The refractive index of a glass convex lens i | vedclass

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(C) The power of a lens is given by $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens,$R_1 = +40 \ cm$ and

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$2$:$5$

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(C) The power of a lens is given by $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a convex lens,$R_1 = +40 \ cm$ and $R_2 = -40 \ cm$.In air $(n_a = 1)$: $P_{air} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight) = 0.5 	imes \frac{2}{40} = \frac{0.5}{20} = \frac{1}{40} \ cm^{-1}$.In liquid $(n_l = 1.25)$: The effective refractive index is $n_{rel} = \frac{n_g}{n_l} = \frac{1.5}{1.25} = 1.2$.$P_{liquid} = (1.2 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight) = 0.2 	imes \frac{2}{40} = \frac{0.4}{40} = \frac{1}{100} \ cm^{-1}$.The ratio of power in liquid to air is $\frac{P_{liquid}}{P_{air}} = \frac{1/100}{1/40} = \frac{40}{100} = \frac{2}{5}$.

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 $40 \ cm$. 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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