Question

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
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

Answer 1

Let a point object O is placed on the principal axis
By lens formula for image formation by lens $L_1$
$\frac{1}{v^{\prime }}-\frac{1}{u}=\frac{1}{f_1}$ ....... (1)
I' serves as a virtual object for the second lens $L_2$ which forms a final image I at a distance 'v'. Then for $L_2$
$\frac{1}{v}-\frac{1}{v^{\prime }}=\frac{1}{f_2}$ ........ (2)
Adding equation (1) and (2)
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$ .......... (3)
If we replace these two lenses by a single lens such that if an object is kept at u image is formed at v then its focal length f is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ ....... (4)
From equation (3) and (4) we get
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$