The intensity of a plane electromagnetic wave is 0.02 W/m^2. What is the radiation pressure exerted by this wave on a perfectly reflecting surface at normal incidence?

B. 1.33 10^-10 Pa — For a perfectly reflecting surface, the radiation pressure is twice the intensity divided by the speed of light. Using P = 2I/c, we calculate the pressure exerted by the incident and reflected momentum.

JEE Advanced · PhysicsHard

$The intensity of a plane electromagnetic wave is I = 0.02 W/m^{2} . What is the radiation pressure P exerted by this wave on a perfectly reflecting surface at normal incidence?$

A. $6.67 \times 1 0^{- 11} Pa$
B. $1.33 \times 1 0^{- 10} Pa$
C. $2.0 \times 1 0^{- 10} Pa$
D. $3.33 \times 1 0^{- 11} Pa$

Show correct answer & step-by-step solution

Correct answer: B — $1.33 \times 1 0^{- 10} Pa$

Solution

P = \frac{2 I}{c} = \frac{2 \times 0.02}{3 \times 1 0 ^{8}} = \frac{0.04}{3 \times 1 0 ^{8}} \approx 1.33 \times 1 0^{- 10} Pa .

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The intensity of a plane electromagnetic wave is I=0.02 W/m2. What is the radiation pressure P exerted by this wave on a perfectly reflecting surface at normal incidence?

Solution

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$The intensity of a plane electromagnetic wave is I = 0.02 W/m^{2} . What is the radiation pressure P exerted by this wave on a perfectly reflecting surface at normal incidence?$