A concave lens (refractive index = 1.5) has both surfaces of same radius of curvature R. If it is immersed in a liquid of refractive index 1.75 it will act as a (a) concave lens of focal length (3.5) R (b) concave lens of focal length 2 R. (c) convex lens of focal length (3.5) R (d) convex lens of focal length 2R

Answer: (c). In air: 1/f = (1.5 - 1)(-2/R) = -1/R (diverging). In liquid: 1/fl = ((1.5)/(1.75) - 1)(-2/R) = (-1/7)(-2/R) = 2/7R fl = +3.5R, positive, so it acts as a convex (converging) lens. So the answer is (c) convex lens of focal length 3.5R.

A concave lens (refractive index = 1.5) has both surfaces of same… | MHT CET 2025 (22 Apr, Shift 2)

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Spherical Surfaces & Thin Lenses

A concave lens (refractive index $= 1.5$ ) has both surfaces of same radius of curvature R. If it is immersed in a liquid of refractive index 1.75 it will act as a

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Answer: (c)

In air: $\frac{1}{f} = (1.5 - 1)\left(-\frac{2}{R}\right) = -\frac{1}{R}$ (diverging).
In liquid: $\frac{1}{f_l} = \left(\frac{1.5}{1.75} - 1\right)\left(-\frac{2}{R}\right) = \left(-\frac{1}{7}\right)\left(-\frac{2}{R}\right) = \frac{2}{7R}$
$f_l = +3.5R$ , positive, so it acts as a convex (converging) lens.
So the answer is (c) convex lens of focal length $3.5R$ .

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