Question

A beam of light with intensity \(10^{-3} \, Nm^{-2}\) and cross sectional area \(20 \, cm^2\) is incident on a fully reflective surface at angle \(45^\circ\). Then the force exerted by the beam on the surface is

• A. \(2.3 \times 10^{-15} \, N\)
• B. \(1.33 \times 10^{-14} \, N\)
• C. \(6.67 \times 10^{-15} \, N\)
• D. \(9.4 \times 10^{-15} \, N\)

Hint:

For a fully reflecting surface, radiation force is twice that for complete absorption because the light reverses its momentum after reflection.

Answer 1

The Correct Option is D

Step 1: Note the data.
Intensity $I = 10^{-3}\ \text{W m}^{-2}$, area $A = 20\ \text{cm}^2 = 2\times10^{-3}\ \text{m}^2$, incidence angle $\theta = 45^\circ$, and $c = 3\times10^{8}\ \text{m s}^{-1}$.
Step 2: Force on a perfect reflector.
A fully reflecting surface returns the photons, so the momentum transfer is doubled. The radiation force is \[ F = \frac{2IA\cos\theta}{c}. \]
Step 3: Combine $I$ and $A$.
\[ IA = (10^{-3})(2\times10^{-3}) = 2\times10^{-6}\ \text{W}. \]
Step 4: Apply the doubling and the angle.
With $\cos45^\circ = \dfrac{1}{\sqrt2}$, \[ 2IA\cos\theta = 2(2\times10^{-6})\frac{1}{\sqrt2} = \frac{4\times10^{-6}}{\sqrt2} = 2.828\times10^{-6}. \]
Step 5: Divide by $c$.
\[ F = \frac{2.828\times10^{-6}}{3\times10^{8}} = 9.4\times10^{-15}\ \text{N}. \]
Step 6: Conclude.
The beam pushes on the surface with a force of about $9.4\times10^{-15}\ \text{N}$.
\[ \boxed{9.4\times10^{-15}\ \text{N}} \]