Question:
A light of intensity \(12\;Wm^{-2}\) incidents on a black surface of area \(4\;cm^2\). The radiation pressure on the surface is
A light of intensity \(12\;Wm^{-2}\) incidents on a black surface of area \(4\;cm^2\). The radiation pressure on the surface is
Show Hint
For radiation pressure:
\[
P=\frac{I}{c}
\]
for a perfectly absorbing surface, and
\[
P=\frac{2I}{c}
\]
for a perfectly reflecting surface.
Updated On: Jun 22, 2026
- \(1\times10^{-8}\;Pa\)
- \(4\times10^{-8}\;Pa\)
- \(1.6\times10^{-7}\;Pa\)
- \(4.8\times10^{-7}\;Pa\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for radiation pressure.
For a perfectly black surface, radiation pressure is \[ P=\frac{I}{c} \] where \[ I=\text{intensity of light} \] and \[ c=3\times10^8\;m s^{-1} \]
Step 2: Substitute the given values.
Given, \[ I=12\;Wm^{-2} \] Therefore, \[ P=\frac{12}{3\times10^8} \] \[ P=4\times10^{-8}\;Pa \]
Step 3: Final conclusion.
Hence, the radiation pressure on the surface is \[ \boxed{4\times10^{-8}\;Pa} \]
For a perfectly black surface, radiation pressure is \[ P=\frac{I}{c} \] where \[ I=\text{intensity of light} \] and \[ c=3\times10^8\;m s^{-1} \]
Step 2: Substitute the given values.
Given, \[ I=12\;Wm^{-2} \] Therefore, \[ P=\frac{12}{3\times10^8} \] \[ P=4\times10^{-8}\;Pa \]
Step 3: Final conclusion.
Hence, the radiation pressure on the surface is \[ \boxed{4\times10^{-8}\;Pa} \]
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