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. Numerical answer type. Work it out, then open the answer.

Answer: 4. F = P/2cP = 24 WR Light hits the hemisphere normally with I = (P)/(4π R²); reflection gives pressure 2I/c along each radius Sideways parts cancel; the axial part is 2I/ccos θ dA, and ∫ cos θ dA = π R² (projected area) F = 2I/cπ R² = (2P)/(4π R² c)π R² = P/2c F = (24)/(2 × 3 × 10⁸) = 4 × 10⁻⁸ N

A point source of light is placed at the centre of curvature of a… | JEE Main 2023 (30 Jan, Shift 1)

Q6JEE Main 2023 · 30 JAN · SHIFT 1Difficult

Momentum & Radiation Pressure

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $24\text{ W}$ . The radius of curvature of hemisphere is $10\text{ cm}$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________ $\times 10^{-8}\text{ N}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Light hits the hemisphere normally with $I = \dfrac{P}{4\pi R^2}$ ; reflection gives pressure $\dfrac{2I}{c}$ along each radius
Sideways parts cancel; the axial part is $\dfrac{2I}{c}\cos\theta\,dA$ , and $\int \cos\theta\, dA = \pi R^2$ (projected area)
$F = \dfrac{2I}{c}\pi R^2 = \dfrac{2P}{4\pi R^2 c}\pi R^2 = \dfrac{P}{2c}$
$F = \dfrac{24}{2 \times 3 \times 10^8} = 4 \times 10^{-8}\ \text{N}$

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