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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)

1. \( \frac{2E}{C} \)
2. \( \frac{2E}{C^2} \)
3. \( \frac{E}{C^2} \)
4. \( \frac{E}{C} \)

Correct answer: \( \frac{2E}{C} \)

Solution

When radiation falls on a perfectly reflecting surface, the momentum transferred is twice the momentum of the incident radiation because the radiation reverses direction. The momentum of the radiation is given by \( p = \frac{E}{c} \), so the total momentum transferred is \( 2p = \frac{2E}{c} \).

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