Question:
Light falls normally on a non-reflecting surface. If the average force exerted on a surface with area \(15\,\text{cm}^2\) during \(20\) minute time interval is \(10^{-6}\,\text{N}\), then energy flux of light is \((c=3\times10^8\,\text{m s}^{-1})\)
Light falls normally on a non-reflecting surface. If the average force exerted on a surface with area \(15\,\text{cm}^2\) during \(20\) minute time interval is \(10^{-6}\,\text{N}\), then energy flux of light is \((c=3\times10^8\,\text{m s}^{-1})\)
Show Hint
For a perfectly absorbing surface,
\[
P_r=\frac{I}{c}.
\]
For a perfectly reflecting surface,
\[
P_r=\frac{2I}{c}.
\]
Always identify whether the surface absorbs or reflects the incident light.
Updated On: Jun 18, 2026
- \(20\times10^4\,\text{W m}^{-2}\)
- \(15\times10^4\,\text{W m}^{-2}\)
- \(25\times10^4\,\text{W m}^{-2}\)
- \(10\times10^4\,\text{W m}^{-2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use radiation pressure for a perfectly absorbing surface.
Since the surface is non-reflecting (perfectly absorbing), the radiation pressure is \[ P_r=\frac{I}{c} \] where \[ I=\text{energy flux (intensity)} \] and \[ c=3\times10^8\,\text{m s}^{-1}. \] Also, \[ P_r=\frac{F}{A} \] Therefore, \[ \frac{F}{A}=\frac{I}{c} \] or \[ I=\frac{Fc}{A}. \]
Step 2: Convert the given area into SI units.
Given, \[ A=15\,\text{cm}^2 \] \[ A=15\times10^{-4}\,\text{m}^2 \] \[ A=1.5\times10^{-3}\,\text{m}^2. \] Also, \[ F=10^{-6}\,\text{N}. \]
Step 3: Calculate the energy flux.
Substituting in \[ I=\frac{Fc}{A}, \] \[ I=\frac{(10^{-6})(3\times10^8)} {1.5\times10^{-3}} \] \[ I=\frac{300}{1.5\times10^{-3}} \] \[ I=2\times10^5\,\text{W m}^{-2}. \]
Step 4: Express in the given form.
\[ 2\times10^5 = 20\times10^4. \] Therefore, \[ I=20\times10^4\,\text{W m}^{-2}. \]
Step 5: Final conclusion.
Hence, the energy flux of light is \[ \boxed{20\times10^4\,\text{W m}^{-2}} \]
Since the surface is non-reflecting (perfectly absorbing), the radiation pressure is \[ P_r=\frac{I}{c} \] where \[ I=\text{energy flux (intensity)} \] and \[ c=3\times10^8\,\text{m s}^{-1}. \] Also, \[ P_r=\frac{F}{A} \] Therefore, \[ \frac{F}{A}=\frac{I}{c} \] or \[ I=\frac{Fc}{A}. \]
Step 2: Convert the given area into SI units.
Given, \[ A=15\,\text{cm}^2 \] \[ A=15\times10^{-4}\,\text{m}^2 \] \[ A=1.5\times10^{-3}\,\text{m}^2. \] Also, \[ F=10^{-6}\,\text{N}. \]
Step 3: Calculate the energy flux.
Substituting in \[ I=\frac{Fc}{A}, \] \[ I=\frac{(10^{-6})(3\times10^8)} {1.5\times10^{-3}} \] \[ I=\frac{300}{1.5\times10^{-3}} \] \[ I=2\times10^5\,\text{W m}^{-2}. \]
Step 4: Express in the given form.
\[ 2\times10^5 = 20\times10^4. \] Therefore, \[ I=20\times10^4\,\text{W m}^{-2}. \]
Step 5: Final conclusion.
Hence, the energy flux of light is \[ \boxed{20\times10^4\,\text{W m}^{-2}} \]
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