Question

An object is lying at a distance of $2\text{m}$ from a spherical interface as shown in the figure. If the radius of the interface is $\frac{2}{3}\text{m}$ and refractive index of the medium is $\mu =1.4$, calculate the magnification of the image formed by the interface.

Answer 1

Given :
Distance of the object from the pole of the spherical interface $\left(u\right)=-2\text{m}$
Radius of the interface $\left(R\right)=\frac{2}{3}\text{m}$
Refractive index of the interface $\left({\mu }_2\right)=1.4$

Position of the image is given by :

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

$\frac{1.4}{v}-\frac{1}{-2}=\frac{1.4-1}{\frac{2}{3}}$

$\Rightarrow \frac{1.4}{v}=\frac{0.4\times 3}{2}-\frac{1}{2}$

$\Rightarrow \frac{1.4}{v}=\frac{0.2}{2}$

$\Rightarrow v=14\text{m}$

Therefore, magnification :

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

$m=\frac{1}{1.4}\times \frac{14}{-2}$

$\Rightarrow m=-5$