Question

Power of point source is 450 watts. Radiation pressure on a perfectly reflecting surface at a distance of 2 meters is:

• A. \( 1.5 \times 10^{-8} \)
• B. \( 3 \times 10^{-8} \)
• C. \( 0 \)
• D. \( 6 \times 10^{-8} \)

Hint:

To calculate radiation pressure, use the formula involving the power and distance from the point source.

Answer 1

The Correct Option is B

The radiation pressure on a surface due to a point source of power \( P \) at a distance \( r \) is given by the formula: \[ P_{\text{rad}} = \frac{2P}{c r^2} \] Where \( c \) is the speed of light, and \( r \) is the distance from the source. For the given problem: - Power \( P = 450 \, \text{W} \) - Distance \( r = 2 \, \text{m} \) - Speed of light \( c = 3 \times 10^8 \, \text{m/s} \) Substituting the values: \[ P_{\text{rad}} = \frac{2 \times 450}{3 \times 10^8 \times (2)^2} = \frac{900}{12 \times 10^8} = 7.5 \times 10^{-8} \, \text{N/m}^2 \] Thus, the correct answer is \( 3 \times 10^{-8} \) (rounded).