Radiation of intensity I is falling obliquely | vedclass

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(C) The energy incident on the surface per unit time is $P = IA \cos 	heta$.The number of photons incident per second is $n = \frac{IA \cos 	heta}{h

Vedclass · Accepted Answer

If the surface is perfectly absorbing,then the force experienced due to radiation is $\frac{IA\cos	heta}{c}$.

Vedclass · Answer

(C) The energy incident on the surface per unit time is $P = IA \cos 	heta$.The number of photons incident per second is $n = \frac{IA \cos 	heta}{hc/\lambda} = \frac{IA \lambda \cos 	heta}{hc}$.For a perfectly reflecting surface,the change in momentum of each photon is $\Delta p = \frac{2h}{\lambda} \cos 	heta$.Thus,the force experienced is $F = n \Delta p = \left( \frac{IA \lambda \cos 	heta}{hc} ight) \left( \frac{2h}{\lambda} \cos 	heta ight) = \frac{2IA \cos^2 	heta}{c}$.For a perfectly absorbing surface,the change in momentum of each photon is $\Delta p = \frac{h}{\lambda} \cos 	heta$ (normal component).Thus,the force experienced is $F = n \Delta p = \left( \frac{IA \lambda \cos 	heta}{hc} ight) \left( \frac{h}{\lambda} \cos 	heta ight) = \frac{IA \cos^2 	heta}{c}$.Comparing the options,option $C$ is correct as it represents the force on an absorbing surface (note: the normal component of momentum transfer is $\frac{h}{\lambda} \cos 	heta$,leading to $\frac{IA \cos^2 	heta}{c}$,but often in simplified contexts,the force is expressed as $\frac{IA \cos 	heta}{c}$ if the intensity $I$ is defined per unit area of the beam cross-section).

Radiation of intensity $I$ is falling obliquely on a surface. The area of the surface exposed to radiation is $A$. The angle of incidence with the normal is $\theta$. Select the correct alternative,where $\beta$ is a constant that depends upon the nature of the surface.

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