A tank is filled with a liquid to a depth of 80 cm. A point source of light is placed at the centre of the bottom. The area of the surface of the liquid through which light from the source can emerge out is:

(Take refractive index of liquid = \(2/\sqrt{3}\) )

Question

Question

This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 2 Official Paper

Answer 1

Concept:

The Refractive index is the ratio between the speed of light in air to the speed in a medium.
Snell’s law clarifies the relation between the angle of incidence(i) and angle of refraction(r).
\(μ = \frac {sin ~r }{sin~ i}\)
Area of circle = πr²

Calculation:

Given that, μ = \(\frac{2}{\sqrt3}\) , depth of tank = 80 cm = 0.8 m

When light is placed at the centre of the bottom then the light come out in conical shape

F1 Vinanti Defence 31.12.22 D9

Now let the angle OBC = i and after refraction angle (r) = 90°

Then from Snell's law \(μ = \frac {sin ~r }{sin~ i}\)

\(\frac 2 {√ 3} = \frac {sin~ 90}{sin~ i}\)

⇒ i = 60°

Now, in triangle OBC

tan 60° = \(\frac {OC}{OB} = \frac {OC}{0.8}\)

⇒ OC = √3 × 0.8

The area of the surface of the liquid through which light from the source can emerge out is circular and radius is equal to OC

Then the area of the surface = πr² = π(√3 × 0.8)² = 6.03 m²