Question bank/JEE Main/Physics

There is a small source of light at some depth below the surface
of water (refractive index = 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 $2\pi rh$]

A50B34C17D21

Show answer and solution▾

Solid angle $d\Omega = 2\pi R^{2}(1 - cos\beta)$
Percentage of light
$= \frac{2\pi R^{2}(1 - cos\beta)}{4\pi R^{2}} \times 100$
$= \frac{1 - cos\beta}{2} \times 100 = (\frac{4 - \sqrt{7}}{8}) \times 100 \approx 17\%$

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx