What is the force exerted by a Light Beam On a Surface ?

Explanation

Case I- `a=1, r=0`
Initial momentum of photon ( in downward direction at an angle `theta` with vertical)`=(h)/(lambda)`

final momentum of photon `=0`
change in momentum ( in upward direction at an angle `theta` with vertical) `=(h)/(lambda)`

energy incident per unit time `= IA cos theta`
Intensity`=`power per unit normal area
`I=(P)/(A cos theta), P=IA cos theta`
No.of photons incident per unit time `=(IA cos theta)/(hc). lambda`
total change in momentum per unit time ( in upward direction at an angel `theta` with vertical)
`-(IAcos thetalambda)/(hc).(h)/(lambda)=(IA cos theta)/(c )`

Force `(F)=` total change in momentum per unit time
`F=(IA cos theta)/(c )` ( direaction on photon and on the plate)
Pressure`=`normal force per unit Area
`"Pressure"=(F cos theta)/(A) P(IA cos^(2)theta)/(cA)=(I)/(c )cos^(2) theta`
CaseII When `r=1,a=0`
`:.` change in momentum of one photon
`=(2h)/(lambda) cos theta` ( upward)

No.of photons incident per unit time
`=("energy incident per unit time")/(hv)`
`=(IA cos theta.lambda)/(hc)`
`:.` total change in momentum per unit time
`=(IA cos theta.lambda)/(hc)xx(2h)/(lambda)cos theta =(2IAcos^(2)theta)/(c )`(upward)
`:.` force on the plate`=(2IA cos ^(2)theta)/(c )` (downward)
Pressure `=(2IAcos^(2)theta)/(cA) P=(2Icos^(2)theta)/(c )`
Case III `0 lt r lt 1, a+r=1`
change in momentum of photon when it is reflected `=(2h)/(lambda) cos theta` (downward)
change in momentum of photon when it is absorbed `=(h)/(lambda)` ( in the opposite direction of incident beam)
energy incident per unit time `IA cos theta`
no. of photns incident per unit time `=(IA cos theta.lambda)/(hc)`
no. of reflected photon `(n_(r ))=(IA cos theta.lambda)/(hc)`
no.of reflected photon`(n_(r ))=(IA cos theta lambda)/(hc) (1-r)`
force on plate due to absorbed photons `F_(a)=n_(a).DeltaP_(a)`
`=(IA cos theta lambda)/(hc) (1-r)(h)/(lambda)`
`=(IA cos theta)/(c )(1-r)` ( at an angle `theta` with vertical)
force on plate due to reflected photons `F_(r )=n_(r )DeltaP_(r )`
`=(IA cos theta.lambda)/(hc)xx(2h)/(lambda)cos theta`(vertically downward)
`=(IAcos^(2) theta)/(c ).2r`
now resultant force is given by `F_(R )=sqrt((1-r)^(2)+(2r)^(2)cos^(2)theta+4r(r-1)cos^(2)theta)`
and pressure, `P=(F_(a)cos theta+F_(r ))/(A)`
`=(IA cos theta(1-r)cos theta)/(cA)+(IA cos^(2)theta.2r)/(cA)`
`=(Icos ^(2)theta)/(c )(1-r)+(Icos^(2)theta)/(c )=(Icos^(2)theta)/(c )(1+r)`