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 respectively

• A. $\frac{f}{2}, 2P$
• B. $2f, \frac{P}{2}$
• C. $-2f, -\frac{P}{2}$
• D. $\frac{2f}{3}, \frac{2}{3}P$

Hint:

Grinding one face of an equi-convex lens flat removes exactly half of its total light-bending capability. Consequently, its optical power drops to half ($\frac{P}{2}$), which mathematically forces its focal length to double ($2f$).

Answer 1

The Correct Option is B

Step 1: Compare two surfaces against one.
A lens's power adds up the bending from each surface. The original equi-convex lens has two equally curved surfaces; grinding one flat removes one bending surface. So we compare a two-surface contribution against a one-surface contribution.
Step 2: Original power per surface.
For the equi-convex lens, $P = (\mu-1)\left(\dfrac{1}{R} + \dfrac{1}{R}\right) = \dfrac{2(\mu-1)}{R}$. The two surfaces contribute equally, so each surface contributes $\dfrac{(\mu-1)}{R}$, exactly half of $P$.
Step 3: After grinding one surface flat.
A flat surface has infinite radius and contributes zero bending. Only one curved surface survives.
Step 4: New power.
The surviving surface gives $P' = \dfrac{(\mu-1)}{R} = \dfrac{1}{2}\,P = \dfrac{P}{2}$.
Step 5: New focal length.
Since focal length is the reciprocal of power, $f' = \dfrac{1}{P'} = \dfrac{1}{P/2} = \dfrac{2}{P} = 2f$.
Step 6: Conclude.
The lens becomes weaker: focal length doubles and power halves. \[ \boxed{f' = 2f,\quad P' = \tfrac{P}{2}} \]