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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
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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 |