Q

# 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)

$\frac{E}{C}$

138 Views

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

Exams
Articles
Questions