Page 417 - Schaum's Outline of Theory and Problems of Applied Physics

402 LENSES [CHAP. 32

O F′
F

Fig. 32-9

SOLVED PROBLEM 32.13
Find the focal length of a magnifying glass that produces an erect image magnified three times of an
object 1.5 in. away.
An erect image magnified three times means a magnification of +3. Since the object distance is p = 1.5 in.
and m =−q/p,

q =−mp = (−3)(1.5 in.) =−4.5in,
The negative image distance signifies a virtual image. The required focal length is
pq (1.5 in.)(−4.5 in.)
f = = =+2.25 in.
p + q 1.5 in. − 4.5 in.

SOLVED PROBLEM 32.14
A 35-mm camera has a telephoto lens whose focal length is 150 mm. What range of adjustments should
the lens have in order to be able to bring to a sharp focus objects as close as 1.5 m from the camera?
The image distance that corresponds to f = 0.15 m and p = 1.5mis
pf (1.5m)(0.15 m)
q = = = 0.167 m = 167 mm
p − f 1.5m − 0.15 m
An object at p =∞ is brought to a focus at q = f = 150 mm (see Fig. 32-10). Hence a range of adjustment of
167 mm − 150 mm = 17 mm will permit objects to be photographed at distances from 1.5 m to infinity.

Rays from same point
on distant object
F′
F

Fig. 32-10

SOLVED PROBLEM 32.15
A slide projector uses a converging lens to produce a real, inverted image of a transparent slide on a
screen. What should be the focal length of the lens of a slide projector if a slide 5 cm by 5 cm is to appear
90 cm by 90 cm on a screen5maway?
Here h = 5cmand h =−90 cm (h is negative because the image is inverted), so the magnification is

h −90 cm
m = = =−18
h 5cm
Since q = 5m = 500 cm and m =−q/p, the object distance is
q 500 cm
p =− =− = 27.8cm
m −18