Embed Size (px)
Citation preview
Physics 112 Homework 9 (solutions) (2004 Fall)
1
Solutions to Homework Questions 9
Chapt23, Problem-1: Does your bathroom mirror show you older or younger than your actual age? Compute
an order-of-magnitude estimate for the age difference, based on data that you specify.
Solut ion:
If you stand 40 cm in front of the mirror, the time required for light scattered from your face to travel
to the mirror and back to your eye is
!t =2 d
c=
2 0.40 m( )3.0"10
8 m s
= 2.7 "10#9
s .
Thus, the image you observe shows you ~10 !9 s younger than your current age.
Chapt23, Problem-8: A dentist uses a mirror to examine a tooth. The tooth is 1.00 cm in front of the
mirror, and the image is formed 10.0 cm behind the mirror. Determine (a) the mirror’s radius of curvature and (b) the
magnification of the image.
Solut ion:
(a) Since the object is in front of the mirror, p > 0 . With the image behind the mirror,
q < 0 . The
mirror equation gives the radius of curvature as
2
R=
1
p+
1
q=
1
1.00 cm!
1
10.0 cm=
10 -1
10.0 cm,
or
R = 210.0 cm
9
! " #
$ % & =
+ 2.22 cm
(b) The magnification is M = !
q
p= !
!10.0 cm( )1.00 cm
= +10.0
Chapt23, Problem-13: A concave makeup mirror is designed so that a person 25 cm in front of it sees an
upright image magnified by a factor of 2. What is the radius of curvature of the mirror?
Solut ion:
The image is upright, so M > 0 and we have
M = !
q
p= +2 .0 , or
q =! 2.0 p =! 2.0 25 cm( )= !50 cm
The radius of curvature is then found to be
2
R=
1
p+
1
q=
1
25 cm!
1
50 cm=
2!1
50 cm, or
R = 20.50 m
+1
!
" #
$
% & =
1.0 m
Physics 112 Homework 9 (solutions) (2004 Fall)
2
Chapt23, Problem-23: A paperweight is made of a solid glass hemisphere of index of refraction 1.50. The
radius of the circular cross section is 4.0 cm. The hemisphere is placed on its flat surface with the center directly over a 2.5-
mm-long line drawn on a sheet of paper. What length of line is seen by someone looking vertically down on the
hemisphere?
Solut ion:
Since the center of curvature of the surface is on the side the light comes from, R< 0 giving
R= ! 4.0 cm . Then,
n1
p+
n2
q=
n2! n
1
R becomes
1.00
q=
1.00 !1.50
! 4.0 cm!
1.50
4.0 cm, or
q = ! 4.0 cm
Thus, the magnification M =
! h
h= "
n1
n2
#
$ %
&
' (
q
p, gives
! h = "n
1q
n2p
#
$ %
&
' ( h ="
1.50 "4.0 cm( )1.00 4.0 cm( )
2.5 mm( ) = 3.8 mm
Chapt23, Problem-27: A contact lens is made of plastic with an index of refraction of 1.50. The lens has an
outer radius of curvature of +2.00 cm and an inner radius of curvature of +2.50 cm. What is the focal length of the lens?
Solut ion:
With R1 =+ 2.00 cm and R2 =+ 2.50 cm , the lens maker’s equation gives the focal length as
1
f= n !1( )
1
R1
!1
R2
"
# $
%
& ' = 1.50 !1( )
1
2 .00 cm!
1
2 .50 cm
"
# $
%
& ' = 0.0500 cm
-1
or f =
1
0.0500 cm-1=
20.0 cm
Chapt23, Problem-34: The nickel’s image in Figure P23.34
has twice the diameter of the nickel when the lens is 2.84 cm from the nickel.
Determine the focal length of the lens.
Solut ion:
We must first realize that we are looking at an upright, magnified,
virtual image. Thus, we have a real object located between a
converging lens and its front-side focal point, so q < 0, p > 0, and f > 0 .
The magnification is M =!
q
p= +2 , giving
q = ! 2 p . Then, from the thin lens equation,
1
p!
1
2 p= +
1
2 p=
1
f or
f = 2 p = 2 2.84 cm( ) =
5.68 cm
Physics 112 Homework 9 (solutions) (2004 Fall)
3
Chapt23, Problem37: A diverging lens is to be used to produce a virtual image one-third as tall as the object.
Where should the object be placed?
Solut ion:
All virtual images formed by diverging lenses are upright images. Thus, M > 0 and the magnification gives
M = !
q
p= +
1
3, or
q =!
p
3
Then, from the thin lens equation,
1
p!
3
p=!
2
p=
1
f or
p =! 2 f = 2 f
The object should be placed at distance 2 f in front of the lens
Chapt23, Problem-40: An object is placed 20.0 cm to the left of a converging lens of focal length 25.0 cm.
A diverging lens of focal length 10.0 cm is 25.0 cm to the right of the converging lens. Find the position and magnification
of the final image.
Solut ion:
With p1 = 20.0 cm and
f1 = 25.0 cm , the thin lens equation gives the position of the image formed by the
first lens as
q1 =
p1f1
p1! f
1
=20.0 cm( ) 25.0 cm( )20.0 cm ! 25.0 cm
= !100 cm
and the magnification by this lens is M1 = !
q1
p1
= !!100 cm( )20.0 cm
=+ 5.00
This virtual image serves as the object for the second lens, so the object distance is
p2 = 25.0 cm + q1 = 125 cm . Then, the thin lens equation gives the final image position as
q2 =
p2f2
p2! f
2
=125 cm( ) !10.0 cm( )
125 cm ! !10.0 cm( )= ! 9.26 cm
with a magnification by the second lens of M2 =!
q2
p2
= !! 9.26 cm( )125 cm
=+ 0.0741
Thus, the final image is located 9.26 cm in front of the second lens and
the overall magnification is M = M1M2 = +5.00( ) + 0.0741( )=
+ 0.370
Physics 112 Homework 9 (solutions) (2004 Fall)
4
Chapt23, Problem-45: Lens L1 in Figure P23.45 has a focal length
of 15.0 cm and is located a fixed distance in front of the film plane of a camera.
Lens L2 has a focal length of 13.0 cm, and its distance d from the film plane can be
varied from 5.00 cm to 10.0 cm. Determine the range of distances for which objects
can be focused on the film.
Solut ion:
Since the final image is to be real and in the film plane, q2 =+ d .
Then, the thin lens equation gives p2 =
q2f2
q2! f
2
=d 13.0 cm( )d !13.0 cm
.
Note from Figure P23.45 that d < 12.0 cm . The above result then
shows that p2 < 0 , so the object for the second lens will be a virtual object.
The object of the second lens L2( ) is the image formed by the first lens
L1( ) , so
q1 = 12.0 cm ! d( )! p2 = 12.0 cm ! d 1 +
13.0 cm
d -13.0 cm
" # $
% & ' = 12.0 cm !
d2
d -13.0 cm
If d = 5.00 cm , then q1 = +15.1 cm ; and when d = 10.0 cm ,
q1 = + 45.3 cm
From the thin lens equation, p1 =
q1f1
q1! f
1
=q1 15.0 cm( )q
1! 15.0 cm
When q1 = +15.1 cm
d = 5.00 cm( ) , then p1 = 1.82 !10
3 cm = 18.2 m
When q1 = + 45.3 cm
d = 10.0 cm( ), then p1 = 22.4 cm = 0.224 m
Thus, the range of focal distances for this camera is 0.224 m to 18.2 m
Chapt23, Conceptual-1: Tape a picture of yourself on a bathroom mirror. Stand several centimetres away
from the mirror. Can you focus your eyes on both the picture taped to the mirror and your image in the mirror at the same
time? So, where is the image of yourself?
Solut ion:
You will not be able to focus your eyes on both the picture and your oown image at the same time. To focus
on the picture, you must adjust your eyes so that an object several centimeters away (the picture) is in
focus. Thus, you are focusing on the mirror surface. But, your image in the mirror is as far behind the
mirror as you are in front of it. Thus, you must focus your eyes beyond the surface of the mirror (twice as
far away as the picture) to bring the image of yourself into focus.
Chapt23, Conceptual-7: A virtual image is often described as one through which the light rays do not
actually travel, as they do for a real image. Can a virtual image be photographed?
Solut ion:
Light rays diverge from the position of a virtual image just as they do from an actual object. Thus, a
virtual image can be photographed as easily as any real object. Of course, the camera would have to be
placed near the axis of the lens or mirror in order to intercept the light rays.
Physics 112 Homework 9 (solutions) (2004 Fall)
5
Chapt23, Conceptual-9: Suppose you want to use
a converging lens to project the image of two trees onto a screen.
One tree is a distance x from the lens; the other is at 2x, as in
Figure Q23.9. You adjust the screen so that the near tree is in focus.
If you now want the far tree to be in focus, do you move the screen
toward or away from the lens?
Solut ion:
We consider the two trees to be separate objects. The
far tree is an object that is further from the lens than the near tree. Thus, the image of the far tree will
be closer to the lens than the image of the near tree. The screen must be moved closer to the lens to put
the far tree in focus.
Chapt23, Conceptual-10: Why does a clear stream always appear to be shallower than it actually is?
Solut ion:
All objects beneath the stream appear to be closer to the surface than they really are because of
refraction. Thus, the pebbles on the bottom of the stream appear to be close to the surface of a shallow
stream.
Chapt23, Conceptual-16: If a cylinder of solid glass or clear plastic is place above the words LEAD
OXIDE and viewed from the side, as shown in Figure Q23.16, the word LEAD appears inverted, but the word OXIDE does
not. Explain.
Solut ion:
Both words are inverted. However OXIDE looks the same right side up and upside down. LEAD does not.
