Question

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f1 = 38.7 cm . Precisely 36.3 cm behind this lens is a concave lens with a focal length f2 = -10.0 cm . The object to be photographed is 3.97 m in front of the objective lens How far behind the concave lens should the film be placed? What is the linear magnification of this lens combination?

Answer 1

(a) For this two-lens system, we can treat the effects of each lens individually.
So, let’sfirst determine the image distance produced by the first, objective lens

1//do1 + 1/di1 = 1/f1 = 1/di1 = 1/f1 - 1/do1
Putting in the numbers we’re given,

1/di1 = 1/38.7cm - 1/397cm = 0.02332cm^-1 = di1 = 42.881cm

Thisrealimage distance is measured from the first lens. Notice thatdi1is actuallylargerthan
the separation between the lenses, so it is beyond the second, concave lens.In other words,
we have a virtual object for the second lens

do2 = (36.3cm)-di1 = -6.581cm
Next, let’s apply the thin-lens equation for the concave lens

1/do2 + 1/di2 = 1/f2 = 1/di2 = 1/f2 - 1/do2
which with our numbers works out to

1/di2 = 1/-10.0cm - 1/-6.581cm = = 0.05195cm^-1 = di2 = 19.249cm

(b) The magnification of multiple lenses is just the product of the individual magnifications

m1 = - di1/do1 = - 42.881cm/397cm = -0.1080

m2 = - di2/do2 = - 19.249cm/-6.581 = 2.924

mtotal = m1*m2 = -0.1080*2.924 = -0.3157