Find force exerted by the beam on sphere if it is perfectly reflecting
Answers
Explanation :
Let O be centre of sphere and OZ be line opposite to incident beam . Consider a radius OP of sphere making an angle θ with Oz to get aciroll on sphere. Charge θtoθ+dθ and rotate radius about OZ to get another circle on sphere . The part of sphere between these circles is a ring of area 2πr2sinθdθ. Consider a small part △A of this right at P . Energy of light falling on this part in time △t is
△u=I△t(△Acosθ)
The momentum of this light falling on △A is △u along QP. The light e is reflected by sphere along PR .Therefore in momentum is
△P=2△uCcosθ=2CI△t(△Acos2θ) (direction along OP−→−)
The force on △A due to light falling on it , is
△P△=2CI△Acos2θ (directed along PO−→−)
The resultant force on ring as well as an sphere is along ZO by force on △A along ZO.
△P△tcosθ=2CI△Acos3θ (along ZO−→−)
The force acting on ring is
dF=2CI(2πr2sinθdθ)cos3θ
The force on entire sphere is
F=∫π204πr2ICcos3θsinθdθ
=−∫π204πr2ICcos3θd(cosθ)
=−∫π204πr2IC[cos4θ4]π20
=πr2IC
Vote that integration is done only for hemisphere that faces the incident beam.