Question

A small bulb is placed at the bottom of a tank containing a liquid of refractive index µ at a depth H. It is observed that light emerges from a circular area of radius r of the surface. Obtain the expression for r in terms of H and µ.

Answer 1

This problem is based on the principle of Total Internal Reflection (TIR). Light escapes the surface only within a specific cone, the edge of which is defined by the critical angle (θ_c).

Derivation:

1. Using Geometry:

Imagine a right-angled triangle formed by the bulb at depth H, the center of the circular area, and the radius r.
From the triangle:

tanθ_c = `r/H`

⇒ r = H tanθ_c

2. Using Snell’s Law:

At the boundary, the critical angle is given by:

`sin theta_c = 1/mu`

3. Converting sinθ_c and tanθ_c:

Since, `tan theta = (sin theta)/(cos theta)`

= `(sin theta)/(sqrt(1 - sin^2theta))`

We substitute the value of sinθ_c:

`tan theta_c = (1/mu)/(sqrt(1 - (1/mu)^2))`

= `(1/mu)/(sqrt(1 - 1/mu^2))`

= `(1/mu)/(sqrt((mu^2 - 1)/mu^2))`

= `(1/mu)/(sqrt((mu^2 - 1))/mu)`

Dividing by a fraction is the same as multiplying by its reciprocal:

= `1/mu xx mu/sqrt(mu^2 - 1)`

= `1/sqrt(mu^2 - 1)`

4. Final Expression:

Substitute the value back into the radius formula:

r = H tanθ_c

`r = H(1/sqrt(mu^2 - 1))`

r = `H/sqrt(mu^2 - 1)`