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crack_iit

HCV-2 PAGE NO.365 Q.NO.7

a beam of white light is incident normally on a plane surface absorbing 70%of the light & reflecting the rest.if the incident beam's power=10w,find force exerted by it on surface?

what is the general method to do such kind of problems?

iitkgp_bipin

Let the energy emitted per second = E

Momentum of photons emitted = E/c

70% of photons are absorbed, initial momentum = 0.7E/c and final momentum of photons = 0
Change in momentum of surface = 0.7E/c - 0 = 0.7E/c

30% of photons are reflected, initial momentum = 0.3E/c and after reflecting momentum will be -0.3E/c
so change in momentum of surface = 0.3E/c-(-0.3E/c) = 0.6E/c

Net change in momentum = 0.7E/c + 0.6E/c = 1.3E/c

Force = Rate of change of momentum = d(1.3E/c)/dt

= (1.3/c)(dE/dt)

= 1.3P/c (since dE/dt is the rate of change of energy which is power)

jus_look

let I=INTENSITY
E=ENERGY POSESSED BY PHOTON
A=PERPENDICULAR AREA OF INTERACTION..
t=TIME
p=MOMENTUM OF PHOTON....P=POWER OF BEAM
C=SPEED OF LIGHT
x=%of light absorbed
f=force
I=dE/(Adt)
>dE/dt=IA
NOW....E=pc
>dE/dt=cdp/dt( c is const)

case 1:complete absorb
f1=xdp/dt=dE/(cdt)=IA/c

case2:complete reflect
f2=(100-x)*2*dp/dt.........(*2 bcoz of momentum conservation)
=2IA/C
now total force exerted....F=f1+f2
={x(IA/C)+2(100-x)IA/C}/100

SO GENERAL FORMULA......
F=P/C{[200-x]/100} here P=IA/C=POWER OF BEAM....NOT MOMRNTUM...

kmmankad

Power of the Radiation=Force Exerted by the radiation x Velocity of Radiation

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