0
qmirror pmirror 40.0 cm p1 40.0 cm 16.7 cm 23.3 cm
to p 0 , the virtual im age m oves from negative
infinity to 0.
N ote the “ jum p” in the im age position as the ball passes through the focal point of
the m irror.
(b) The ball and its im age coincid e w hen p 0 and w hen
1 p 1 p 2 p 2 R , or n1 1.55
From n2 , w ith v0 y
0 , the tim es for the ball to fall from p 3.00 m to these
positions are found to be
2
t
y
ay
1
f water
2
2.00 m
0.639 s and
9.80 m s2
1.55 1.33
1.33
1
R1
1
R2
398
23.21
CH APTER 23
From
f water
f air
1.33
1.00
1.55 1.00
1.55 1.33
1.33
0.55
, w ith R
0.22
, the im age position is found
to be
q
n2
p
n1
1.00
1.309
50.0 cm
38.2 cm
or the virtual im age is 38.2 cm below the upper surface of the ice
23.22
The center of curvature of a convex surface is located behind the surface, and the sign
convention for refracting surfaces (Table 23.2 in the textbook) states that R 0 , giving
R
8.00 cm . The object is in front of the surface ( p 0 ) and in air ( n1 1.00 ), w hile the
second m ed ium is glass ( n2 1.50 ). Thus, n1 p n2 q
1.00 1.50
p
q
23.23
1.50 1.00
8.00 cm
(a) If p 20.0 cm ,
q
(b) If p 8.00 cm ,
q
(c) If p 4.00 cm ,
q
(d ) If p 2.00 cm ,
q
and red uces to q
24.0 cm 20.0 cm
20.0 cm 16.0 cm
24.0 cm 8.00 cm
8.00 cm 16.0 cm
24.0 cm 4.00 cm
4.00 cm 16.0 cm
24.0 cm 2.00 cm
2.00 cm 16.0 cm
n2 n1 R becom es
24.0 cm p
p 16.0 cm
120 cm
24.0 cm
8.00 cm
3.43 cm
Since the center of curvature of the surface is on the sid e the light com es from , R 0
n n2 n2 n1
giving R
becom es
4.0 cm . Then, 1
p q
R
1.00
q
1.00 1.50
4.0 cm
1.50
, or q
4.0 cm
Thus, the m agnification M
h
n1q
h
n2 p
1.50
h
h
4.0 cm
n1 q
, gives
n2 p
4.0 cm
1.00 4.0 cm
2.5 mm
3.8 mm
M irrors and Lenses
23.24
399
For a plane refracting surface R
n1
p
n2
q
n2
n1
R
n2
p
n1
becom es q
(a) When the pool is full, p 2.00 m and
q
1.00
1.333
2.00 m
1.50 m
or the pool appears to be 1.50 m d eep
(b) If the pool is half filled , then p 1.00 m and q
0.750 m . Thus, the bottom of the
pool appears to be 0.75. m below the w ater surface or 1.75 m below ground level.
23.25
As parallel rays from the Sun object distance, p
enter the transparent sphere from
air n1 1.00 , the center of curvature of the surface is on the sid e the light is going
tow ard (back sid e). Thus, R 0 . It is observed that a real im age is form ed on the surface
opposite the Sun, giving the im age d istance as q
2R .
Then
n1
p
n2
q
n2
w hich red u ces to
n1
R
becom es
n 2n 2.00
0
and gives
n
2R
n 1.00
R
n
2.00
400
23.26
CH APTER 23
Light scattered from the bottom of the plate und ergoes tw o refractions, once at the top
of the plate and once at the top of the w ater. All surfaces are planes R
, so the
im age d istance for each refraction is q
nwater
p1B
nglass
q1B
n2 n1 p . At the top of the plate,
1.333
8.00 cm
1.66
6.42 cm
or the first im age is 6.42 cm below the top of the plate. This im age serves as a real object
for the refraction at the top of the w ater, so the final im age of the bottom of the plate is
form ed at
nair
p2B
nwater
q2B
nair
nwater
1.00
18.4 cm
1.333
12.0 cm
q1B
13.8 cm or 13.8 cm below the w ater surface.
N ow , consid er light scattered from the top of the plate. It und ergoes a single refraction,
at the top of the w ater. This refraction form s an im age of the top of the plate at
qT
nair
pT
nwater
1.00
12.0 cm
1.333
9.00 cm
or 9.00 cm below the w ater surface.
The apparent thickness of the plate is then
y
23.27
q2B
qT
13.8 cm 9.00 cm
4.8 cm
In the d raw ing at the right, object O (the jelly fish)
is located d istance p in front of a plane w ater-glass
interface. Refraction at that interface prod uces a
virtual im age I at d istance q in front it. This
im age serves as the object for refraction at the
glass-air interface. This object is located d istance
p q t in front of the second interface, w here t
is the thickness of the layer of glass. Refraction at
the glass-air interface prod uces a final virtual
im age, I, located d istance q in front of this
interface.
M irrors and Lenses
401
From n1 p n2 q (n2
for a plane, the relation betw een the object
n1 ) R w ith R
and im age d istances for refraction at a flat surface is q
n2 n1 p . Thus, the im age
d istance for the refraction at the w ater-glass interface is q
ng nw p . This gives an
object d istance for the refraction at the glass-air interface of p
(ng nw ) p t and a final
im age position (m easured from the glass-air interface) of
na
p
ng
q
na
ng
ng
nw
na
p
nw
p t
na
t
ng
(a) If the jelly fish is located 1.00 m (or 100 cm ) in front of a 6.00 cm thick pane of glass,
then p
100 cm and t 6.00 cm and the position of the final im age relative to the
glass-air interface is
1.00
100 cm
1.333
q
1.00
1.50
6.00 cm
(b) If the thickness of the glass is negligible (t
from the glass-air interface is
q
ng
na
ng
nw
na
p
nw
p 0
79.0 cm
0.790 m
0) , the d istance of the final im age
1.00
100 cm
1.333
75.0 cm
0.750 m
so w e see that the 6.00 cm thickness of the glass in part (a) m ad e a 4.00 cm
d ifference in the apparent position of the jelly fish.
(c) The thicker the glass, the greater the d istance betw een the final im age and the outer
surface of the glass.
23.28
The w all of the aquarium (assum ed to be of negligible thickness) is a plane ( R
n n2
refracting surface separating w ater (n1 1.333) and air (n2 1.00) . Thus, 1
p q
gives the im age position as q
n2
p
n1
p , the change in the im age position is
p
. When the object position changes by
1.333
q
p
. The apparent speed of the fish is
1.333
then given by
vimage
q
t
p
t
1.333
2.00 cm s
1.333
)
n2 n1
R
1.50 cm s
402
23.29
CH APTER 23
With R1
as
1
f
or
23.30
f
n 1
1
R1
1
0.050 0 cm
1
R2
1
1.50 1
1
2.00 cm
1
2.50 cm
0.050 0 cm
1
20.0 cm
The lens m aker’ s equation is used to com pute the focal
length in each case.
(a)
(b)
23.31
2.50 cm , the lens m aker’ s equation gives the focal length
2.00 cm and R2
1
f
n 1
1
R1
1
R2
1
f
1.44 1
1
12.0 cm
1
18.0 cm
1
f
1.44 1
1
18.0 cm
1
12.0 cm
f
16.4 cm
f
The focal length of a converging lens is positive, so f
then yield s a focal length of
q
f
(a) When p
q
10.0 cm . The thin lens equation
p 10.0 cm
pf
p
16.4 cm
p 10.0 cm
20.0 cm ,
20.0 cm 10.0 cm
20.0 cm 10.0 cm
20.0 cm
and
M
q
p
20.0 cm
20.0 cm
1.00
so the im age is located 20.0 cm beyond the lens , is real (q 0) , is inverted (M
and is the same size as the object M
(b) When p
f
1.00 .
10.0 cm , the object is at the focal point and no image is formed .
Instead , parallel rays emerge from the lens .
0) ,
M irrors and Lenses
403
(c) When p 5.00 cm ,
q
5.00 cm 10.0 cm
10.0 cm
5.00 cm 10.0 cm
and
M
q
p
10.0 cm
5.00 cm
2.00
so the im age is located 10.0 cm in front of the lens , is virtual (q 0) , is
upright (M
23.32
0) , and is the twice the size of the object M
2.00 .
(a) and (b) Your scale d raw ings should look sim ilar to those given below :
Figure (a)
Figure (b)
A carefully d raw n-to-scale version of Figure (a) should yield a real, inverted im age
that is located 20 cm in back of the lens and the sam e size as the object. Sim ilarly, a
carefully d raw n-to-scale version of Figure (b) should yield an upright, virtual im age
located 10 cm in front of the lens and tw ice the size of the object.
(c) The accuracy of the graph d epend s on how accurately the ray d iagram s are d raw n.
Sources of uncertainty: a parallel line from the tip of the object m ay not be exactly
parallel; the focal points m ay not be exactly located ; lines through the focal points
m ay not be exactly the correct slope; the location of the intersection of tw o lines
cannot be d eterm ined w ith com plete accuracy.
23.33
From the thin lens equation,
q
f p
p f
p
1
p
1
q
1
, the im age d istance is found to be
f
20.0 cm p
20.0 cm p
20.0 cm
p 20.0 cm
(a) If p 40.0 cm , then q
13.3 cm and M
q
p
13.3 cm
40.0 cm
The im age is virtual, upright, and 13.3 cm in front of the lens
13
404
CH APTER 23
(b) If p 20.0 cm , then q
M
q
p
10.0 cm and
10.0 cm
20.0 cm
12
The im age is virtual, upright, and 10.0 cm in front of the lens
(c) When p 10.0 cm , q
6.67 cm and M
q
p
6.67 cm
10.0 cm
2 3
The im age is virtual, upright, and 6.67 cm in front of the lens
23.34
(a) and (b) Your scale d raw ings should look sim ilar to those given below :
Figure (a)
Figure (b)
A carefully d raw n-to-scale version of Figure (a) should yield an upright, virtual
im age located 13.3 cm in front of the lens and one-third the size of the object.
Sim ilarly, a carefully d raw n -to-scale version of Figure (b) should yield an upright,
virtual im age located 6.7 cm in front of the lens and tw o-third s the size of the object.
(c) The results of the graphical solution are consistent w it h the algebraic answ ers found
in problem 23.33, allow ing for sm all d eviances d ue to uncertainties in
m easurem ent. Graphical answ ers m ay vary, d epend ing on the size of the graph
paper and accuracy of the d raw ing.
M irrors and Lenses
23.35
405
(a) The real im age case is show n in the ray
d iagram . N otice that p q 12.9 cm , or
q 12.9 cm p . The thin lens equation, w ith
f 2.44 cm , then gives
1
p
or p2
1
12.9 cm
p
1
2.44 cm
12.9 cm p 31.5 cm2
0
Using the quad ratic form ula to solve gives
p 9.63 cm or p 3.27 cm
Both are valid solutions for the real im age case.
(b) The virtual im age case is show n in the second
d iagram . N ote that in this case, q
12.9 cm p ,
so the thin lens equation gives
1
p
or p2
1
12.9 cm
p
1
2.44 cm
12.9 cm p 31.5 cm2
0
The quad ratic form ula then gives p 2.10 cm or p
15.0 cm
Since the object is real, the negative solution m ust be rejected leaving
23.36
p 2.10 cm .
We m ust first realize that w e are looking at an upright, m agnified , virtual im age. Thus,
w e have a real object located betw een a converging lens and its front -sid e focal point, so
q 0, p 0, and f 0 .
The m agnification is M
1
p
1
2p
1
2p
1
or f
f
q
p
2 , giving q
2 p 2 2.84 cm
2 p . Then, from the thin lens equation,
5.68 cm
406
23.37
CH APTER 23
It is d esired to form a m agnified , real im age on the screen using a single thin lens. To d o
this, a converging lens m ust be used and the im age w ill be inv erted . The m agnification
then gives
M
h
h
1.80 m
24.0 10-3 m
q
, or q 75.0 p
p
Also, w e know that p q 3.00 m . Therefore, p 75.0 p 3.00 m giving
(b)
p
3.00 m
76.0
3.95 10
2
m
39.5 mm
(a) The thin lens equation th en gives
or f
23.38
75.0
76.0
39.5 mm
1
75.0 p
76.0
75.0 p
1
f
39.0 mm
To have a m agnification of M
q p 3.00 , it is necessary that q 3.00 p . The thin
lens equation, w ith f
18.0 cm for the convergent convex lens, gives the required
object d istance as
1
p
23.39
75.0
p
76.0
1
p
1
3.00 p
2
3.00 p
1
18.0 cm
p
or
2 18.0 cm
3.00
Since the light rays incid ent to the first lens are parallel, p1
equation gives q1 f1
10.0 cm .
12.0 cm
and the thin lens
The virtual im age form ed by the first lens serves as the object for the second lens, so
p2 30.0 cm q1 40.0 cm . If the light rays leaving the second lens are parallel, then
q2
23.40
and the thin lens equation gives f 2
p2
40.0 cm .
(a) Solving the thin lens equation for the im age d istance q gives
1
q
1
f
1
p
p f
pf
or
q
pf
p
f
407
M irrors and Lenses
(b) For a real object, p 0 and p p . Also, for a d iverging lens, f
The result of part (a) then becom es
q
p
f
p
0 and f
f .
p f
p
f
f
Thus, w e see that q 0 for all numeric values of p and f . Since negative im age
d istances m ean virtual im ages, w e conclud e that a d iverging lens w ill alw ays form
virtual im ages of real objects.
(c) For a real object, p 0 and p p . Also, for a converging lens, f
The result of part (a) then becom es
q
p f
p
f
0
if
p
f
0 and f
f .
0
Since q m ust be positive for a real im age, w e see that a converging lens w ill form
real im ages of real objects only w hen p
f (or p f since both p and f are
positive in this situation).
23.41
The thin lens equation gives the im age position for the first lens as
q1
p1 f1
p1 f1
30.0 cm 15.0 cm
30.0 cm 15.0 cm
30.0 cm
and the m agnification by this lens is M 1
q1
p1
30.0 cm
30.0 cm
1.00
The real im age form ed by the first lens serves as the object for the second lens, so
p2 40.0 cm q1
10.0 cm . Then, the thin lens equation gives
q2
p2 f 2
p2 f 2
10.0 cm 15.0 cm
10.0 cm 15.0 cm
30.0 cm
and the m agnification by the second lens is
M2
q2
p2
30.0 cm
10.0 cm
3.00
Thus, the final, virtual im age is located 30.0 cm in front of the second lens
408
CH APTER 23
and the overall m agnification is M
23.42
M1M 2
1.00
3.00
3.00
(a) With p1
15.0 cm , the thin lens equation gives the position of the im age form ed
by the first lens as
q1
p1 f1
p1 f1
15.0 cm 10.0 cm
30.0 cm
15.0 cm 10.0 cm
This im age serves as the object for the second lens, w ith an object d istance of
p2 10.0 cm q1 10.0 cm 30.0 cm
20.0 cm (a virtual object). If the im age form ed
by this lens is at the position of O1 , the im age d istance is
q2
10.0 cm
p1
10.0 cm 15.0 cm
25.0 cm
The thin lens equation then gives the focal length of the second lens as
p2 q2
p2 q2
f2
20.0 cm
25.0 cm
20.0 cm 25.0 cm
11.1 cm
(b) The overall m agnification is
M
M1 M 2
(c) Since q2
q1
p1
q2
p2
30.0 cm
15.0 cm
25.0 cm
20.0 cm
2.50
0 , the final im age is virtual ; and since M 0 , it is upright
M irrors and Lenses
23.43
4.00 cm 8.00 cm
p1 f1
p1 f1
From the thin lens equation, q1
8.00 cm
4.00 cm 8.00 cm
8.00 cm
q1
p1
The m agnification by the first lens is M 1
409
2.00
4.00 cm
The virtual im age form ed by the first lens is the ob ject for the second lens, so
p2 6.00 cm q1
14.0 cm and the thin lens equation gives
q2
p2 f 2
p2 f 2
14.0 cm
14.0 cm
16.0 cm
7.47 cm
16.0 cm
The m agnification by the second lens is M 2
m agnification is M
M1M 2
2.00
0.533
q2
p2
7.47 cm
14.0 cm
0.533 , so the overall
1.07
The position of the final im age is 7.47 cm in front of the second lens , and its height is
h
Mh
1.07 1.00 cm
Since M
23.44
1.07 cm
0 , the final im age is upright ; and since q2
0 , this im age is virtual
(a) We start w ith the final im age and w ork backw ard . From Figure P23.44, observe that
q2
50.0 cm 31.0 cm
19.0 cm . The thin lens equation
then gives
p2
q2 f 2
q2 f 2
19.0 cm 20.0 cm
19.0 cm 20.0 cm
9.74 cm
The im age form ed by the first lens serves as the object for the second lens and is
located 9.74 cm in front of the second lens.
Thus, q1 50.0 cm 9.74 cm 40.3 cm and the thin lens equation gives
p1
q1 f1
q1 f1
40.3 cm 10.0 cm
40.3 cm 10.0 cm
13.3 cm
The original object should be located 13.3 cm in front of the first lens.
410
CH APTER 23
(b) The overall m agnification is
M
q1
p1
M1 M 2
q2
p2
19.0 cm
40.3 cm
13.3 cm
5.90
9.74 cm
(c) Since M 0 , the final im age is inverted ;
and since q2
23.45
0 , it is virtual
N ote: Final answ ers to this problem are highly sensitive to round -off error. To avoid
this, w e retain extra d igits in interm ed iate answ ers and round only the final answ ers to
the correct num ber of significant figures.
Since the final im age is to be real and in the film plane, q2
Then, the thin lens equation gives p2
q2 f 2
q2 f 2
d
d 13.0 cm
d 13.0 cm
From Figure P23.45, observe that d 12.0 cm . The above result then show s that p2
so the object for the second len s w ill be a virtual object.
The object of the second lens L2 is the im age form ed by the first lens L1 , so
q1
12.0 cm d
If d 5.00 cm , then q1
p2 12.0 cm d 1
13.0 cm
d 13.0 cm
12.0 cm
15.125 cm ; and w hen d 10.0 cm , q1
From the thin lens equation, p1
q1 f1
q1 f1
d2
d 13.0 cm
45.333 cm
q1 15.0 cm
q1 15.0 cm
5.00 cm , then p1 1.82 103 cm 18.2 m
When q1
15.125 cm d
When q1
45.333 cm d 10.0 cm , then p1
22.4 cm 0.224 m
Thus, the range of focal d istances for this cam era is 0.224 m to 18.2 m
0,
M irrors and Lenses
23.46
411
(a) From the thin lens equation, the im age d istance for the first lens is
15.0 cm 10.0 cm
p1 f1
p1 f1
q1
30.0 cm
15.0 cm 10.0 cm
(b) With q1
30.0 cm , the im age of the first lens is located 30.0 cm in back of that lens.
Since the second lens is only 10.0 cm beyond the first lens, this m eans that the first
lens is trying to form its im age at a location 20.0 cm beyond the second lens .
(c) The im age the first lens form s (or w ould form if allow ed to d o so) serves as the
object for the second lens. Consid ering the answ er to part (b) above, w e see that this
w ill be a virtual object, w ith object d istance p2
20.0 cm .
(d ) From the thin lens equation, the im age d istance for the second lens is
q2
p2 f 2
p2 f 2
(e)
M1
q1
p1
(f)
M2
q2
p2
(g) M
23.47
Since q
20.0 cm 5.00 cm
30.0 cm
15.0 cm
2.00
1
p
0.200
0.200
0.400
0 , the final im age is real , and since M 0 , that im age is inverted .
8.00 cm w hen p
1
f
4.00 cm
2.00
4.00 cm
20.0 cm
M1M 2
(h) Since q2
20.0 cm 5.00 cm
1
q
1
10.0 cm
10.0 cm , w e find that
1
8.00 cm
18.0
80.0 cm
1
20.0 cm
18.0 4.00
80.0 cm
Then, w hen p 20.0 cm ,
1
q
or q
1
f
80.0 cm
14.0
1
p
18.0
80.0 cm
14.0
80.0 cm
5.71 cm
Thus, a real im age is form ed 5.71 cm in front of the mirror
412
23.48
CH APTER 23
(a) We are given that p 5 f , w ith both p and f being positive. The thin lens equation
then gives
q
5f f
pf
p
f
5f
5f 4
q
p
(b) M
5f
4
f
1
4
5f
(c) Since q 0 , the im age is real . Because M 0 , the im age is inverted . Since the
object is real, it is located in front of the lens, and w ith q 0 , the im age is located in
back of the lens. Thus, the im age is on the opposite side of the lens from the object.
23.49
Since the object is very d istant p
, the im age d istance equals the focal length, or
q 50.0 mm . N ow consid er tw o rays that pass und eviated through the center of the
thin lens to opposite sid es of the im age as show n in the sketch below .
From the sketch, observe that
tan
1
35.0 mm
2
50.0 mm
2
0.350
Thus, the angular w id th of the im age is
2tan
23.50
1
0.350
38.6
(a) Using the sign convention from Table 23.2, the rad ii of curvature of the surfaces are
R1
15.0 cm and R2
10.0 cm . The lens m aker’ s equation then gives
1
f
(b) If p
n 1
1
R1
, then q
1
R2
1.50 1
f
12.0 cm
The thin lens equation gives, q
1
1
15.0 cm 10.0 cm
p
pf
p
f
12.0 cm
p 12.0 cm
or f
12.0 cm
and the follow ing results:
M irrors and Lenses
(c) If p 3 f
23.51
36.0 cm , q
(d ) If p
f
(e) If p
f 2
413
9.00 cm
12.0 cm , q
6.00 cm
6.00 cm , q
4.00 cm
As light passes left-to-right through the lens, the im age position is given by
q1
p1 f1
p1 f1
100 cm 80.0 cm
400 cm
100 cm 80.0 cm
This im age serves as an object for the m irror w ith an object d istance of
p2 100 cm q1
300 cm (virtual object). From the m irror equation, the position of the
im age form ed by the m irror is
q2
300 cm
p2 f 2
p2 f 2
300 cm
50.0 cm
50.0 cm
60.0 cm
This im age is the object for the lens as light now passes through it going right -to-left.
The object d istance for the lens is p3 100 cm q2 100 cm
60.0 cm , or p3 160 cm .
From the thin lens equ ation,
q3
p3 f 3
p3 f 3
160 cm 80.0 cm
160 cm
160 cm 80.0 cm
Thus, the final im age is located 160 cm to the left of the lens
The overall m agnification is M
M
400 cm
100 cm
M 1M 2 M 3
60.0 cm
300 cm
Since M 0 , the final im age is inverted
q1
p1
160 cm
160 cm
q2
p2
0.800
q3
, or
p3
414
23.52
CH APTER 23
Since the object is m id w ay betw een the lens and m irror, the object d istance for the
m irror is p1
12.5 cm . The m irror equation gives the im age position as
1
q1
2
R
1
p1
2
1
20.0 cm 12.5 cm
5 4
50.0 cm
1
, or q1
50.0 cm
50.0 cm
This im age serves as the object for the lens, so p2 25.0 cm q1
25.0 cm . N ote that
since p2 0 , this is a virtual object. The thin lens equation gives the im age position for
the lens as
Since q2
25.0 cm
p2 f 2
p2 f 2
q2
25.0 cm
16.7 cm
16.7 cm
50.3 cm
0 , this is a virtual im age that is located 50.3 cm in front of the lens or
25.3 cm behind the mirror . The overall m agnification is
M
Since M
23.53
q1
p1
M1 M 2
q2
p2
50.0 cm
12.5 cm
0 , the final im age is upright
A hem isphere is too thick to be d escribed
as a thin lens. The light is und eviated on
entry into the flat face. We next consid er
the light’ s exit from the curved surface,
for w hich R
6.00 cm .
The incid ent rays are parallel, so p
Then,
n1
p
n2
q
from w hich
n2
n1
R
becom es 0
q 10.7 cm
.
1.00
q
1.00 1.56
6.00 cm
50.3 cm
25.0 cm
+8.05
M irrors and Lenses
23.54
415
(a) The thin lens equation gives the im age d istance for the first lens as
q1
p1 f1
p1 f1
40.0 cm 20.0 cm
40.0 cm 20.0 cm
40.0 cm
q1
p1
The m agnification by this lens is then M 1
40.0 cm
40.0 cm
1.00
The real im age form ed by the first lens is the object for the second lens. Thus,
p2 50.0 cm q1
10.0 cm and the thin lens equation gives
q2
p2 f 2
p2 f 2
10.0 cm 5.00 cm
10.0 cm 5.00 cm
10.0 cm
The final im age is 10.0 cm in back of the second lens
(b) The m agnification by the second lens is M 2
q2
p2
10.0 cm
10.0 cm
1.00 , so the overall
m agnification is M M1M 2
1.00 1.00
1.00 . Since this m agnification has a
value of unity, the final im age is the sam e size as the original object, or
h M h1
1.00 2.00 cm
2.00 cm
The im age d istance for the second lens is positive, so the final im age is real .
(c) When the tw o lenses are in contact, the focal length of the com bination is
1
f
1
f1
1
f2
1
20.0 cm
1
, or f
5.00 cm
4.00 cm
The im age position is then
q
5.00 cm 4.00 cm
pf
p
f
5.00 cm 4.00 cm
20.0 cm
416
23.55
CH APTER 23
With light going through the piece of glass from left to right, the rad ius of the first
surface is positive and that of the secon d surface is negative accord ing to the sign
convention of Table 23.2. Thus, R1
2.00 cm and R2
4.00 cm .
Applying
n1
p
n2
q
1.00
1.00 cm
n2
n1
R
1.50
q1
to the first surface gives
1.50 1.00
2.00 cm
w hich yield s q1
2.00 cm . The first surface form s a virtual im age 2.00 cm to the left of
that surface and 16.0 cm to the left of the second surface.
The im age form ed by the first surface is the object for the second surface, so
n n2 n2 n1
gives
p2
16.0 cm and 1
p q
R
1.50
16.0 cm
1.00
q2
1.00 1.50
or q2
4.00 cm
32.0 cm
The final im age form ed by the piece of glass is a real image located
32.0 cm to the right of the second surface
23.56
Consid er an object O1 at d istance p1 in
front of the first lens. The thin lens equation
gives the im age position for this lens as
1 1 1
.
q1 f1 p1
The im age, I1 , form ed by the first lens serves as the object, O2 , for the second lens. With
the lenses in contact, this w ill be a virtual object if I1 is real and w ill be a real object if I1
is virtual. In either case, if the thicknesses of the lenses m ay be ignored ,
p2
q1 and
1
p2
1
q1
1
f1
1
p1
Applying the thin lens equation to the second lens,
1
p2
1
q2
1
becom es
f2
M irrors and Lenses
1
f1
1
p1
1
q2
1
1
or
f2
p1
1
q2
1
f1
417
1
f2
Observe that this result is a thin lens typ e equation relating the position of the original
object O1 and the position of the final im age I2 form ed by this tw o lens com bination.
Thus, w e see that w e m ay treat tw o thin lenses in contact as a single lens having a focal
1 1 1
length, f, given by
f
f1 f 2
23.57
From the thin lens equation, the im age d istance for the first lens is
q1
p1 f1
p1 f1
40.0 cm 30.0 cm
40.0 cm 30.0 cm
and the m agnification by this lens is M 1
120 cm
q1
p1
120 cm
40.0 cm
3.00
The real im age form ed by the first lens serves as the object for the second lens, w ith
object d istance of p2 110 cm q1 10.0 cm (a virtual object). The thin lens equation
gives the im age d istance for the second lens as
q2
(a) If f2
M2
p2 f 2
p2 f 2
10.0 cm f 2
10.0 cm
20.0 cm , then q2
q2 p2
20.0 cm
f2
20.0 cm and the m agnification by the second lens is
10.0 cm
2.00
The final im age is located 20.0 cm to the right of the second lens
and the overall m agnification is M
M1M 2
(b) Since M 0 , the final im age is inverted
3.00
2.00
6.00
418
CH APTER 23
(c) If f2
20.0 cm , then q2
q2
p2
M2
and
6.67 cm
6.67 cm
-10.0 cm
0.667
The final im age is 6.67 cm to the right of the second lens
and the overall m agnification is M
M1M 2
3.00
0.667
Since M 0 , the final im age is inverted
23.58
The object is located at the focal point of the upper
m irror. Thus, the upper m irror creates an im age at
infinity (that is, parallel rays leave this m irror). The
low er m irror focuses these parallel rays at its focal
point, located at the hole in the upper m irror. Thus, the
image is real, inverted, and actual size
For the upper m irror:
1
p
1
q
1
f
1
7.50 cm
1
q1
1
:
7.50 cm
q1
For the low er m irror:
1
1
q2
1
:
7.50 cm
q2
7.50 cm
Light d irected into the hole in the upper m irror
reflects as show n, to behave as if it w ere reflecting
from the hole.
23.59
(a) The lens m aker’ s equation,
1
5.00 cm
n 1
1
9.00 cm
w hich sim plifies to n 1
1
f
n 1
1
R1
1
, gives
R2
1
11.0 cm
1
99.0
5.00 11.0 9.00
1.99
2.00
M irrors and Lenses
419
(b) As light passes from left to right through the lens, the thin lens equation gives the
im age d istance as
q1
p1 f
p1 f
8.00 cm 5.00 cm
13.3 cm
8.00 cm 5.00 cm
This im age form ed by the lens serves as an object for the m irror w ith object d istance
p2 20.0 cm q1
6.67 cm . The m irror equation then gives
q2
p2 R
2 p2 R
6.67 cm 8.00 cm
2 6.67 cm
10.0 cm
8.00 cm
This real im age, form ed 10.0 cm to the left of the m irror, serves as an object for the
lens as light passes through it from right to left. The object d istance is
p3 20.0 cm q2
10.0 cm , and the thin lens equation gives
q3
10.0 cm 5.00 cm
p1 f
p3
f
10.0 cm
10.0 cm 5.00 cm
The final im age is located 10.0 cm to the left of the lens and its overall
m agnification is
M
q1
p1
M 1M 2 M 2
q3
p3
q2
p2
13.3
8.00
10.0
6.67
(c) Since M 0 , the final im age is inverted
23.60
From the thin lens equation, the object d istance is p
(a) If q
4 f , then p
(b) When q
4f f
4f
4f
3
f
3f f
3 f , w e find p
(c) In case (a), M
and in case (b), M
q
p
3f
4f
4f 3
q
p
3f
3f 4
3f
4
f
3
4
qf
q
f
10.0
10.0
2.50
420
23.61
CH APTER 23
If R1
3.00 m and R2
1
f
f
or
n1
n2
1
6.00 m , the focal length is given by
1
3.00 m
n1
1
6.00 m
n2
n2
1
6.00 m
6.00 m n2
n2
[1]
n1
(a) If n1 1.50 and n2 1.00 , then f
The thin lens equation gives q
6.00 m 1.00
1.00 1.50
pf
p f
10.0 m
12.0 m
12.0 m
10.0 m 12.0 m
5.45 m
A virtual im age is form ed 5.45 m to the left of the lens
(b) If n1 1.50 and n2 1.33 , the focal length is
f
and
6.00 m 1.33
1.33 1.50
q
pf
p f
46.9 m
10.0 m
46.9 m
10.0 m 46.9 m
8.24 m
The im age is located 8.24 m to the left of the lens
(c) When n1 1.50 and n2
and
q
pf
p f
2.00 , f
6.00 m 2.00
2.00 1.50
10.0 m 24.0 m
10.0 m 24.0 m
24.0 m
17.1 m
The im age is 17.1 m to the left of the lens
(d ) Observe from Equation [1] that f 0 if n1 n2 and f 0 w hen n1 n2 . Thus, a
d iverging lens can be changed to converging by surround ing it w ith a m ed ium
w hose ind ex of refraction exceed s that of the lens m aterial.
M irrors and Lenses
23.62
421
The inverted im age is form ed by light that leaves the object and goes d irectly through
the lens , never having reflected from the m irror. For the form ation of this inverted
im age, w e have
q1
p1
M
giving
1.50
q1
1.50 p1
The thin lens equation then gives
1
p1
1
1.50 p1
1
or
10.0 cm
p1
1
1.50
10.0 cm 1
16.7 cm
The upright im age is form ed by light that passes through the lens after reflecting from
the m irror. The object for the lens in this upright im age form ation is the im age form ed
by the m irror. In ord er for the lens to form the upright im age at the sam e location as the
inverted im age, the im age form ed by the m irror m ust be located at the position of the
original object (so the object d istances, and hence im age d istances, are the sam e for both
the inverted and upright im ages form ed by the lens). Therefore, the object d istance and
the im age d istance for the m irror are equal, and their com m on value is
qmirror
pmirror
The m irror equation,
1
40.0 cm p1
1
1
pmirror
qmirror
1
23.3 cm
f mirror
1
23.3 cm
40.0 cm 16.7 cm
2
R
1
f mirror
23.3 cm
, then gives
2
or
23.3 cm
f mirror
23.3 cm
2
11.7 cm
23.63. (a) The lens m aker’ s equation for a lens m ad e of m aterial w ith refractive ind ex
n1 1.55 and im m ersed in a m ed ium having refractive ind ex n2 is
1
f
n1
n2
1
1
R1
1
R2
1.55 n2
n2
Thus, w hen the lens is in air, w e have
and w hen it is im m ersed in w ater,
1
R1
1
R2
1
f air
1
f water
1.55 1.00
1.00
1.55 1.33
1.33
1
R1
1
R1
1
R2
1
[2]
R2
[1]
422
CH APTER 23
Divid ing Equation [1] by Equation [2] gives
f water
f air
If fair
f water
1.33
1.00
1.55 1.00
1.55 1.33
1.33
0.55
0.22
79.0 cm , the focal length w hen im m ersed in w ater is
79.0 cm 1.33
0.55
0.22
263 cm
(b) The focal length for a m irror is d eterm ined by the law of reflection, w hich is
ind epend ent of the m aterial of w hich the m irror is m ad e and of the surround ing
m ed ium . Thus, the focal length d epend s only on the rad ius of curvature and not on
the m aterial m aking up the m irror or the surround ing m ed ium . This m eans that, for
the m irror,
f water
fair
79.0 cm