Question

A small bulb is placed at the bottom of a tank, containing a transparent liquid of refractive index `sqrt 2`, to a depth of 1 m. Calculate the area of the surface of the liquid through which light from the bulb emerges.

Answer 1

Given: Depth (h) = 1 m

Refractive index (µ) = `sqrt 2`

Critical angle (C) = `1/mu`

= `1/sqrt 2`

= 45°

Radius of the circle (r) = h tan C

= 1 m × tan 45°

= 1 m × 1

= 1 m

Area (A) = πr²

= π × (1)²

= π m²

≈ 3.14159 m²

∴ The area of the surface through which light emerges is π m².