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Electromagnetic Waves question

2015 · Q113
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Electromagnetic Waves question

2015 · Q113

NEETPhysicsElectromagnetic WavesMCQ+4 / −1
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)
  1. A
    2EC2{{2E} \over {{C^2}}}C22E​
  2. B
    EC2{E \over {{C^2}}}C2E​
  3. C
    EC{E \over C}CE​
  4. D
    2EC{{2E} \over C}C2E​
View written solutionFree

Correct answer: D

Momentum transferred to the surface

= change in momentum

= Pf - Pi

= +EC−(−EC) + {E \over C} - \left( { - {E \over C}} \right)+CE​−(−CE​)

= 2EC{{2E} \over C}C2E​

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