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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2023
  • Medium
  • Numerical

The radius of curvature of each surface of a convex lens having refractive index \(1.8\) is \(20\,cm\). The lens is now immersed in a liquid of refractive index \(1.5\). The ratio of power of lens in air to its power in the liquid will be \(x : 1\). The value of \(x\) is \(.....\)

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Answer & Solution

Correct Answer

(B) \(4\)

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Detailed explanation

\(P=(1.8-1)\left(\frac{1}{20}+\frac{1}{20}\right)\) by lens maker's formula

\(P^{\prime}=\left(\frac{1.8}{1.5}-1\right)\left(\frac{1}{20}+\frac{1}{20}\right)\)

Dividing \(\frac{ P }{ P ^{\prime}}=\frac{0.8}{1.2-1}=4\)

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