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.

Explanation

For refraction at the first surface, `mu_(1) = 1, mu_(2) = 1.50, u = -8 cm`
`R = - cm`
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
`(1.5)/(v') - (100)/(-8cm) = (1.50 - 1.00)/(-8 cm)`
v = -8 cm
It means that due to the first surface the image is formed at the centre of the sphere.
For the second surface, `mu_(1) = 1.50, mu_(2) = 1, u = -9 cm, R = -9 cm`
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2)-mu_(1))/(R)`
`(1)/(v) - (1.50)/(-9 cm) = (1-1.50)/(-8 cm)`
v = -9 cm
Hence, the final image is formed at the centre of the sphere.