Question

There is a small source of light at some depth below the surface of water (refractive index = \(\frac{4}{3}\)) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) :
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is \(2\pi rh\)]:

• A. 17%
• B. 34%
• C. 50%
• D. 21%

Hint:

Only rays inside the cone of half-angle equal to the critical angle escape; this cone cuts a spherical cap on a sphere centred at the source.

Answer 1

The Correct Option is A

Critical angle C: \(\sin C = \frac{1}{\mu} = \frac{3}{4}\), so \(\cos C = \frac{\sqrt{7}}{4} \approx 0.661\).
Light escapes only within the cone of half-angle C about the vertical. On a sphere of radius r centred at the source, this cone cuts a cap of height \(h = r - r\cos C = r(1 - \cos C)\).
Cap area \(= 2\pi r h = 2\pi r^2(1 - \cos C)\); total area \(= 4\pi r^2\).
Fraction escaping \(= \frac{1 - \cos C}{2} = \frac{1 - 0.661}{2} \approx 0.17\), i.e. about 17%.
Answer: (1) 17%