JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Geometrical Optics

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 spherical cap of height

h

and radius of curvature

r

2 π rh

]

(A) $21%$
(B) $17%$ ✓ Correct
(C) $50%$
(D) $34%$

Answer: (B)

Solution

Light escapes only through the cone bounded by the critical angle

β

, where

sin β = \frac{1}{n} = \frac{3}{4}

, giving

cos β = \frac{7}{4}

. The fraction of the source's output falling within this cone is the solid angle

2 π (1 - cos β)

divided by the full

4 π

, that is

\frac{1 - cos β}{2} = \frac{1 - \frac{7}{4}}{2} = \frac{4 - 7}{8}

. Numerically this is about

\frac{4 - 2.65}{8} \approx 0.17

, so nearly

17%

of the light emerges.

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Geometrical Optics

Solution

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