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(B) Using the lens maker's formula,the power $P$ of a lens in air is given by $P = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a conve

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$4$

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(B) Using the lens maker's formula,the power $P$ of a lens in air is given by $P = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a convex lens,$R_1 = 20 \ cm$ and $R_2 = -20 \ cm$. Thus,$P = (1.8 - 1) \left( \frac{1}{20} - \frac{1}{-20} ight) = 0.8 	imes \frac{2}{20} = 0.8 	imes 0.1 = 0.08 \ cm^{-1}$.When immersed in a liquid of refractive index $n_l = 1.5$,the power $P'$ is given by $P' = \left( \frac{n_g}{n_l} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.$P' = \left( \frac{1.8}{1.5} - 1 ight) \left( \frac{1}{20} + \frac{1}{20} ight) = (1.2 - 1) 	imes 0.1 = 0.2 	imes 0.1 = 0.02 \ cm^{-1}$.The ratio of power in air to power in liquid is $\frac{P}{P'} = \frac{0.08}{0.02} = 4$.Therefore,$x = 4$.

The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 \ cm$. The lens is now immersed in a liquid of refractive index $1.5$. The ratio of power of the lens in air to its power in the liquid will be $x : 1$. The value of $x$ is $.....$

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