For a system of lenses the net power of the system will be

P_{eq} = P_{1} + P_{2} + P_{3} + P_{4} + ……..

Provided all the thin lens all in close contact. There focal length of the net system can be written as

$ \displaystyle \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + ……. $

f should be taken with proper sign.

When a convex and a concave lens of equal focal length are placed in contact, the equivalent focal length is equal to infinite. Therefore the power becomes zero.

If the lenses are kept at a separation d, then the effective focal length f is given as

$ \displaystyle \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} – \frac{d}{f_1 f_2} $

If the lens is converging put +ve for its focal length & -ve for the diverging lens.

⇒ $ \displaystyle P = P_1 + P_2 – d (P_1 P_2) $

Note: The overall magnification M of the system is given as the product of individual magnification m; M = m_{1} m_{2} m_{3} …….

### Silvering of the Surface

If a surface of a thin spherical lens is silvered it behaves like a mirror and we can calculate its focal length . Suppose a thin spherical lens is polished at its right face . The radii of curvature of left and right face are R_{1} & R_{2} .

When a ray of light is incident on this silvered lens it will be first refracted by the lens , then reflected from mirror and again refracted by lens .

Hence Power of equivalent Mirror ,

$\large P_{eq} = P_{lens} + P_{mirror} + P_{lens}$

$\large P_{eq} = 2P_{lens} + P_{mirror} $

$\large – \frac{1}{f_{eq}} = 2 \frac{1}{f_{lens}} – \frac{1}{f_{mirror}} $

$\large \frac{1}{f_{eq}} = – \frac{2}{f_{lens}} + \frac{1}{f_{mirror}} $

**Note :** (i) f is +ve for converging Lens

(ii) f is -ve for converging mirror