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QJEEM23Phf719: JEE Main 2023 Physics - Class 12 JEEM23Phf719 Solution: A point source of light is placed at the centre of curvature of …

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 W24 \mathrm{~W} . The radius of curvature of hemisphere is 10 cm10 \mathrm{~cm} and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________ × 10−8 N\times~10^{-8} \mathrm{~N} .

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JEE Main 2023 (Online) 30th January Morning Shift Physics - Electromagnetic Waves Question 41 English Explanation



amp; Force =∫PdAcos⁡θamp;=2IC∫dAcos⁡θ=2ICπR2=2p04πR2⋅πR2Camp;=p02C=242×3×108=4×10−8 N\begin{aligned}& \text { Force }=\int P d A \cos \theta \\\\& =\frac{2 \mathrm{I}}{\mathrm{C}} \int \mathrm{dA} \cos \theta=\frac{2 \mathrm{I}}{\mathrm{C}} \pi \mathrm{R}^2=2 \frac{\mathrm{p}_0}{4 \pi \mathrm{R}^2} \cdot \frac{\pi \mathrm{R}^2}{\mathrm{C}} \\\\& =\frac{\mathrm{p}_0}{2 \mathrm{C}}=\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} \mathrm{~N}\end{aligned}

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