Question

A point source of light is placed at the centre of curvature of a hemispherical surface. The radius of curvature is \(r\), and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it, if the source emits a power \(W\), is:

• A. \(Wc\)
• B. \(\dfrac{W}{c}\)
• C. \(\dfrac{W}{2c}\)
• D. \(\dfrac{2W}{c}\)

Hint:

Normal-incidence reflection gives pressure \(2I/c\); integrate the axial component \(\cos\theta\) over the hemisphere.

Answer 1

The Correct Option is C

Step 1: The source is isotropic, so the intensity at any point on the hemisphere (all at distance \(r\)) is \(I = \dfrac{W}{4\pi r^2}\). Because every ray leaves the centre radially, it strikes the spherical surface at normal incidence.

Step 2: For a perfectly reflecting surface at normal incidence, the radiation pressure is \(P = \dfrac{2I}{c}\), directed along the outward normal (the radial direction).

Step 3: By symmetry only the component along the axis of the hemisphere survives. Taking a surface element \(dA = r^2\sin\theta\, d\theta\, d\phi\) at polar angle \(\theta\), its axial force is \(dF_z = \dfrac{2I}{c}\cos\theta\, dA\).

Step 4: Integrate over the hemisphere (\(0\le\theta\le\tfrac{\pi}{2}\), \(0\le\phi\le 2\pi\)):
\[F = \frac{2}{c}\cdot\frac{W}{4\pi r^2}\int_0^{2\pi}\!\!d\phi\int_0^{\pi/2}\!\!\cos\theta\sin\theta\, r^2\, d\theta\]
\[F = \frac{2W}{4\pi c}\,(2\pi)\left(\tfrac{1}{2}\right) = \frac{W}{2c}\]

Step 5: Hence the force on the hemisphere is
\[\boxed{F = \dfrac{W}{2c}}\]