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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 ________ × 10⁻⁸ N.

Asked in JEE Main 30th Jan 1st Shift 2023 · Force and radiation pressure

Figure: Force and radiation pressure
Answer: 4

Step-by-step solution

The hemisphere subtends half the total solid angle, so the power falling on it is P' = 24/2 = 12 W

Every ray leaves the centre of curvature, so it hits the mirror at normal incidence and returns along itself: an element carries dF = 2(dP)/c directed along its own radius

The radial components cancel in pairs; only the axial component survives: F = ∫2(P/4π c)cos θ dΩ over the hemisphere = P/(2c)

F = 24/(2 × 3 × 10⁸) = 4 × 10⁻⁸ N — i.e. P'/c = 12/(3 × 10⁸), not 2P'/c

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