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(D) The radiation pressure $P$ on a surface is given by $P = \frac{I}{c}$ for absorption and $P = \frac{2I}{c}$ for reflection,where $I$ is the intens

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(D) The radiation pressure $P$ on a surface is given by $P = \frac{I}{c}$ for absorption and $P = \frac{2I}{c}$ for reflection,where $I$ is the intensity and $c$ is the speed of light.Since the inner surface is completely reflecting,the force $dF$ on a small area element $dA$ is $dF = (2 \frac{I}{c}) dA \cos 	heta$,where $	heta$ is the angle with the normal.The intensity $I$ at a distance $R$ from a point source of power $P_0$ is $I = \frac{P_0}{4 \pi R^2}$.The force component along the axis of symmetry is $dF_z = dF \cos 	heta = \frac{2I}{c} dA \cos^2 	heta$.Integrating over the hemisphere,the total force $F = \int \frac{2}{c} (\frac{P_0}{4 \pi R^2}) \cos^2 	heta (R^2 \sin 	heta d	heta d\phi)$.$F = \frac{P_0}{2 \pi c} \int_0^{2 \pi} d\phi \int_0^{\pi/2} \cos^2 	heta \sin 	heta d	heta = \frac{P_0}{2 \pi c} (2 \pi) [-\frac{\cos^3 	heta}{3}]_0^{\pi/2} = \frac{P_0}{c} (\frac{1}{3})$.Wait,for a hemispherical shell,the force is $F = \frac{P_0}{2c}$.Given $P_0 = 24 \, W$ and $c = 3 	imes 10^8 \, m/s$,$F = \frac{24}{2 	imes 3 	imes 10^8} = 4 	imes 10^{-8} \, N$.

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