A perfectly absorbing surface is illuminated by an electromagnetic wave of intensity $I = 300 W m^{- 2}$ . What is the radiation pressure $P$ on the surface?

medium MCQ 2 marks Class 12 · Physics · Electromagnetic Waves

Answer

1.0 \times 1 0^{- 6} N m^{- 2}

The correct answer is $1.0 \times 1 0^{- 6} N m^{- 2}$ — Thus the pressure is $1.0 \times 1 0^{- 6} N m^{- 2}$ .

Step-by-step solution

For a perfectly absorbing surface the radiation pressure is $P = \frac{I}{c}$ .
Substitute $I = 300 W m^{- 2}$ and $c = 3.0 \times 1 0^{8} m s^{- 1}$ .
Calculate $P = \frac{300}{3.0 \times 1 0 ^{8}} = 1.0 \times 1 0^{- 6} N m^{- 2}$ .
Thus the pressure is $1.0 \times 1 0^{- 6} N m^{- 2}$ .

Common mistake: Using

2 I / c

(the formula for a reflecting surface) instead of

I / c

Concept tested: Radiation pressure on an absorbing surface