Question

An electromagnetic radiation of intensity 360 W /cm² is incident normally on a non-reflecting surface having area A. Average force on the surface is found to be 2.4 x 10^-4 N. Find the value of A.

• A. 0.02 m²
• B. 0.2 m²
• C. 2 m²
• D. 20 m²

Answer 1

The Correct Option is A

To solve the problem, we need to determine the area \( A \) of a surface on which electromagnetic radiation is incident. The given values are:

• Intensity of radiation \( I = 360 \, \text{W/cm}^2 \)
• Average force on the surface \( F = 2.4 \times 10^{-4} \, \text{N} \)

We know that intensity \( I \) is related to force \( F \) and area \( A \) by the following relation for a non-reflecting surface:

F = \dfrac{I \cdot A}{c}

Where:

• I is the intensity.
• A is the area of the surface in question.
• c is the speed of light in vacuum, approximately 3 \times 10^8 \, \text{m/s}.

First, convert the intensity from \(\text{W/cm}^2\) to \(\text{W/m}^2\):

360 \, \text{W/cm}^2 = 360 \times 10^4 \, \text{W/m}^2 = 3.6 \times 10^6 \, \text{W/m}^2

Substitute the values in the formula:

2.4 \times 10^{-4} = \dfrac{3.6 \times 10^6 \cdot A}{3 \times 10^8}

Simplify to find A:

2.4 \times 10^{-4} = \dfrac{3.6 \cdot A}{300}

2.4 \times 10^{-4} \times 300 = 3.6 \cdot A

72 \times 10^{-4} = 3.6 \cdot A

A = \dfrac{72 \times 10^{-4}}{3.6} = 20 \times 10^{-4} = 0.02 \, \text{m}^2

Therefore, the area \( A \) is 0.02 \, \text{m}^2.

Hence, the correct answer is 0.02 m².