Question:

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

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Use the formula for radiation pressure on a perfectly reflecting surface. Remember to calculate the intensity of the light at the given distance.

Updated On: Jan 14, 2026

$1.5 \times 10^{-4} \text{ Pascals}$
0
$6 \times 10^{-5} \text{ Pascals}$
$3 \times 10^{-5} \text{ Pascals} $

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The Correct Option is C

Approach Solution - 1

To determine the radiation pressure exerted by a light source on a perfectly reflecting surface, we can use the formula for radiation pressure on a reflecting surface:

$P = \frac{2I}{c}$

where:

$P$ is the radiation pressure,
$I$ is the intensity of the light, and
$c$ is the speed of light, approximately $3 \times 10^8 \text{ m/s}$.

First, we must calculate the intensity $I$ of the light at a distance 2 meters away from a point source with power 450 W. Intensity is given by the formula:

$I = \frac{P_{\text{source}}}{A}$

where:

$P_{\text{source}}$ is the power of the light source (450 W), and
$A$ is the area over which the power is spread, given by the surface area of a sphere: $A = 4\pi r^2$.

Since the distance $r = 2 \text{ m}$, we have:

$A = 4\pi (2)^2 = 16\pi \text{ m}^2$

Substitute the values into the equation for intensity:

$I = \frac{450}{16\pi}$

Calculating this, we get:

$I \approx \frac{450}{50.27} \approx 8.95 \text{ W/m}^2$

Now, we can calculate the radiation pressure:

$P = \frac{2 \times 8.95}{3 \times 10^8} \approx \frac{17.9}{3 \times 10^8} \approx 5.97 \times 10^{-8} \text{ N/m}^2$

Given that we have used approximations, the closest given option is:

$6 \times 10^{-5} \text{ Pascals}$

Thus, the radiation pressure exerted by the 450 W light source is approximately $6 \times 10^{-5} \text{ Pascals}$.

This confirms the correct answer is: