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(D) For a biconvex lens with radii of curvature $R$ and refractive index $\mu_g = 1.5$, the lens maker's formula in air is:$\frac{1}{f} = (\mu_g - 1)

Vedclass · Accepted Answer

$40$

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(D) For a biconvex lens with radii of curvature $R$ and refractive index $\mu_g = 1.5$, the lens maker's formula in air is:$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R} - \frac{1}{-R} ight) = (1.5 - 1) \left( \frac{2}{R} ight) = \frac{1}{R}$Given $f = 10 \ cm$, we have $R = 10 \ cm$.When the lens is cut perpendicular to the principal axis, we get two plano-convex lenses, each with one flat surface $(R_1 = \infty)$ and one curved surface $(R_2 = -10 \ cm)$.The focal length $f'$ of each plano-convex lens is given by:$\frac{1}{f'} = (\mu_g - 1) \left( \frac{1}{\infty} - \frac{1}{-10} ight) = 0.5 	imes \frac{1}{10} = \frac{1}{20} \implies f' = 20 \ cm$.When these two pieces are glued such that the convex surfaces touch each other, the combination acts as a single lens with two curved surfaces of radii $R_1 = 10 \ cm$ and $R_2 = -10 \ cm$ (biconvex again).When immersed in water $(\mu_w = 4/3)$, the new focal length $F'$ is given by:$\frac{1}{F'} = \left( \frac{\mu_g}{\mu_w} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$$\frac{1}{F'} = \left( \frac{1.5}{4/3} - 1 ight) \left( \frac{1}{10} - \frac{1}{-10} ight) = \left( \frac{4.5}{4} - 1 ight) \left( \frac{2}{10} ight) = \left( \frac{0.5}{4} ight) \left( \frac{1}{5} ight) = \frac{1}{8} 	imes \frac{1}{5} = \frac{1}{40}$Therefore, $F' = 40 \ cm$.

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