There is a small source of light at some depth below the surface of water (refractive index = 4 \over 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]:

Geometrical Optics PYQ — JEE Main 2020 · 9 Jan · Shift 2 · Q57 with Solution

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Correct answer: A

Idea of the problem

A point source inside water emits light uniformly in all directions. Only those rays that strike the water-air surface with angle of incidence less than or equal to the critical angle can emerge.

So, the fraction of light emerging out is equal to:

\frac{\text{solid angle of escaping rays}}{\text{total solid angle }4\pi}

Since only the upper hemisphere can even reach the surface, and within that hemisphere only a cone of semi-vertical angle equal to the critical angle escapes, we compute the corresponding solid angle.

Critical angle at water-air interface

Given refractive index of water:

\mu = \frac{4}{3}

For water to air,

\sin c = \frac{1}{\mu} = \frac{3}{4}

Thus,

c = \sin^{-1}\left(\frac{3}{4}\right)

and

\cos c = \sqrt{1-\sin^2 c} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt7}{4}

Geometrical interpretation using spherical cap

Imagine a sphere of radius $r$ centered at the source. The total light emitted is uniformly spread over the sphere of area:

4\pi r^2

The rays that can emerge form a cone in the upward direction with semi-angle $c$ . On the sphere, this corresponds to a spherical cap.

Height of the cap:

h = r(1-\cos c)

Area of spherical cap is given:

A_{\text{cap}} = 2\pi r h = 2\pi r^2(1-\cos c)

Therefore, fraction of total light emerging is:

\frac{A_{\text{cap}}}{4\pi r^2} = \frac{2\pi r^2(1-\cos c)}{4\pi r^2} = \frac{1-\cos c}{2}

Substitute $\cos c = \dfrac{\sqrt7}{4}$ :

\text{fraction} = \frac{1-\frac{\sqrt7}{4}}{2}

Using $\sqrt7 \approx 2.646$ ,

\frac{\sqrt7}{4} \approx 0.6615

So,

\text{fraction} \approx \frac{1-0.6615}{2} = \frac{0.3385}{2} \approx 0.169

Thus percentage emerging is approximately:

0.169 \times 100 \approx 16.9\% \approx 17\%

Option check

A: $17\%$ ✅
B: $34\%$ ❌
C: $50\%$ ❌
D: $21\%$ ❌

So the correct option is A.

More from Geometrical Optics

View written solutionFree

Correct answer: A

Idea of the problem

A point source inside water emits light uniformly in all directions. Only those rays that strike the water-air surface with angle of incidence less than or equal to the critical angle can emerge.

So, the fraction of light emerging out is equal to:

\frac{\text{solid angle of escaping rays}}{\text{total solid angle }4\pi}

Critical angle at water-air interface

Given refractive index of water:

\mu = \frac{4}{3}

For water to air,

\sin c = \frac{1}{\mu} = \frac{3}{4}

Thus,

c = \sin^{-1}\left(\frac{3}{4}\right)

and

\cos c = \sqrt{1-\sin^2 c} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt7}{4}

Geometrical interpretation using spherical cap

Imagine a sphere of radius $r$ centered at the source. The total light emitted is uniformly spread over the sphere of area:

4\pi r^2

The rays that can emerge form a cone in the upward direction with semi-angle $c$ . On the sphere, this corresponds to a spherical cap.

Height of the cap:

h = r(1-\cos c)

Area of spherical cap is given:

A_{\text{cap}} = 2\pi r h = 2\pi r^2(1-\cos c)

Therefore, fraction of total light emerging is:

\frac{A_{\text{cap}}}{4\pi r^2} = \frac{2\pi r^2(1-\cos c)}{4\pi r^2} = \frac{1-\cos c}{2}

Substitute $\cos c = \dfrac{\sqrt7}{4}$ :

\text{fraction} = \frac{1-\frac{\sqrt7}{4}}{2}

Using $\sqrt7 \approx 2.646$ ,

\frac{\sqrt7}{4} \approx 0.6615

So,

\text{fraction} \approx \frac{1-0.6615}{2} = \frac{0.3385}{2} \approx 0.169

Thus percentage emerging is approximately:

0.169 \times 100 \approx 16.9\% \approx 17\%

Option check

A: $17\%$ ✅
B: $34\%$ ❌
C: $50\%$ ❌
D: $21\%$ ❌

So the correct option is A.

Geometrical Optics question

Geometrical Optics question

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