Question

An object is lying in the $\text{medium}1$ of referactive index ${\mu }_1=1.2$, is at a distance of $10\text{cm}$ from a spherical interface as shown in the figure. The radius of the interface is $20\text{cm}$ and refractive index of the $\text{medium}2$ is ${\mu }_2=1.8$. Calculate the magnification of the image formed by the interface.

• A. $\frac{4}{3}$
• B. $\frac{3}{2}$
• C. $\frac{4}{5}$
• D. $-\frac{4}{3}$

Answer 1

The correct option is A $\frac{4}{3}$

Given :
Distance of the object from the pole of the spherical interface $\left(u\right)=-10\text{cm}$
Radius of the interface $\left(R\right)=20\text{cm}$
Refractive index of the medium 1 $\left({\mu }_1\right)=1.2$
Refractive index of the medium 2 $\left({\mu }_1\right)=1.8$

Position of the image is given by :

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{1.8}{v}-\frac{1.2}{-10}=\frac{1.8-1.2}{20}$

$\Rightarrow \frac{1.8}{v}=\frac{0.6}{20}-\frac{1.2}{10}$

$\frac{1.8}{v}=\frac{-1.8}{20}$

$\Rightarrow v=-20\text{cm}$

Therefore, magnification :

$m=\frac{{\mu }_1}{{\mu }_2}\times \frac{v}{u}$

$m=\frac{1.2}{1.8}\times \frac{-20}{-10}$

$\Rightarrow m=\frac{4}{3}$