Question

A thin liquid convex lens is formed in glass. Refractive index of liquid is $\dfrac{4}{3}$ and that of glass is $\dfrac{3}{2}$. If f is the focal length of the liquid lens in air its focal length and nature in the glass is
A. f, convex
B. f, concave
C. 2f, concave
D. 3f, concave

Answer 1

First calculate the refractive index of the liquid in with respect to air and write the expression for the focal length of the lens in air. Then calculate the refractive of liquid in glass and then calculate the focal length of the lens in glass.

Formula used:
$\dfrac{1}{f}=\left( {{\mu }_{1,2}}-1 \right)\dfrac{2}{R}$,
where $f$ is the focal length of the lens made of medium 1 and placed in medium 2. $R$ is the radius of curvature of both the surface of the lens. ${{\mu }_{1,2}}$ is a refractive index of medium 1 with respect to medium 2.
${{\mu }_{1,2}}=\dfrac{{{\mu }_{1}}}{{{\mu }_{2}}}$,
where ${{\mu }_{1}}$ is refractive index of medium 1 and ${{\mu }_{2}}$ is that of medium 2.

Complete step by step answer:
It is given that the refractive index of the liquid is ${{\mu }_{l}}=\dfrac{4}{3}$. The refractive index of air is ${{\mu }_{a}}=1$. Therefore, the refractive index of the liquid with respect to air is equal to,
${{\mu }_{l,a}}=\dfrac{\dfrac{4}{3}}{1}=\dfrac{4}{3}$.
With this the focal length of the liquid convex lens, when it is placed in the medium of air is equal to,
$\dfrac{1}{f}=\left( {{\mu }_{l,a}}-1 \right)\dfrac{2}{R}$
$\Rightarrow \dfrac{1}{f}=\left( \dfrac{4}{3}-1 \right)\dfrac{2}{R}$
$\Rightarrow \dfrac{1}{f}=\dfrac{2}{3R}$
$\Rightarrow R=\dfrac{2f}{3}$
It is given that the refractive index of the glass is ${{\mu }_{g}}=\dfrac{3}{2}$.
Therefore, the refractive index of the liquid with respect to the glass is equal to,
${{\mu }_{l,g}}=\dfrac{\dfrac{4}{3}}{\dfrac{3}{2}}=\dfrac{8}{9}$
The focal length of the convex lens of liquid, when it is placed in the medium of the glass is equal to,
$\dfrac{1}{f'}=\left( {{\mu }_{l,g}}-1 \right)\dfrac{2}{R}$
$\Rightarrow \dfrac{1}{f'}=-\dfrac{2}{9R}$
$\Rightarrow f'=-\dfrac{9R}{2}$
Substitute the value of R in the above equation.
$\therefore f'=-\dfrac{9}{2}\left( \dfrac{2f}{3} \right)=-3f$.
Since the value of f’ is negative, the nature of the lens in glass is that of a concave lens and its focal length is equal to 3f.

Hence, the correct option is D.

Note: The actual formula of the focal length of a lens is given by $\dfrac{1}{f}=\left( {{\mu }_{1,2}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$, where ${{R}_{1}}$ and ${{R}_{2}}$ are the radii of curvature of the two spherical surfaces. In this case, we considered that both radii are equal as it does not matter on the values of the radii.