Question

Energy of radiation incident on a perfect absorbing surface per unit are in unit time is 3.6J. The radiation pressure is:

Hint:

Pay attention to the surface type:
- Perfectly absorbing (black body): $P = I/c$
- Perfectly reflecting (mirror): $P = 2I/c$

Answer 1

Step 1: Understanding the Concept:
Electromagnetic radiation carries momentum. When it strikes a surface, it exerts a force, and consequently a pressure, called radiation pressure. The magnitude depends on the intensity of the radiation and whether the surface absorbs or reflects it.

Step 2: Key Formula or Approach:
"Energy per unit area per unit time" is the definition of Intensity ($I$).
For a perfectly absorbing surface, the radiation pressure ($P$) is given by:
\[ P = \frac{I}{c} \]
Where $c$ is the speed of light in vacuum.

Step 3: Detailed Explanation:
Given:
Intensity, $I = 3.6$ J/(m$^2\cdot$s) = $3.6$ W/m$^2$
Speed of light, $c = 3 \times 10^8$ m/s
Substitute these values into the formula:
\[ P = \frac{3.6}{3 \times 10^8} \]
\[ P = 1.2 \times 10^{-8} \text{ N/m}^2 \]

Step 4: Final Answer:
The radiation pressure is $1.2 \times 10^{-8}$ N/m$^2$.