Question

A point object lies inside a transparent solid sphere of radius $20$cm and of refractive index n = $2$. When the object is viewed from air through the nearest surface it is seen at a distance $5$cm from the surface. Find the apparent distance of object when it is seen through the farthest curved surface.

Answer 1

$\frac{-{\mu }_1}{u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}...1$

Here object is placed in the medium of refrective index ${\mu }_1$ and a spherical surface of radius of curvature'R' seperate it from medium of refrective index ${\mu }_2$

since,

${\mu }_2=2,{\mu }_1=1$

$u=?,v=5cmR=20cm$

from eq 1 we get

$\frac{1-2}{20}=\frac{1}{5}-\frac{2}{u}$

$\frac{-1}{20}-\frac{1}{5}=-\frac{2}{u}$

$u=8cm$

Now in second case,

$\mu =40-8=32cmv=?$

Thus,

$\frac{1-2}{20}=\frac{1}{v}-\frac{2}{32}$

$=\frac{1}{v}=\frac{-1}{20}+\frac{2}{32}$

$=\frac{1}{v}=\frac{=}{1}80$

$v=80cm$