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  • A plane light wave of intensity I=0.20Wcm −2 falls on a plane mirror surface with reflection coefficient ρ=0.8. The angle of incident is 45o. In terms of corpus

A plane light wave of intensity I=0.20Wcm −2 falls on a plane mirror surface with reflection coefficient ρ=0.8. The angle of incident is 45o. In terms of corpuscular theory, what is  magnitude of the normal pressure (N/cm2) exerted on that surface:

Option: 1

2


Option: 2

1


Option: 3

0.5


Option: 4

1.5


Answers (1)

best_answer

\\ \text{Momentum corresponding to incident photons normal to the surface.} \\ \\ \left(\frac{d p}{d t}\right)_{\text {incident }}=\frac{I}{c} d A \cos ^{2} \theta$ \\ \\ Since reflection coefficient is $\rho$, so the momentum of the reflected photons per second normal to surface \\ \\ $\left(\frac{d p}{d t}\right)_{\text {reflected }}=\frac{I}{c} d A \rho \cos ^{2} \theta

\\ \text{Hence, rate of change of momentum of the} \\ \\ \left(\frac{d p}{d t}\right)_{p h o t o n s}=\frac{I}{c} d A(\rho-1) \cos ^{2} \theta$ \\ \\ From Newton's thirt law, $F=\left(\frac{d p}{d t}\right)_{\text {surface }}=\frac{I}{c} d A(\rho+1) \cos ^{2} \theta$ \\ \\ Hence, pressure exerted on surface, $P=\frac{d F}{d A}=\frac{I}{c}(\rho+1) \cos ^{2} \theta$ \\ \\ On substituting values, we get $P=0.5 Ncm ^{-2}

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Divya Prakash Singh

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