CHAPTER 23
MIRRORS AND LENSES
35.
−4.5 cm
R
f= 2 =
= −2.25 cm (convex surface),
2
M = + 12 (virtual image).
do
di = − Mdo = − 2 .
1
1
1
1
1
1
−
−
=
+
=
=
f
do
di
do
do/2
do ,
So
54.
(a) di =
)
do = − f = 2.3 cm .
do f
(15 cm)(22 cm)
=
= −47 cm .
do − f
15 cm − 22 cm
−47.1 cm
di
M = − d = − 15 cm = +3.14.
o
So
hi = Mho = +3.14 (4.0 cm) = 13 cm‚ virtual and upright .
(b) di =
So
55.
(36 cm)(22 cm)
= 57 cm .
36 cm − 22 cm
hi = −1.57 (4.0 cm) = −6.3 cm = 6.3 cm‚ real and inverted .
(a) f = −18.0 cm,
di =
56.6 cm
M = − 36 cm = −1.57.
do = 10 cm.
do f
(10 cm)(−18.0 cm)
=
= −6.4 cm .
10 cm − (−18.0 cm)
do − f
−6.43 cm
di
M = − d = − 10 cm =
o
0.64‚ virtual and upright .
(b) di =
M=−
(25 cm)(−18.0 cm)
= −10.5 cm .
25 cm − (−18.0 cm)
−10.5 cm
25 cm = 0.42‚ virtual and upright .
56.
For a diverging lens, the focal length is negative.
di =
− do |f|
do f
= d + |f| < 0, so the image is always virtual.
o
do − f
− |f|
di
|f|
M = − d = − d + |f| = d + |f| < 1, since do + |f| > |f|.
o
o
o
Therefore 0 < M < 1, i.e., the image is upright and reduced.
57.
M = −2.0 (real).
(a) f = 12 cm,
di = − Mdo = 2.0do.
1
1
1
1
1
3
f = do + di = do + 2.0do = 2do ,
(b) M = +2.0 (virtual).
3
3
do = 2 f = 2 (12 cm) = 18 cm .
di = − Mdo = −2.0do.
1
1
1
1
=
+
=
f
do
2do ,
−2.0do
58.
)
)
1
1
do = 2 f = 2 (12 cm) = 6.0 cm .
1
1
1
1
1
(a) f = d + d = 6.0 cm + 400 cm = 0.169 m-1,
o
i
) f = 5.9 cm .
400 cm
di
(b) M = − d = − 6.0 cm = −66.7.
o
So hi = Mho = −66.7 (1.0 cm) = −66.7 cm = 67 cm‚ inverted .
59.
do = 4.0 m,
−35 mm
hi
M= h =
= −0.0206 (inverted).
1.7 × 103 mm
o
di = − Mdo = 0.0206 (4.0 m) = 0.0824 m.
1
1
1
1
1
−1
=
+
=
+
f
do
di
4.0 m
0.0824 m = 12.4 m ,
64.
do = 3.0 cm,
M = +3.5 (virtual image).
)
di = − Mdo = −3.5 (3.0 cm) = −10.5 cm.
1
1
1
1
1
=
+
=
+
= 0.238 cm−1,
f
do
di
3.0 cm
−10.5 cm
65.
(a) do = 30 cm,
(b) di =
f = −45 cm.
di =
(30 cm)(45 cm)
= −90 cm .
30 cm − 45 cm
CHAPTER 25
f = 0.081 m = 8.1 cm .
)
f = 4.2 cm .
do f
(30 cm)(−45 cm)
=
= −18 cm .
30 cm − (−45 cm)
do − f
OPTICAL INSTRUMENTS
5.
The pre-flash occurs before the aperture is open and the film exposed. The bright light
causes the iris to reduce down (giving a small pupil) so that when the second flash comes
momentarily, you don’t have a wide opening through which you get the red-eye reflection from
the retina.
10.
(a) Nearsighted .
(b) do = ∞,
di = −12.5 m (image on object side).
1
1
1
1
1
P= f = d + d =
+ –12.5 m = −0.080 D‚ diverging .
∞
o
i
11.
do = ∞,
di = −200 cm = −2.00 m (image on object side).
1
1
1
1
1
P= f = d + d =
+
= −0.50 D‚ diverging .
∞
−2.00 m
o
i
13.
(a) do = 25 cm = 0.25 m,
di = −100 cm = −1.0 m (image on object side).
1
1
1
1
1
P = f = d + d = 0.25 cm +
= +3.0 D .
−1.0 m
o
i
1
1
1
1
(b) d = f − d = 3.0 D −
= 3.0 D,
∞
i
o
)
1
di = 3.0 D = 0.33 m = 33 cm.
This image is way behind the retina and so she has to take them out .
15.
do = 25 cm = 0.25 m,
di = −1.5 m (image on object side).
1
1
1
1
1
P = f = d + d = 0.25 m +
= +3.3 D.
−1.5 m
o
i
(a) Converging .
(b) P = +3.3 D .
16.
(a) do = ∞,
di = −150 cm = −1.50 m (image on object side).
1
1
1
1
1
P= f = d + d =
+
= −0.67 D .
∞
−1.50 m
o
i
1
1
1
1
(b) d = f − d = −0.667 D − 0.25 m = −4.67 D,
i
o
)
di = −0.21 m < − 0.25 m.
Therefore the answer is yes and the near point is 21 cm .
(c) 30–40 years old from Table 25.1.
17.
First find his new near point.
do = 33 cm = 0.33 m,
1
1
1
1
−
−
−
=
=
+2.0
D
f
do
0.33 m = 1.0 D,
di
)
1
P = f = +2.0 D.
di = −1.0 m.
So the near point is
1.0 m.
To bring this near point to 25 cm, the power of the new lenses must be
1
1
1
P' = f' = 0.25 m +
= +3.0 D .
−1.0 m
18.
Top:
do = ∞,
di = −500 cm = −5.0 m (image on object side).
1
1
1
1
1
P= f = d + d =
+
= −0.20 D .
∞
−5.0 m
o
i
Bottom:
do = 25 cm = 0.25 m,
1
1
P = 0.25 m +
= +2.6 D .
−0.70 m
di = −70 cm = −0.70 m (image on object side).