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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. |
Answer» Solution : 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 momentumis ` (Delta p = 2 (Delta U/ c ) cos theta = (2/c) I Delta t (Delta A (cos^2) theta)).` along the inward normal. Theforce 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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