When two thin lenses of focal lengths f_{1} and f_{2} are kept coaxially and in contact, prove that their combined focal length ‘f’ is given by: `1/f = 1/f_1 + 1/f_2`

#### Solution

Let two thin lenses L_{1} and L_{2} of focal lengths f_{1} and f_{2} be put in contact. O is a point object at a distance u from the lens L_{1} Its image is formed at I after refraction through the two lenses at a distance v from the combination. The lens L_{1} forms the image of O at I’. I’, then serves as a virtual object for the lens L_{2} which forms a real image at I.

Now, we know that by lens formula

`1/v - 1/u = 1/f`

Applying this relation for refraction at the lens

L_{1} of focal length f1 can be written as

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

For refraction at the lens L_{2}

u = v', v = v

∴ For this lens `1/v - 1/v' = 1/f_2` ...(ii)

And if F is the focal length of the combination, then

`1/v - 1/u = 1/F`

Adding (i) and (ii), we get

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

or `1/v -1/u = 1/f_1 + 1/f_2`

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