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MHT CET 2026 (19 May, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2026 (19 May, Shift 1)

Q7MHT CET 2026 · 19 MAY · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

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

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Answer: (c)

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

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