Q128 / 180 · NEET 2015 (Phase 1) · Physics

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

A
$\frac{E}{C}$
Correct answer
B
$\frac{2 E}{C}$
C
$\frac{2 E}{C^{2}}$
D
$\frac{E}{C^{2}}$

Solution

$E = \frac{h C}{λ} P = \frac{h}{λ} \Rightarrow P = \frac{E}{C}$

Initial momentum of light when falls on a perfectly reflecting surface.

$P_{i} = \frac{E}{C}$

When it is reflected momentum is in opposite direction

$P_{f} = - \frac{E}{C}$

Net momentum $= P_{f} - P_{i}$

$= - \frac{E}{C} - \frac{E}{C}$

$= - \frac{2 E}{C}$

$Δ P = \frac{2 E}{C}$

Hence, the correct answer is option: 'B'.

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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)

Solution

Relevant Questions

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