In the path of a uniform light beam of large cross-sectional area and intensity I. a solid sphere of redius R which is perfectly reflecting is placed . Find the force exerted on this sphere due to the light beam.

Explanation

Let `O` be the centre of the sphere and `OZ` be the line opposite to the incident beam (figure). Consider a radius `OP` of the sphere making 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 `2pir^(2) sin theta d theta`. Consider a small part `DeltaA` of this ring at `P`. Energy of the light falling on this part in time `Delta t` is
`DeltaU=IDeltat(DeltaA cos theta)`

The momentum of this light falling on `DeltaU//c` along `QP`. The light is reflected by the sphere along `PR`. The change in momentum is
`DeltaP=2(DeltaU)/(c)cos theta=(2)/(c)I Deltat(DeltaA cos^(2)theta)` (direction along `vec(OP)`)
The force on `DeltaA` due to the light faling on it, is
`(DeltaP)/(Deltat)=(2)/(c )IDeltaA cos^(2)theta`. ( direction along `vec(PO)`)
The resultant force on the ring as well as on the sphere is along `ZO` by symmetry. The component of the force on `DeltaA` along `ZO`
`(Deltarho)/(Deltat)cos =(2)/(c )IDeltaA cos^(3)theta`. (along `vec(ZO)`)
The force acting on the ring is `df=(2)/(c )I(2 pir^(2) sin theta d theta) cos^(3) theta`
The force on the entire sphere is `F=int_(0)^(pi//2)(4pir^(2)l)/(c )cos^(3) theta sin theta d theta`
`=- underset(0)overset(pi//2)int(4pir^(2)l)/(c )cos ^(3) theta d(cos theta)=underset(theta=0)overset(pi//2)int(4pir^(2)l)/(c )[(cos^(4)theta)/(4)]^(pi//2)=(pir^(2)I)/(c )`
Note that intergraion is done only for the hemisphere that faces the incident beam.