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. Find the force on the hemisphere due to the light falling on its if the source emits a power W.

Explanation

The energy emitted by the source per unit time, i.e., W
falls on an area ` 4pi (r^2)` at a distance r in unit time. Thus,
the energy falling per

unit area per unit time is
`(W/(4pi (r^2))).`
Consider a small area dA at the point P of
the hemisphere . The energy falling per
unit time on it is `(WdA/ (4 pi(r^2)))`. The corresponding momentum
incident on this area per unit time is ` (WdA/(4pi(r^2)c)). As the light
is reflected back, the change in momentum per unit
time, i.e., the force on dA is
(dF= (2 Wd A/ (4 pi (r^2) c)). `
Suppose the radius OP through the area dA makes an
angle theta with the symmetry axis OX. The force on dA is
along this radius. By symmetry, the resultant force on
the hemisphere is along OX. The component of dF along
OX is
` dF cos theta= (2 WdA/ (4 pi (r^2)c)) cos theta.`
If we project the area dA on the plane containing the
rim, the projectin is dA cos theta. Thus, the component of
dF along OX is,
` dF cos theta = (2W/ (4pi (r^2)c))(projection of dA).`
The net force along OX
` F= (2W/ (4pi(r^2)c))(sum projection of dA).`
When all the small areas dA are projected, we get the
area enclosed by the rim which is (`pi (r^2). Thus,
` (F= (2W/ (4pi (r^2)c)) xx pi (r^2)= (W/2c).)`