A thin equiconvex lens is made of glass of re | vedclass

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(B) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For an equiconvex lens in air,$R_1 = R

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$\frac{15}{8}$

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(B) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For an equiconvex lens in air,$R_1 = R$ and $R_2 = -R$,so $\frac{1}{f_a} = (_a\mu_g - 1) \frac{2}{R}$.Given $f_a = 0.2 \ m$ and $_a\mu_g = 1.5$,we have $\frac{1}{0.2} = (1.5 - 1) \frac{2}{R} \implies 5 = 0.5 	imes \frac{2}{R} \implies R = 0.2 \ m$.When dipped in a liquid of refractive index $_a\mu_l$,the focal length becomes $f_l = -0.5 \ m$ (concave behavior).The new focal length is $\frac{1}{f_l} = (_l\mu_g - 1) \frac{2}{R}$,where $_l\mu_g = \frac{_a\mu_g}{_a\mu_l}$.Substituting the values: $\frac{1}{-0.5} = \left( \frac{1.5}{_a\mu_l} - 1 ight) \frac{2}{0.2}$.$-2 = \left( \frac{1.5}{_a\mu_l} - 1 ight) 	imes 10$.$-0.2 = \frac{1.5}{_a\mu_l} - 1$.$0.8 = \frac{1.5}{_a\mu_l} \implies _a\mu_l = \frac{1.5}{0.8} = \frac{15}{8}$.

$A$ thin equiconvex lens is made of glass of refractive index $1.5$ and its focal length is $0.2 \ m$. If it acts as a concave lens of $0.5 \ m$ focal length when dipped in a liquid,the refractive index of the liquid is:

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