Chapter 44: Problem 5

Question: A short linear object of length L lies on the axis of a spherical mirror, a distance O from the mirror. Show that its image will have length $L^\prime$

$$L^\prime~=~L\left({f\over O-f}\right)^2$$

Show that the longitudinal magnification $m^\prime~=~L^\prime/L$ is equal to m² where m is the lateral magnification.

Solution: From previous problem,

$$I~=~{fO\over O-f}$$

and by taking small variations, we have that

$$\delta I~=~-{f^2\over (O-f)^2} \delta O$$

and hence

$$\vert L^\prime\vert~=~{f^2\over (O-f)^2} L$$

Now we know that m = -I/O, which is

$$m~=~{f\over O-f}$$

from which it is clear that $m^\prime~=~m^2$.