The focal length of a thin converging lens in air is 20 cm . When the lens is immersed in a liquid, it…
The focal length of a thin converging lens in air is 20 cm . When the lens is immersed in a liquid, it behaves like a concave lens of power 1 D . If the refractive index of the material of the lens is 1.5 , the refractive index of the liquid is
$\frac{5}{3}$
$\frac{4}{3}$
$\frac{5}{4}$
$\frac{7}{4}$
Solution
In air, $\frac{1}{\mathrm{f}_1}=\left(\mu_2-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\ldots$ (i)
In medium, $P_2=-1 D \Rightarrow f_2=\frac{100}{P_2} \mathrm{~cm}=-100 \mathrm{~cm}$
$\frac{1}{\mathrm{f}_2}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\ldots$ (ii)
From eqns (i) and (ii), we get
$\frac{f_2}{f_1}=\frac{\left(\mu_2-1\right)}{\left(\frac{\mu_2}{\mu_1}-1\right)}$
$\begin{aligned} & \Rightarrow \frac{-100}{20}=\frac{(1.5-1)}{\left(\frac{1.5}{\mu_1}-1\right)}=\frac{0.5}{\left(\frac{1.5}{\mu_1}-1\right)} \\ & \therefore \mu_1=\frac{5}{3}\end{aligned}$