Radiation Pressure (Modern Physics)
im getting 2IRH/3c
im on PC so i cant take a pic but heres what i did :
divide the cone into disks of height dh. take a disk of radius r. consider a small area on it at angle theta as rd(theta) * dh. consider over a time diff dt on this momentum change = IAcos^2(theta)/c * dt
dP/dt = F = IA cos^2(theta)/c
now take component along direction of radiation cuz vectors will cancel and only the one in the direction of Incident ray remain.
dF(net) = IA cos^3(theta)/c
A = rdtheta * dh
integrate wrt theta from -pi/2 to pi/2
F (for small height dh at specific r) = 4Ir/3c * dh
r/h = R/H => r = Rh/H
F = 4IR/3Hc * h dh
integrate wrt h from 0 to H
F = 4IR/3Hc * H^2/2 = 2IRH/3c
what step is wrong and why
im on PC so i cant take a pic but heres what i did :
divide the cone into disks of height dh. take a disk of radius r. consider a small area on it at angle theta as rd(theta) * dh. consider over a time diff dt on this momentum change = IAcos^2(theta)/c * dt
dP/dt = F = IA cos^2(theta)/c
now take component along direction of radiation cuz vectors will cancel and only the one in the direction of Incident ray remain.
dF(net) = IA cos^3(theta)/c
A = rdtheta * dh
integrate wrt theta from -pi/2 to pi/2
F (for small height dh at specific r) = 4Ir/3c * dh
r/h = R/H => r = Rh/H
F = 4IR/3Hc * h dh
integrate wrt h from 0 to H
F = 4IR/3Hc * H^2/2 = 2IRH/3c
what step is wrong and why
