When unpolarized light passes through a polarizer, the intensity is reduced by a factor of ½. ... When polarized light of intensity I0 is incident on a polarizer, the transmitted intensity is given by I = I0cos2θ, where θ is the angle between the polarization direction of the incident light and the axis of the filter.
because each polarizer allows 1/2 cos^2Theta where theta is some angle from the vertical. So if Theta for P1 is zero, then
Itrans=1/2 Io
Then, the light getting thru P2 is.5Iocos^2Theta, and finally, the light getting thru the final P3 is .5 Io (cos^2Theta *cos^2(90-Theta)
If Io is 1.0, then
.5(cosTheta*sin(90-Theta)=3/32
cosTheta * (sin90sinTheta-cos90sinTheta)=6/32
but cos90=0 so
sinTheta*cosTheta=1/2 sin 2*theta=6/32
sin Theta= sqrt (12/32)=.612
Theta=37 deg
Three polarizers are placed in succession with their axes of polarization as follows: P1 vertical, P2 at angle theta to the vertical, and P3 at 90 degrees to the vertical. Unpolarized light of intensity I.0 is incident on the first polarizer. The final intensity after the passing through the 3 polarizers is measured to be (3/32)I.0
Determine the angle theta between the first polarizer P1 and second polarizer P2.
1 answer
1 answer