Refractive index of a glass convex lens is 1. | vedclass

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(B) The power of a lens is given by $P = \frac{1}{f} = (\mu_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, where $\mu_l$ is the relative refrac

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$2 : 5$

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(B) The power of a lens is given by $P = \frac{1}{f} = (\mu_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$, where $\mu_l$ is the relative refractive index of the lens material with respect to the surrounding medium.For a convex lens in air, $\mu_l = \frac{\mu_g}{\mu_a} = \frac{1.5}{1} = 1.5$. The radii are $R_1 = 40 	ext{ cm}$ and $R_2 = -40 	ext{ cm}$.$P_{air} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight) = 0.5 	imes \left( \frac{2}{40} ight) = 0.5 	imes 0.05 = 0.025 	ext{ cm}^{-1}$.When immersed in a liquid of refractive index $\mu_m = 1.25$, the relative refractive index is $\mu_l' = \frac{\mu_g}{\mu_m} = \frac{1.5}{1.25} = 1.2$.$P_{liquid} = (1.2 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight) = 0.2 	imes 0.05 = 0.01 	ext{ cm}^{-1}$.The ratio of power is $\frac{P_{liquid}}{P_{air}} = \frac{0.01}{0.025} = \frac{10}{25} = \frac{2}{5}$.

Refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \text{ cm}$. 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

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