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A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture . If the beam carries an intensity I find force exerted by beam on sphere .

$(a)\;\large\frac{\pi r^2 I}{C}\qquad(b)\;\large\frac{\pi r^2 I}{2C}\qquad(c)\;\large\frac{\pi r^2 I}{3C}\qquad(d)\;\large\frac{4}{3}\;\large\frac{\pi r^2 I}{C}$

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Answer : (a) $\;\large\frac{\pi r^2 I}{C}$
Explanation :
Let O be centre of sphere and OZ be line opposite to incident beam . Consider a radius OP of sphere making an angle $\;\theta\;$ with Oz to get aciroll on sphere. Charge $\;\theta\;to\;\theta+d \theta\;$ and rotate radius about OZ to get another circle on sphere . The part of sphere between these circles is a ring of area $\;2 \pi r^2 sin \theta \; d \theta\;.$ Consider a small part $\;\bigtriangleup A\;$ of this right at P . Energy of light falling on this part in time $\;\bigtriangleup t\;$ is
$\;\bigtriangleup u=I \bigtriangleup t\;(\bigtriangleup A cos \theta)$
The momentum of this light falling on $\;\bigtriangleup \;A\;$ is $\;\bigtriangleup u\;$ along QP. The light e is reflected by sphere along PR .Therefore in momentum is
$\bigtriangleup P=\large\frac{2 \bigtriangleup u}{C}\;cos \theta=\large\frac{2}{C}\;I \bigtriangleup t \;(\bigtriangleup A cos^{2} \theta) \quad \; $ (direction along $\;\overrightarrow{OP}$)
The force on $\;\bigtriangleup A\;$ due to light falling on it , is
$\large\frac{\bigtriangleup P}{\bigtriangleup}=\large\frac{2}{C}\;I \bigtriangleup A cos^{2} \theta \quad \;$ (directed along $\;\overrightarrow{PO}$)
The resultant force on ring as well as an sphere is along ZO by force on $\;\bigtriangleup A\;$ along ZO.
$\large\frac{\bigtriangleup P}{\bigtriangleup t }\;cos \theta=\large\frac{2}{C}\;I \bigtriangleup A cos^{3} \theta\;$ (along $\;\overrightarrow{ZO}$)
The force acting on ring is
$d F=\large\frac{2}{C} I \;(2 \pi r^2 sin \theta d \theta)\;cos^{3} \theta$
The force on entire sphere is
$F=\int_{0}^{\large\frac{\pi}{2}}\;\large\frac{4 \pi r^2 I}{C} cos^3 \theta sin \theta \; d\theta$
$=-\int_{0}^{\large\frac{\pi}{2}}\;\large\frac{4 \pi r^2 I}{C}\;cos^3 \theta d(cos \theta)$
$=-\int_{0}^{\large\frac{\pi}{2}}\;\large\frac{4 \pi r^2 I}{C} \; [\large\frac{cos^{4} \theta}{4}]_{0}^{\large\frac{\pi}{2}}$
$=\large\frac{\pi r^2 I}{C}$
Vote that integration is done only for hemisphere that faces the incident beam.
answered Feb 22, 2014 by yamini.v

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