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

Light from a point source at depth h below the surface can emerge only within a cone of half-angle equal to the critical angle i_c. Radius of the illuminated circle: r = h · tan i_c (as i = i_c). sin i_c = $\displaystyle 1$/μ = $\displaystyle 1$/√$\displaystyle 2$ ⟹ i_c = $\displaystyle 45$°, so tan i_c = $\displaystyle 1$ and r = h. Area = πr² = $\displaystyle 3.14$ × ($\displaystyle 1$)² = $\displaystyle 3.14$ m².