An equiconvex lens made of glass of refractiv | vedclass

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(D) Given,${ }^a \mu_g = \frac{3}{2}$,$f_{	ext{air}} = f$.Refractive index of water,${ }^a \mu_w = \frac{4}{3}$.Using the lens maker's formula,when t

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$300 \%$ increase

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(D) Given,${ }^a \mu_g = \frac{3}{2}$,$f_{	ext{air}} = f$.Refractive index of water,${ }^a \mu_w = \frac{4}{3}$.Using the lens maker's formula,when the lens is in air:$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \left( \frac{3}{2} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \quad \dots(i)$When the lens is immersed in water,the focal length $f_w$ is given by:$\frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \left( \frac{3/2}{4/3} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \left( \frac{9}{8} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \quad \dots(ii)$Dividing $(i)$ by $(ii)$:$\frac{f_w}{f} = \frac{1/2}{1/8} = 4 \implies f_w = 4f$.The percentage change in focal length is:$\frac{f_w - f}{f} 	imes 100 = \frac{4f - f}{f} 	imes 100 = 300 \%$ increase.

An equiconvex lens made of glass of refractive index $\frac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\frac{4}{3}$. The percentage change in the focal length is

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