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(B) The focal length of the biconvex lens is $f$. Since the radii are equal, let $R_1 = R$ and $R_2 = -R$.The refractive index of glass with respect t

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$4f$

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(B) The focal length of the biconvex lens is $f$. Since the radii are equal, let $R_1 = R$ and $R_2 = -R$.The refractive index of glass with respect to air is ${}_a\mu_g = 3/2$.Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Substituting the values: $\frac{1}{f} = (\frac{3}{2} - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\frac{1}{2}) \left( \frac{2}{R} \right) = \frac{1}{R}$.Thus, $f = R$.When the lens is dipped in water, the refractive index of glass with respect to water is ${}_w\mu_g = \frac{{}_a\mu_g}{{}_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.Using the Lens Maker's formula for the lens in water:$\frac{1}{f_w} = ({}_w\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (\frac{9}{8} - 1) \left( \frac{1}{R} + \frac{1}{R} \right)$.$\frac{1}{f_w} = (\frac{1}{8}) \left( \frac{2}{R} \right) = \frac{1}{4R}$.Since $R = f$, we get $\frac{1}{f_w} = \frac{1}{4f}$, which implies $f_w = 4f$.

The focal length of a biconvex lens made of glass with equal radii of curvature is $f$. If the lens is dipped in water, what will be its new focal length? (Take the refractive index of glass as $3/2$ and water as $4/3$)

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