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A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface

$\begin {array} {1 1} (a)\;\large\frac{E}{c} & \quad (b)\;\large\frac{2E}{c} \\ (c)\;Ec & \quad (d)\;\large\frac{E}{c^2} \end {array}$

 

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Ans : (b)
Initial momentum of surface, $p_i = \large\frac{E}{c}$
Since surface is perfectly reflecting, so, final momentum, $p_f = \large\frac{E}{c}$
So, change in momentum, $ \Delta p = p_f - p_i = -\large\frac{2E}{c}$
So, momentum transferred to the surface is $ \Delta pā€™ = | \Delta p| = \large\frac{2E}{c}$

 

answered Dec 30, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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