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 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
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For a fully reflecting surface, radiation force is twice that for complete absorption because the light reverses its momentum after reflection.
Updated On: Jun 15, 2026
- \(2.3 \times 10^{-15} \, N\)
- \(1.33 \times 10^{-14} \, N\)
- \(6.67 \times 10^{-15} \, N\)
- \(9.4 \times 10^{-15} \, N\)
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The Correct Option is D
Solution and Explanation
Step 1: Identify the given quantities.
Intensity of light is
\[ I=10^{-3} \, Wm^{-2} \] Area of cross-section is
\[ A=20 \, cm^2 \] Convert area into square metre,
\[ A=20\times10^{-4}=2\times10^{-3} \, m^2 \] Angle of incidence is
\[ \theta=45^\circ \] Speed of light is
\[ c=3\times10^8 \, ms^{-1} \]
Step 2: Use force formula for a fully reflecting surface.
For a fully reflective surface, force due to radiation pressure is given by
\[ F=\frac{2IA\cos\theta}{c} \]
Step 3: Substitute the values.
\[ F=\frac{2\times10^{-3}\times2\times10^{-3}\times\cos45^\circ}{3\times10^8} \] Since,
\[ \cos45^\circ=\frac{1}{\sqrt{2}} \] \[ F=\frac{4\times10^{-6}\times \frac{1}{\sqrt{2}}}{3\times10^8} \] \[ F=\frac{2.828\times10^{-6}}{3\times10^8} \] \[ F=9.4\times10^{-15} \, N \]
Step 4: Final conclusion.
Hence, the force exerted by the beam on the surface is
\[ \boxed{9.4\times10^{-15} \, N} \]
Intensity of light is
\[ I=10^{-3} \, Wm^{-2} \] Area of cross-section is
\[ A=20 \, cm^2 \] Convert area into square metre,
\[ A=20\times10^{-4}=2\times10^{-3} \, m^2 \] Angle of incidence is
\[ \theta=45^\circ \] Speed of light is
\[ c=3\times10^8 \, ms^{-1} \]
Step 2: Use force formula for a fully reflecting surface.
For a fully reflective surface, force due to radiation pressure is given by
\[ F=\frac{2IA\cos\theta}{c} \]
Step 3: Substitute the values.
\[ F=\frac{2\times10^{-3}\times2\times10^{-3}\times\cos45^\circ}{3\times10^8} \] Since,
\[ \cos45^\circ=\frac{1}{\sqrt{2}} \] \[ F=\frac{4\times10^{-6}\times \frac{1}{\sqrt{2}}}{3\times10^8} \] \[ F=\frac{2.828\times10^{-6}}{3\times10^8} \] \[ F=9.4\times10^{-15} \, N \]
Step 4: Final conclusion.
Hence, the force exerted by the beam on the surface is
\[ \boxed{9.4\times10^{-15} \, N} \]
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