"The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is"

Question

Accepted Answer

"

First, the intensity $I$ at a distance $r$ from a point light source spreading its energy isotropically is given by

$I = \frac{P}{4\pi r^2},$

where

$P = 450 \, \text{W}$

$r = 2 \, \text{m}.$

So,

$$I = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi} \approx \frac{450}{50.27} \approx 8.96 \, \text{W/m}^2.$$

For a perfectly reflecting surface, the radiation pressure $p$ is given by

$p = \frac{2I}{c},$

where $c \approx 3 \times 10^8 \, \text{m/s}$ is the speed of light.

Substituting the value of $I$:

$$p = \frac{2 \times 8.96}{3 \times 10^8} \approx \frac{17.92}{3 \times 10^8} \, \text{Pa}.$$

Calculating the final value:

$p \approx 5.97 \times 10^{-8} \, \text{Pa},$

which is approximately $6 \times 10^{-8} \, \text{Pa}.$

Thus, the correct answer is:

Option D: $6 \times 10^{-8}$ Pascals.

"

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is