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 \((c=3\times10^8\,\text{m s}^{-1})\)

• A. \(20\times10^4\,\text{W m}^{-2}\)
• B. \(15\times10^4\,\text{W m}^{-2}\)
• C. \(25\times10^4\,\text{W m}^{-2}\)
• D. \(10\times10^4\,\text{W m}^{-2}\)

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.

Answer 1

The Correct Option is A

Step 1: Use radiation pressure for perfect absorber.
P_r = I/c = F/A → I = Fc/A.

Step 2: Convert area to SI.
A = 15 cm² = 1.5×10⁻³ m².

Step 3: Compute intensity.
I = (10⁻⁶ × 3×10⁸)/(1.5×10⁻³) = 300/(1.5×10⁻³) = 2×10⁵ W/m² = 20×10⁴ W/m².

Step 4: Final Answer:
20×10⁴ W/m².