Physics - Geometrical Optics Question with Solution | TestHub

Given, the refractive index of the material of the lens is ${\mu }_1=1.5$, the refractive index of the medium is ${\mu }_m=1.6$ and the focal length of the lens in air is $f_a=20 \mathrm{cm}$.

In air:

$\frac{1}{f_a}=\left({\mu }_1-1\right)\left[\frac{2}{R}\right]$

In liquid:

$\frac{1}{f_m}=\left(\frac{{\mu }_1}{{\mu }_m}-1\right)\left[\frac{2}{R}\right]$

The focal length as measured in air and that as measured inside the liquid medium are related by, 

$\frac{f_m}{f_a}=\frac{\left({\mu }_1-1\right){\mu }_m}{\left({\mu }_1-{\mu }_m\right)} ...\left(1\right)$

Equation (1) implies that

$$\begin{gathered} \frac{f_m}{20}=\frac{(1.5-1)1.6}{(1.5-1.6)} \\ \Rightarrow f_m=-160 \mathrm{cm} \end{gathered}$$