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

It is the minimum angle for total internal reflection.

sin C = `1/mu`

Circular Patch Formation Concept:

• Light emitted from the bulb spreads in all directions.
• Rays with an incidence angle smaller than the critical angle will emerge from the liquid surface.
• Rays with an incidence angle greater than the critical angle will be totally internally reflected.

Thus, only rays within a cone of semi-vertical angle C will emerge, creating a circular patch on the surface.

Geometry of the Situation:

• The bulb is at a depth H.
• The ray that emerges at the critical angle will strike the surface at a distance r from the center.

Using trigonometry, we have:

tan C = `r/H`

r = H tan C

Expressing tanC in terms of μ:

From Snell’s Law:

sin C = `1/mu`

Using trigonometric identities, we get:

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

tan C = `(sin C)/(cos C)`

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

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

Substituting tan C into the formula for r:

`r = H  .  1/sqrt(mu^2 - 1)`

This expression relates the radius of the circular patch to the depth of the bulb H and the refractive index μ of the liquid.