Unit 2 Electromagnetic Waves
2.4 Energy and Momentum in Electromagnetic Waves
2.4.2 Radiation pressure
Electromagnetic waves carry energy, therefore they possess linear momentum. In a 'collision' between an electromagnetic wave and a surface, momentum wiil be transferred in a similar manner to an moving object colliding with a surface.The force The force per unit area exerted on the surface as a result of the collision by the EMR is called radiation pressure. The existence of radiation pressure was predicted by Maxwell and it is evidence of the waveparticle duality.
He showed that if a beam of EMR is completely absorbed, then
From Newton's second law Force = ΔP/Δt
Work done or energy expended in stopping the EMR (ΔU) in unit time = F X distance moved in unit time,
ΔU = Fs where s = vΔt
ΔU = FvΔt
ΔU = F.vΔt but F = ΔP/Δt and v = c (speed of light)
ΔU = [^{ΔP}/_{Δt}].cΔt
ΔU = ΔP.c
ΔP = ^{ΔU}/_{c}
If the beam is reflected then the force required to change the velocity from c to c is doubled (Force required to change c to 0 + Force required to change 0 to c)
ΔP = ^{2ΔU}/_{c}
Radiation pressure P_{rad} = ^{Force}/_{Area}
P_{rad} =^{[ΔP/Δt]}/_{A}
P_{rad} =^{[ΔU/c.Δt]}/_{A}
P_{rad} =^{[ΔU]}/_{[c.AΔt]}
If we define I_{avg} as as energy per unit time per unit area, then
P_{rad} = ^{[Iavg]}/_{c} for total absorption and
P_{rad} = ^{[2Iavg]}/_{c} for total reflection
So we can generalize into P_{rad} = [kI_{avg}]_{/c} where 1 >= k <=2. k is 1 for perfect absorption and 2 for perfect reflection.
Example 1:
Radiation from a bright lamp falls on your face at a rate of about 1000 Wm^{2}. Estimate the radiation pressure and force exerted
by the the lamp on your face.
P_{rad} = [kI_{avg}]_{/c} where 1 >= k <=2.
3 X 10^{6} Wm^{6} >= [P_{rad}] <=6 X 10^{6} Wm^{6}
If the face area is about 0.08m^{2} then
Force = P_{rad} X Area
3 X 10^{7} N >= Force <=6 X 10^{7} N
