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A convex lens of focal length $0.15 \mathrm{~m}$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \mathrm{~m}$. The refractive index of the liquid is
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Verified Answer
The correct answer is:
$\frac{5}{4}$
$$
f_a=0.15 \mathrm{~m}, \mu_g=3 / 2, f_e=0.225+0.15
$$
From lens formula,
For air, $\frac{1}{f_a}=\left(\frac{\mu_g}{\mu_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For liquid, $\frac{1}{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \begin{aligned} \frac{1 / f_a}{1 / f_l} & =\frac{\left(\mu_g / \mu_a-1\right)}{\left(\frac{\mu_g}{\mu_l}-1\right)} \\ \frac{f_l}{f_a} & =\frac{\left(\frac{3 / 2}{1}-1\right)}{\left(\frac{3 / 2}{\mu_l}-1\right)} \\ \frac{0.375}{0.15} & =\frac{1 / 2}{\left(\frac{3}{2 \mu_l}-1\right)} \\ \left(\frac{3}{2 \mu_l}-1\right) & =\frac{1 / 2}{5 / 2} \\ \frac{3}{2 \mu_l}-1 & =\frac{1}{5} \\ \frac{3}{2 \mu_l} & =\frac{1}{5}+1 \Rightarrow \frac{3}{2 \mu_l}=\frac{6}{5} \\ \mu_l & =\frac{5}{4}\end{aligned}$
f_a=0.15 \mathrm{~m}, \mu_g=3 / 2, f_e=0.225+0.15
$$
From lens formula,
For air, $\frac{1}{f_a}=\left(\frac{\mu_g}{\mu_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For liquid, $\frac{1}{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \begin{aligned} \frac{1 / f_a}{1 / f_l} & =\frac{\left(\mu_g / \mu_a-1\right)}{\left(\frac{\mu_g}{\mu_l}-1\right)} \\ \frac{f_l}{f_a} & =\frac{\left(\frac{3 / 2}{1}-1\right)}{\left(\frac{3 / 2}{\mu_l}-1\right)} \\ \frac{0.375}{0.15} & =\frac{1 / 2}{\left(\frac{3}{2 \mu_l}-1\right)} \\ \left(\frac{3}{2 \mu_l}-1\right) & =\frac{1 / 2}{5 / 2} \\ \frac{3}{2 \mu_l}-1 & =\frac{1}{5} \\ \frac{3}{2 \mu_l} & =\frac{1}{5}+1 \Rightarrow \frac{3}{2 \mu_l}=\frac{6}{5} \\ \mu_l & =\frac{5}{4}\end{aligned}$
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