Question

The radii of curvature of both the surfaces of a convex lens of focal length ' f' and focal power' P' are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens is -

Answer 1

New focal length: $2f$

New focal power: $\frac{P}{2}$

Answer 2

For a thin double convex lens with equal radii $R$ on both sides, the lens-maker's formula is

$$\frac{1}{f} = (n-1)\left(\frac{1}{R}+\frac{1}{R}\right)= \frac{2(n-1)}{R}.$$

Thus,

$$R = 2(n-1)f.$$

After one surface is made plane (i.e., $R_2=\infty$), the lens becomes plano-convex. Its focal length $f'$ is given by

$$\frac{1}{f'} = (n-1)\left(\frac{1}{R}-0\right) = \frac{n-1}{R}.$$

Substitute $R$ from above:

$$\frac{1}{f'} = \frac{n-1}{2(n-1)f} = \frac{1}{2f} \quad \Rightarrow \quad f' = 2f.$$

The focal power is the reciprocal of the focal length. Thus, the new focal power $P'$ is

$$P' = \frac{1}{f'} = \frac{1}{2f} = \frac{P}{2} \quad (\text{since } P = \frac{1}{f}).$$