Question

A small filament is at the centre of a hollow glass sphere of inner and outer radii of 8 cm and 9 cm respectively. The refractive index of glass is 1.50. Calculate the position of the image of the filament when viewed from outside the sphere.

Answer 1

Given: A small filament is at the centre of a hollow glass sphere of inner and outer radii of 8 cm and 9 cm respectively. The refractive index of glass is 1.50. To find the position of the image of the filament when viewed from outside the sphere.

Solution:

For refraction at the first surface,

and as per given criteria,

object distance, $u=-8cm$

the radius of the curvature of inner glass sphere, $R_1=-8cm$

the refractive index of air, ${\mu }_1=1$

refractive index of the glass, ${\mu }_2=1.5$

Using this values in the following formula, we get

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}⟹\frac{1.5}{v}-\frac{1}{-8}=\frac{1.5-1}{-8}⟹\frac{1.5}{v}=\frac{0.5}{(-8)}-\frac{1}{8}⟹\frac{1.5}{v}=\frac{-0.5-1}{8}⟹v=\frac{1.5\times 8}{-1.5}⟹v=-8cm$

It means due to the first surface the image is formed at the centre. For the second surface.

For refraction at the second surface,

and as per given criteria,

object distance, $u=-9cm$

the radius of the curvature of outer glass sphere, $R_2=-9cm$

the refractive index of air, ${\mu }_1=1.5$

refractive index of the glass, ${\mu }_2=1$

Using this values in the following formula, we get

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}⟹\frac{1}{v}-\frac{1.5}{-9}=\frac{1-1.5}{-9}⟹\frac{1}{v}=\frac{0.5}{9}-\frac{1.5}{9}⟹\frac{1}{v}=\frac{0.5-1.5}{9}⟹v=\frac{1\times 9}{-1}⟹v=-9cm$

Thus, the final image is formed at the centre of the sphere.