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(C) Given: $n_1 = 1.3$,$n_2 = 1.4$,$u = -13 \ 	ext{cm}$,$m = -2$. The meniscus is convex towards $A$,so the center of curvature is in $B$,making $R$

Vedclass · Accepted Answer

$\frac{2}{3} \ 	ext{cm}$

Vedclass · Answer

(C) Given: $n_1 = 1.3$,$n_2 = 1.4$,$u = -13 \ 	ext{cm}$,$m = -2$. The meniscus is convex towards $A$,so the center of curvature is in $B$,making $R$ positive $(R > 0)$.Using the refraction formula at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Substituting the values: $\frac{1.4}{v} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R} \implies \frac{1.4}{v} + 0.1 = \frac{0.1}{R} \implies \frac{1.4}{v} = \frac{0.1}{R} - 0.1 = \frac{0.1(1-R)}{R}$.Thus,$v = \frac{1.4R}{0.1(1-R)} = \frac{14R}{1-R}$.The magnification formula for a spherical refracting surface is $m = \frac{n_1 v}{n_2 u}$.Substituting $m = -2$: $-2 = \frac{1.3 	imes v}{1.4 	imes (-13)} \implies -2 = \frac{1.3 	imes v}{-18.2} \implies v = \frac{-2 	imes -18.2}{1.3} = \frac{36.4}{1.3} = 28 \ 	ext{cm}$.Now,equate the two expressions for $v$: $28 = \frac{14R}{1-R} \implies 28(1-R) = 14R \implies 2 - 2R = R \implies 3R = 2 \implies R = \frac{2}{3} \ 	ext{cm}$.

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