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					 A radiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)
 
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2E C  -   
2E C²  -  
E C²  -  
E C  
 -  
 
Correct Option: A
Momentum of light falling on reflecting surface 
| p = | ||
| C | 
As surface is perfectly reflecting so momentum reflect
| p1 = - | ||
| C | 

So, momentum transferred
| = P - P1 = | - | ![]()  | - | ![]()  | ||
| C | C | 
| = | ||
| C | 

