Question

On the axis of a transparent sphere of refractive index $(n=2)$ & radius $8$cm, an object is kept at the distance $8$cm from the surface of the sphere. Find the minimum distance of the image after all possible refraction from the surface of the sphere in cm.

Answer 1

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

$u=-8cm$

$R=$ for Ist surface $+8cm$

${\mu }_2=2$

${\mu }_1=1$

$\frac{2}{v}-\frac{1}{(-8)}=\frac{1}{8}$

$\frac{2}{v}=0$

$v=\infty$

for 2nd surface, the parallel rays are converged to focus$=R=8cm$

$\therefore$ Total distance $=\left(8\right)\left(4\right)=32cm$ from object