JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 28)
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $$24 \mathrm{~W}$$. The radius of curvature of hemisphere is $$10 \mathrm{~cm}$$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________ $$\times~10^{-8} \mathrm{~N}$$.
Answer
4
Explanation
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$$ \begin{aligned} & \text { Force }=\int P d A \cos \theta \\\\ & =\frac{2 \mathrm{I}}{\mathrm{C}} \int \mathrm{dA} \cos \theta=\frac{2 \mathrm{I}}{\mathrm{C}} \pi \mathrm{R}^2=2 \frac{\mathrm{p}_0}{4 \pi \mathrm{R}^2} \cdot \frac{\pi \mathrm{R}^2}{\mathrm{C}} \\\\ & =\frac{\mathrm{p}_0}{2 \mathrm{C}}=\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} \mathrm{~N} \end{aligned} $$
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