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Image Formation by
Mirrors and Lenses
CHAPTER OUTLINE
26.1
Images Formed by Flat
Mirrors
26.2
Images Formed by
Spherical Mirrors
26.3
Images Formed by
Refraction
26.4
Thin Lenses
26.5
Context
Connection
Medical
Fiberscopes
ANSWERS TO QUESTIONS
Q26.1
With a concave spherical mirror, for objects beyond the focal length
the image will be real and inverted. For objects inside the focal
length, the image will be virtual, upright, and magnified. Try a
shaving or makeup mirror as an example.
Q26.2
With a convex spherical mirror, all images of real objects are upright, virtual, and smaller than the
object. As seen in Question 26.1, you only get a change of orientation when you pass the focal
point—but the focal point of a convex mirror is on the nonreflecting side!
Q26.3
The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is
made flat by increasing its radius of curvature without bound, so that its focal length goes to infinity.
From
111
0
pq f
+==
we have
11
pq
=−
; therefore,
. The virtual image is as far behind the mirror
as the object is in front. The magnification is
M
q
p
p
p
=− = =
1. The image is right side up and actual
size.
Q26.4
Stones at the bottom of a clear stream always appear closer to the surface because light is refracted
away from the normal at the surface. Example 26.5 in the textbook shows that its apparent depth is
three quarters of its actual depth.
Q26.5
For definiteness, we consider real objects (
p
>
0).
(a)
For
M
q
p
to be negative,
q
must be positive. This will happen in
qfp
if
p
f
>
, if the
object is farther than the focal point.
(b)
For
M
q
p
to be positive,
q
must be negative.
From
we need
p
f
<
.
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Image Formation by Mirrors and Lenses
(c)
For a real image,
q
must be positive.
As in part (a), it is sufficient for
p
to be larger than
f
.
(d)
For
q
<
0 we need
p
f
<
.
(e)
For
M
>
1, we consider separately
M
<−
1 and
M
>
1.
If
M
q
p
=− <−
1, we need
q
p
>
1
or
qp
>
or
11
<
.
From
111
pq f
+=
,
pp f
+>
or
21
pf
>
or
p
f
2
<
or
p
f
<
2
.
Now if
−>
q
p
or
or
we may require
q
<
0, since then
pfq
=−
with
1
0
f
>
gives
pq
>−
as required
or
.
For
q
<
0 in
qfp
we need
p
f
<
.
Thus the overall condition for an enlarged image is simply
p
f
<
2.
(f)
For
M
<
1, we have the reverse of part (e), requiring
p
f
>
Q26.6
Using the same analysis as in Question 26.5 except
f
<
0.
(a)
Never.
(b)
Always.
(c)
Never, for light rays passing through the lens will always diverge.
(d)
Always.
(e)
Never.
(f)
Always.
Chapter 26
715
Q26.7
We assume the lens has a refractive index higher than its surroundings. For the biconvex lens in
Figure 26.20(a),
R
1
0
>
and
R
2
0
<
. Then all terms in
n
RR
−−
F
H
G
I
K
J
1
11
12
a
f
are positive and
f
>
0. For the
other two lenses in part (a) of the figure,
R
1
and
R
2
are both positive but
R
1
is less than
R
2
. Then
>
and the focal length is again positive.
For the biconcave lens and the planoconcave lens in Figure 26.20(b),
R
1
0
<
and
R
2
0
>
. Then
both terms are negative in
−
and the focal length is negative. For the middle lens in part (b)
of the figure,
R
1
and
R
2
are both positive but
R
1
is greater than
R
2
. Then
<
and the focal
length is again negative.
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This note was uploaded on 10/18/2008 for the course PHYS 3Q2341234 taught by Professor Dafsf during the Spring '08 term at UCLA.
 Spring '08
 DAFSF

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