A perfectly reflecting solid hemisphere of radius R is placed in the path of a parallel beam of light of large aperture. If the beam carrries an intensity l, find the force exerted by the beam on the hemisphere.

Explanation

Let O be the centre of the sphere and OZ be the line opposite to the incident beam ( figure). Consider a radius about OZ to get amaking an angle `theta `with OZ. Rotate this radius about OZ to get a circle on the sphere. Change `theta ` to `theta + d theta` and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area `2 pi r^(2) sin theta d theta`. Consider a small part `Delta A` of this ring at P.Energyof light falling on this part in time `Delta t ` is

The momentum of this light falling on `Delta A` is `DeltaU //c` along QP.The light is reflected by the sphere along PR. The change in momentum is
`Delta p = 2 "" ( Delta U )/( c ) cos theta = ( 2)/( c ) Delta ( Delta A cos^(2) theta )` ( direction along `vec( O)P)`
The force on `Delta A ` due to the light falling on it, is
`( Delta p )/( Delta t ) = ( 2)/( c ) Delta A cos^(2)theta` ( direction along `vec( O) P)`
The resultant force on the ring as well as on the sphere is along ZO by symmetry . The component of the force on `Delta A` along ZO
`( Delta p )/( Delta t ) cos theta = ( 2) /( c ) I Delta cos^(2) theta` ( along `( vec( Z ) O)`
The force acting on the ring is `dF = ( 2)/( c) I ( 2pi r^(2) sin theta d theta ) cos^(3) theta `
The force on the entire sphere is `F= int_(0)^(pi//2) ( 4pi r^(2) I)/( c ) cos^(3) theta d theta`
`F= int_(0)^(pi//2) ( 4pi r^(2) I)/( c ) cos^(3) theta d ( cos theta )`
`= - int_(0-0)^(pi//2) ( 4 pi r^(2) I)/( c ) [ ( cos^(4) theta)/( 4)]_(0)^(pi//2)= ( pi r^(2) I )/( c )`
Note that integration is done only for the hemisphere that faces the incident beam.