Question

Electromagnetic waves are incident normally on a perfectly reflecting surface having surface area A. If I is the intensity of the incident electromagnetic radiation and c is the speed of light in vacuum, the force exerted by the electromagnetic wave on the reflecting surface is

• A. \(\frac{2IA}{c}\)
• B. \(\frac{IA}{c}\)
• C. \(\frac{IA}{2c}\)
• D. \(\frac{I}{2Ac}\)

Answer 1

The Correct Option is A

Given:

• Surface area = \( A \)
• Intensity of electromagnetic wave = \( I \)
• Speed of light in vacuum = \( c \)
• Surface is perfectly reflecting

Step 1: Radiation Pressure

For a **perfectly reflecting** surface, the radiation pressure \( P \) is given by:

\[ P = \frac{2I}{c} \]

Step 2: Force on the Surface

Force exerted by radiation is pressure multiplied by area:

\[ F = P \cdot A = \frac{2I}{c} \cdot A = \frac{2IA}{c} \]

The force exerted by the wave is \({\frac{2IA}{c}} \), so the correct answer is (A).

Answer 2

When an electromagnetic wave is incident on a perfectly reflecting surface, the pressure exerted by the wave on the surface is given by: \[ P = \dfrac{I}{c} \] Where: - \(P\) is the pressure, - \(I\) is the intensity of the incident electromagnetic radiation, - \(c\) is the speed of light in vacuum. The force exerted by the wave on the surface area \(A\) is given by: \[ F = P \times A = \dfrac{I}{c} \times A = \dfrac{IA}{c} \] However, because the surface is perfectly reflecting, the total momentum change will be doubled (since the wave is reflected and thus its momentum is reversed), and the force is therefore: \[ F = 2 \times \dfrac{IA}{c} \] Thus, the force exerted by the electromagnetic wave on the reflecting surface is \(\dfrac{2IA}{c}\).