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Ch. 34
p. 1
CHAPTER 34 – Lenses and Optical Instruments
1. (
a
) From the ray diagram, the object distance is about 3
&
focal lengths, or
250 mm
.
O
I
F
F
(
b
) We find the object distance from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
(
1
/
d
o
) + (1/88.0 mm) = 1/65.0 mm, which gives
d
o
= 249 mm =
24.9 cm
.
2.
To form a real image from a real object requires a
converging lens
.
We find the focal length of the lens from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
(1/285 cm) + (1/48.3 cm) = 1/
f
, which gives
f
=
+ 41.3 cm
.
Because
d
i
> 0, the image is
real
.
3. (
a
) The power of the lens is
P
= 1/
f
= 1/0.275 m =
3.64 D, converging
.
(
b
) We find the focal length of the lens from
P
= 1/
f
;
– 6.25 D = 1/
f
, which gives
f
= – 0.160 m =
– 16.0 cm, diverging
.
4. (
a
) We locate the image from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
(1/18 cm) + (1/
d
i
) = 1/24 cm, which gives
d
i
= – 72 cm.
The negative sign means the image is
72 cm behind the lens (virtual).
(
b
) We find the magnification from
m
= –
d
i
/
d
o
= – (– 72 cm)/(18 cm) =
+ 4.0
.
5. (
a
) Because the Sun is very far away, the image will be at the focal point.
We find the size of
the image from
m
=
h
i
/
h
o
= –
d
i
/
d
o
;
h
i
/(1.4
×
10
6
km) = – (28 mm)/(1.5
×
10
8
km), which gives
h
i
=
– 0.26 mm
.
(
b
) For a 50 mm lens, we have
h
i
/(1.4
×
10
6
km) = – (50 mm)/(1.5
×
10
8
km), which gives
h
i
=
– 0.47 mm
.
(
c
) For a 200 mm lens, we have
h
i
/(1.4
×
10
6
km) = – (200 mm)/(1.5
×
10
8
km), which gives
h
i
=
– 1.9 mm
.
6.
We find the object distance from the required magnification (which is negative for a real object and a real
image):
m
=
h
i
/
h
o
= –
d
i
/
d
o
;
–
(
2
.
7
0
×
10
3
mm)/(36 mm) = – (8.00 m)/
d
o
, which gives
d
o
= 0.107 m.
We find the focal length of the lens from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
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(1/0.107 m) + (1/8.00 m) = 1/
f
, which gives
f
= 0.105 m =
+ 10.5 cm
.
7. (
a
) We find the image distance from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
(1/10.0
×
10
3
mm) + (1/
d
i
) = 1/80 mm, which gives
d
i
=
81 mm
.
(
b
) For an object distance of 3.0 m, we have
(
1
/
3
.
0
×
10
3
mm) + (1/
d
i
) = 1/80 mm, which gives
d
i
=
82 mm
.
(
c
) For an object distance of 1.0 m, we have
(
1
/
1
.
0
×
10
3
mm) + (1/
d
i
) = 1/80 mm, which gives
d
i
=
87 mm
.
(
d
) We find the smallest object distance from
(
1
/
d
omin
) + (1/120 mm) = 1/80 mm, which gives
d
omin
= 240 mm =
24 cm
.
8.
We find the image distance from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
(1/0.125 m) + (1/
d
i
) = – 6.0 D,
which gives
d
i
= – 0.071 m =
– 7.1 cm (virtual image behind the lens).
We find the height of the image from
m
=
h
i
/
h
o
= –
d
i
/
d
o
;
h
i
/(1.0 mm) = – (– 7.1 cm)/(12.5 cm), which gives
h
i
=
0.57 mm (upright).
9. (
a
) We see that the image is behind the lens,
so it is
virtual
.
(
b
) From the ray diagram we see that we need a
converging
lens
.
(
c
) We find the image distance from the magnification:
m
= –
d
i
/
d
o
;
+ 2.5 = –
d
i
/(8.0 cm), which gives
d
i
= – 20 cm.
We find power of the lens from
(
1
/
d
o
) + (1/
d
i
) = 1/
f
=
P
, when
f
is in meters;
(1/0.080 m) + [1/(– 0.20 m)] =
P
=
7.5 D
.
10. We can relate the image and object distance from the magnification:
m
= –
d
i
/
d
o
,
or
d
o
= –
d
i
/
m
.
We use this in the lens equation:
(
1
/
d
o
) + (1/
d
i
) = 1/
f
;
–
(
m
/
d
i
) + (1/
d
i
) = 1/
f
, which gives
d
i
= (1 –
m
)
f
.
(
a
) If the image is real,
d
i
> 0.
With
f
> 0, we see that
m
< 1; thus
m
= – 2.00.
The image distance is
d
i
= [1 – (– 2.00)](50.0 mm) = 150 mm.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics

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