A small filament is at the centre of a hollow glass sphere of inner and outer radii 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.

-9 cm

For refraction at the first surface,

$u = –8 cm, R_1 = –8 cm, μ_1 = 1, μ_2 = 1.5$

$\frac{m_2}{v}-\frac{m_1}{u}=\frac{m_2-m_1}{R_1}$

$\frac{1.5}{v'}+\frac{1}{8}=\frac{0.5}{-8}$

$v' = –8 cm$

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

$u = –9 cm, μ_1 = 1.5, R_2 = –9 cm$

$\frac{m_2}{v}-\frac{m_1}{u}=\frac{m_2-m_1}{R_2}$

$\frac{1}{v}+\frac{1.9}{5}=\frac{1-1.5}{-9}$

$v = –9 cm$

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