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A radiation of energy 'E'falls normally ...

A radiation of energy 'E'falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)

A

`(2E)/( C )`

B

`(2 E)/( C^2)`

C

`(E )/( C^2)`

D

`(E )/( C)`

Text Solution

Verified by Experts

The correct Answer is:
A

Momentum of light `p = ( E )/( C)`
So, momentum transferred to the surface
`= p_(f) - p_(i) = (2E)/( C )`
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