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(D) The focal length of a lens in air is given by the Lens Maker's Formula:$\frac{1}{f} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad

Vedclass · Accepted Answer

$-45$

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(D) The focal length of a lens in air is given by the Lens Maker's Formula:$\frac{1}{f} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \quad \dots(1)$where $n_g = \frac{3}{2}$ and $f = 15 \,cm$.The focal length of the same lens in a liquid of refractive index $n_l$ is given by:$\frac{1}{f'} = \left( \frac{n_g}{n_l} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \quad \dots(2)$Dividing equation $(1)$ by equation $(2)$, we get:$\frac{f'}{f} = \frac{(n_g - 1)}{\left( \frac{n_g}{n_l} - 1 ight)}$Substituting the given values $n_g = \frac{3}{2}$ and $n_l = \frac{9}{5}$:$\frac{f'}{15} = \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{9/5} - 1)} = \frac{1/2}{(\frac{3}{2} 	imes \frac{5}{9} - 1)} = \frac{1/2}{(\frac{5}{6} - 1)} = \frac{1/2}{-1/6} = -3$Therefore, $f' = -3 	imes 15 = -45 \,cm$.

Inside a vessel filled with liquid, a converging lens is placed as shown in the figure. The lens has a focal length of $15 \,cm$ when in air and has a refractive index of $\frac{3}{2}$. If the liquid has a refractive index of $\frac{9}{5}$, the focal length of the lens in the liquid is (in $\,cm$)

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