Refractive index of a glass convex lens is 1.5 . The radius of curvature of each of the two surfaces of the…

Refractive index of a glass convex lens is 1.5 . The radius of curvature of each of the two surfaces of the lens is $20 \mathrm{~cm}$. The ratio of the power of the lens when immersed in a liquid of refractive index 1.25 to that when placed in air is
  1. $2:3$
  2. $2:5$
  3. $3:5$
  4. $5:2$

Solution

Given data: $\mu_{\mathrm{g}}=1.5, \mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}, \mu_l=1.25$ $\begin{aligned} \mathrm{P}_1=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) & =(1.5-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ \therefore \quad \mathrm{P}_2=\left(\frac{\mu_{\mathrm{g}}}{\mu_l}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) & =\left(\frac{1.5}{1.25}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & =(1.2-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \end{aligned}$ Taking ratio, $\frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{1.2-1}{1.5-1}=\frac{2}{5}$ ~

Asked in: MHT CET 2023 (14 May Shift 2)

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