Question

A)    A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f₁ = +40.3 cm. Precisely 35.4 cm behind this lens is a concave lens with a focal length f₂ = -13.9 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?

B)    What is the lateral magnification of this lens combination?

Answer 1

a) For this two lens system we can treat the efect of each lens individually so let first determine
the image distance produced by the first objective lens

1/do +1/di = 1/f1 = 1/di=1/f1-1/do
putting in the number we,re given

1/di = 1/40.3cm - 1/397cm = 0.02229cm^-1 = 44.86cm

This real image is measured from the first lens that di is actually large then the sepration b/t the lenses
so it is beyond the second concave lens in the word we havea virtual object for the second lens
do2 = (35.4cm)-di1 = -9.46cm
lets apply the thin equation
1/do2 +1/di2 = 1/f2 = 1/di2 = 1/f2 - 1/do2
1/di = 1/-13.9cm - 1/-9.46cm = 0.00.03376cm^-1 = 29.62cm

b) m1 = di1/do1 = - 40.3/397 =-0.1015

m2 = di2/do2 = -(29.62/-9.46) = 3.131

mtot = m1m2 = 0.3177