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MHT CET 2021 (24 Sep, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2021 (24 Sep, Shift 1)

Q76MHT CET 2021 · 24 SEP · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

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

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Answer: (b)

Equiconvex: $\frac{1}{f} = (\mu - 1)\frac{2}{R}$ .
One face plane: $\frac{1}{f'} = (\mu - 1)\frac{1}{R} = \frac{1}{2f}$ , so power halves.
So the answer is (b) $2f,\ \frac{P}{2}$ .

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