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

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