A lens of refractive index ' μ ' has focal length ' f '. When the lens is immersed in a liquid of refractive index ' μ₀ ', its focal length becomes ' f₀ '. Then ' f₀ ' is given by (a) ((μ₀ - μ)f)/(μ(μ₀ - 1)) (b) (μ(μ₀ - 1)f)/((μ₀ - μ)) (c) ((μ - μ₀)f)/(μ₀(μ - 1)) (d) (μ₀(μ - 1)f)/((μ - μ₀))

Answer: (d). In air: 1/f = (μ - 1)K, where K = 1/R₁ - 1/R₂. In liquid: 1/f₀ = ((μ)/(μ₀) - 1)K = (μ - μ₀)/(μ₀)K Divide: f₀ = (μ₀(μ - 1)f)/(μ - μ₀) So the answer is (d).

A lens of refractive index ' μ ' has focal length ' f '. When the… | MHT CET 2026 (17 Apr, Shift 1)

Q13MHT CET 2026 · 17 APR · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

A lens of refractive index ' $\mu$ ' has focal length ' $f$ '. When the lens is immersed in a liquid of refractive index ' $\mu_0$ ', its focal length becomes ' $f_0$ '. Then ' $f_0$ ' is given by

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Answer: (d)

In air: $\frac{1}{f} = (\mu - 1)K$ , where $K = \frac{1}{R_1} - \frac{1}{R_2}$ .
In liquid: $\frac{1}{f_0} = \left(\frac{\mu}{\mu_0} - 1\right)K = \frac{\mu - \mu_0}{\mu_0}K$
Divide: $f_0 = \frac{\mu_0(\mu - 1)f}{\mu - \mu_0}$
So the answer is (d).

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Spherical Surfaces & Thin Lenses question from MHT CET 2026 (17 Apr, Shift 1)

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