A double convex thin lens made of glass of re | vedclass

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a double convex lens in air,$R_1 =

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$-407.5$

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a double convex lens in air,$R_1 = 15 \ cm$ and $R_2 = -15 \ cm$. Given $\mu_g = 1.6$,the focal length in air $(f_a)$ is:$\frac{1}{f_a} = (1.6 - 1) \left( \frac{1}{15} - \frac{1}{-15} ight) = 0.6 	imes \frac{2}{15} = \frac{1.2}{15} = \frac{1}{12.5}$.Thus,$f_a = 12.5 \ cm$.When immersed in a liquid of refractive index $\mu_l = 1.63$,the relative refractive index is $\mu_{gl} = \frac{\mu_g}{\mu_l} = \frac{1.6}{1.63}$.The focal length in liquid $(f_l)$ is given by:$\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Dividing the two equations:$\frac{f_l}{f_a} = \frac{(\mu_g - 1)}{(\frac{\mu_g}{\mu_l} - 1)} = \frac{1.6 - 1}{\frac{1.6}{1.63} - 1} = \frac{0.6}{\frac{1.6 - 1.63}{1.63}} = \frac{0.6 	imes 1.63}{-0.03} = -20 	imes 1.63 = -32.6$.$f_l = -32.6 	imes 12.5 = -407.5 \ cm$.

$A$ double convex thin lens made of glass of refractive index $1.6$ has radii of curvature $15 \ cm$ each. The focal length of this lens when immersed in a liquid of refractive index $1.63$ is.......$cm$.

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