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

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

  • Option 1)

    \frac{2E}{C}

  • Option 2)

    \frac{2E}{C^{2}}

  • Option 3)

    \frac{E}{C^{2}}

  • Option 4)

    \frac{E}{C}

 

Answers (1)

best_answer

As we learnt in

Momentum of EM wave -

p = frac{u}{c}

- wherein

u = Energy of EM wave

c = Speed of light in vacuum

 

 

 For a photon E=PC

P=\frac{E}{C}\rightarrow 1

in case of reflection, change in momentum is 2p

\therefore momentum transferred =2p=\frac{2E}{C} 


Option 1)

\frac{2E}{C}

This option is correct 

Option 2)

\frac{2E}{C^{2}}

This option is incorrect 

Option 3)

\frac{E}{C^{2}}

This option is incorrect 

Option 4)

\frac{E}{C}

This option is incorrect 

Posted by

Aadil

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