Page 419 - Schaum's Outline of Theory and Problems of Applied Physics

404 LENSES [CHAP. 32

(a) The object distance of the objective is
q f (160 mm)(6mm)
p = = = 6.23 cm
q − f 160 mm − 6mm
and so its magnification is
q −160 mm
m 1 =− = =−25.7
p 6.23 mm
The minus sign means an inverted image.
(b) From the result of Prob. 32.16,
250 mm 250 mm
m 2 = + 1 = + 1 = 11
f 25 mm
Alternatively the same figure can be obtained by the procedure of part (a).
(c) The total magnification is the product of m 1 and m 2 :

m = m 1 m 2 = (−25.7)(11) =−283

SOLVED PROBLEM 32.18
The angular magnification of a telescope when used to view a distant object is

θ f o focal length of objective
m ang = = =
θ f e focal length of eyepiece
See Fig. 32-12. In the diagram, the rays used to locate the first image are not the same as those used to
locate the final image.

Rays from the same point
on a distant object
Objective
Eyepiece
First image

q q′
Final image

Fig. 32-12

(a) The objective of a telescope has a focal length of 120 cm. What should the focal length of the
eyepiece be to produce an angular magnification of 40? (b) The telescope is used to examine a boat 20 m
long which is 600 m away. If the image distance of the telescope eyepiece is 25 cm, what is the apparent
length of the boat?
f o 120 cm
(a) f e = = = 3cm
m ang 40
(b) The angle subtended by the boat from the location of the telescope is
object length 20 m
θ = = = 0.033 rad
object distance 600 m
If L is the length of the boat’s image as seen through the telescope, the angle that image subtends is
image length L

θ = =
image distance 25 cm