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(B) Let the refractive index of the lens material be $\mu_l$ and the refractive index of the liquid be $\mu_m$. Given $\mu_m = 0.8 \mu_l = \frac{4}{5}

Vedclass · Accepted Answer

$1.2$

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(B) Let the refractive index of the lens material be $\mu_l$ and the refractive index of the liquid be $\mu_m$. Given $\mu_m = 0.8 \mu_l = \frac{4}{5} \mu_l$,which implies $\mu_l = 1.25 \mu_m = \frac{5}{4} \mu_m$.Using the Lens Maker's Formula for air $(f_a)$:$\frac{1}{f_a} = (\mu_l - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.Using the Lens Maker's Formula for the liquid medium $(f_m)$:$\frac{1}{f_m} = (\frac{\mu_l}{\mu_m} - 1) K$.Given that the focal length increases by $100 \%$,$f_m = f_a + 100 \% f_a = 2 f_a$.Dividing the two equations:$\frac{f_m}{f_a} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_m} - 1} = 2$.Substituting $\mu_l = \frac{5}{4} \mu_m$:$\frac{\frac{5}{4} \mu_m - 1}{\frac{5}{4} - 1} = 2$.$\frac{\frac{5}{4} \mu_m - 1}{0.25} = 2 \Rightarrow \frac{5}{4} \mu_m - 1 = 0.5$.$\frac{5}{4} \mu_m = 1.5 \Rightarrow \mu_m = 1.5 	imes 0.8 = 1.2$.

When a convex lens is immersed in a liquid of refractive index equal to $80 \%$ of the refractive index of the material of the lens,the focal length of the lens increases by $100 \%$. The refractive index of the liquid is

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