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  • A radiation of energy  falls normally on a perfectly reflecting surface. The momentum transferred to the surface is<d

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

Option: 1

\mathrm{\frac{E}{c}}


Option: 2

\frac{2E}{c}


Option: 3

\mathrm{E\, c }


Option: 4

\mathrm{\frac{E}{c^2} }


Answers (1)

best_answer

Initial momentum of surface
\mathrm{P_i=\frac{E}{C} }
Where, \mathrm{ c } = velocity of light (constant).
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely. Final momentum
\mathrm{P_f=\frac{E}{c} }
(negative value)
\mathrm{\therefore } Change in momentum
\mathrm{ \begin{aligned} \Delta p & =p_f-p_i \\ & =-\frac{E}{c}-\frac{E}{c}=-\frac{2 E}{c} \end{aligned} }
Thus, momentum transferred to the surface is
\mathrm{ \Delta p^{\prime}=|\Delta p|=\frac{2 E}{c} }
 

Posted by

himanshu.meshram

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