When two thin lenses are kept in contact, prove that their combined or effective focal length F is

given by :

`1/F = 1/f_1 + 1/f_2`

where the terms have their usual meaning.

#### Solution

1) The image is formed in two steps :

In the first step, the lens L_{1} produces the image I' of the object A. It is a real image at a distance v' from L_{1}

Then for the lens L_{1} , we have,

`1/(v') + 1/(-u) = 1/f_1`

`:. 1/(v') - 1/u = 1/f_1` ... (1)

2) In the second step, the image I' acts as an object for L_{2} and the final image I is formed. The image I' is not observed, hence it acts a virtual object for lens L_{2} it is assumed to be kept on left, we can write for L_{2}.

`1/v + 1/(-v') = 1/f_2`

`:. 1/v - 1/(v') = 1/f_2` ...(2)

3) Adding equation (1) and (2), we get

`1/(v') - 1/u + 1/v - 1/(v') = 1/f_1 + 1/f_2`

`:. 1/v - 1/u = 1/f_1 + 1/f_2` ....(3)

4) If a single lens of focal length f of equivalent lens is used for object A and corresponding image I is formed then we have,

`1/v + /(-u) = /1/f` ...(4)

5) From equations (3) and (4),

`1/f = 1/f_1 + 1/f_2` ...(5)