JEE Main · 2023 · 30 Jan 2023 · Shift 1 · EM Waves

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A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 W. The radius of curvature of hemisphere is 10 cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is

____

\times 1 0^{- 8} N

Answer: 4

Solution

A point source of power

P_{0}

at the centre of curvature sends rays onto the hemisphere along its normals, and a perfectly reflecting surface feels a radiation pressure

\frac{2 I}{c}

. Summing only the components along the axis,

F = \frac{2 I}{c} π R^{2}

, and the intensity at radius

R

I = \frac{P _{0}}{4 π R ^{2}}

. Hence

F = \frac{P _{0}}{2 c} = \frac{24}{2 \times 3 \times 1 0 ^{8}} = 4 \times 1 0^{- 8} N

, so the value is

4

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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