Question

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f1 = 38.8 cm . Precisely 36.0 cm behind this lens is a concave lens with a focal length f2 = -10.0 cm . The object to be photographed is 4.02 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

Given is:-

focal length of the objective lens f₁ = 38.8cm

f₂ = -10cm

Distance between both the lenses = 36cm

Distance of the object from objective lens = 4.02m

Now,

part-a

The image from the first (convex) lens is formed at a distance v from the object

which gives  

This image is formed 42.945-36 = 6.945 cm behind the second lens. That becomes the virtual image of the convex lens at  . The object for this image is

by plugging all the values we get

which gives

This u is actually the image of the lens system (interchange object and image in the concave lens formula).

Part-b

First image has magnification of  

and the second image has a magnification of  

Hence overall magnification is 3.0733*0.107 = 0.3502431