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A perfectly reflecting solid sphere of r...

A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, find the force exerted by the beam on the sphere.

Text Solution

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Let O be the centre of the sphere and OZ be the line
opposite to the incident beam. 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 (`2 pi (r^2) sin theta d theta)`.
Consider a small part `Delta A` of this ring at P. Energy of
the light falling on this part in time `(Delta t )` is
` Delta U= I (Delta t) (Delta A cos theta).`
` The momentum of this light falling on Delta A is (Delta U/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) I Delta t (Delta A (cos^2) theta)).`
along the inward normal. The force on (Delta A ) due to the
light falling on it, is
` (Delta p/ Delta t ) = 2/c I (Delta A(cos^2)theta)`
This force is along PO, 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 is.
` (Delta p/ Delta t) cos theta = 2/c (I (Delta A)cos ^3) theta. `
` The force acting on the ring is ,
` dF= (2/c) I (2 pi (r^2)(sin theta) d theta) cos^3 theta.`
The force on the entire sphere is
` F= (int_(0)^(pi/2) (4 pi (r^2) I/c) (cos^3)theta sin theta dtheta)`
` = - (int_(theta =0)^(pi/2)(4 pi (r^2)I/c) cos^3 theta d (cos theta). `
` = - (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.
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