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  • An electromagnetic wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure <img alt="\mathrm{P}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%7D" /

An electromagnetic wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure \mathrm{P} on it. Which of the following statement is not true?
 

Option: 1

Radiation pressure is  \mathrm{I} / \mathrm{c} if the wave is totally absorbed

 


Option: 2

Radiation pressure is \mathrm{I} / \mathrm{c} if the wave is totally reflected
 


Option: 3

Radiation pressure is 2 \mathrm{I} / \mathrm{c} if the wave is totally reflected
 


Option: 4

Radiation pressure is in the range \mathrm{I/c <p>2 \mathrm{I} / \mathrm{c}} for real surfaces
 


Answers (1)

best_answer

\mathrm{Momentum\: per \: unit \: time \: per\: unit \: area \: intensity =\frac{\text { int ensity }}{\text { speed of wave }}=\frac{I}{c}}

Change in momentum per unit time per unit area \mathrm{=\Delta \mathrm{I} / \mathrm{c}=} radiation pressure

\mathrm{(\mathrm{P}), i.e., \mathrm{P}=\Delta \mathrm{I} / \mathrm{c}}
Momentum of incident wave per unit time per unit area \mathrm{=\mathrm{I} / \mathrm{c}}
When wave is fully absorbed by the surface, the momentum of the reflected wave per unit tie per unit area \mathrm{ =0}
Radiation pressure \mathrm{ (\mathrm{P})=} change in momentum per unit time
Per unit area \mathrm{=\frac{\Delta \mathrm{I}}{\mathrm{c}}=\frac{\mathrm{I}}{\mathrm{c}}-0=\frac{\mathrm{I}}{\mathrm{c}}}
When wave is totally reflected, then momentum of the reflected wave per unit time per unit area \mathrm{=-\mathrm{I} / \mathrm{c}}
Radiation pressure \mathrm{(\mathrm{P})=\frac{\mathrm{I}}{\mathrm{c}}-\left(-\frac{\mathrm{I}}{\mathrm{c}}\right)=\frac{2 \mathrm{I}}{\mathrm{c}}}
Here, \mathrm{\mathrm{P}} lies between \mathrm{\frac{\mathrm{I}}{\mathrm{c}} \: and \: \frac{2 \mathrm{I}}{\mathrm{c}}}.

Hence option 2 is correct.

Posted by

Deependra Verma

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