JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Geometrical Optics

A small bulb is placed at the bottom of a tank containing water to a depth of

7

m. The refractive index of water is

\frac{4}{3}

. The area of the surface of water through which light from the bulb can emerge out is

x π

^{2}

. The value of x is ______.

Answer: 9

Solution

Light can emerge only within a cone defined by the critical angle

C

, where:

sin C = \frac{1}{μ} = \frac{3}{4}

From this:

cos C = \frac{7}{4}

and

tan C = \frac{3}{7}

The radius of the circular area on the surface is:

r = h tan C = 7 \times \frac{3}{7} = 3 m

The area is:

A = π r^{2} = π \times 3^{2} = 9 π m^{2}

Therefore,

x = 9

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer

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