Question

A thin equiconvex lens is made up of glass of refractive index 1.5 and its focal length in air is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the velocity of light in the liquid is

Answer 1

Dear Student,
From the lens maker's formula: $\frac{1}{f_a}=(n_g-n)(1/R1-1/R2)$ where f_a = focal length in air; ng = refractive index of the material of the lens; n = refractive index of the medium; R1 &R2 are the radii of curvatures near and far from the source.
$$\begin{gathered} \frac{1}{0.2}=(1.5-1)(1/R1-1/R2) \\ wheninwater: \\ \frac{-1}{0.5}=(1.5-n_l)(1/R1-1/R2) \\ Dividing,weget: \\ -\frac{0.5}{0.2}=\frac{{\displaystyle (1.5-1)}}{(1.5-n_l)}=\frac{{\displaystyle 0.5}}{(1.5-n_w)} \\ {n}_l=1.7;nl=\frac{{\displaystyle c}}{v},where,c=velocityoflightinvacuum;v=velocityoflightintheliquid. \\ v=1.76\times {10}^{8}m/s \end{gathered}$$

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