Question

Unpolarized light of intensity 32Wm⁻² passes through three polarizers such that transmission axes of the first and second polarizer makes and angle 30° with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of final emerging light will be(a) 32 Wm⁻²
(b) 3 Wm⁻²
(c) 8 Wm⁻²(d) 4 Wm⁻²

Answer 1

Since the light is unpolarized, after crossing the first polarizer the intensity will be 322=16W/m2.
Now if the second polarizer is at an angle θ with first polarizer then the output from the second polarizer will be I^2=I1cos^2θ=16cos^2θ.
Now the angle between the second and third polarizer will be 90°−θ.
Now After passing through the third polarizer the intensity be will I3=I2cos^2(90−θ)=16 cos^2θ sin^2θ

Now we have,
16 cos^2θ sin^2θ = 3
16cos^2θ(1−cos^2θ)=3
16cos^2θ−16cos^4θ=3
16cos^4θ−16cos^2θ+3=0
16x2−16x+3=0 (substituting x=cos^2θ)
x=1/4, 3/4  Hence cos^2θ=1/4 or 3/4
cosθ=1/2 or √3/2,
θ=60° or 30°
So the angle can be either 60° or 30°.

Cos theta = 1/2
cos^2 theta = 1/4

Now since the intensity of light was 16 W/m^2

So, the new intensity will become = I1^2 * (1/4) = 4 W/m^2
and if we take cos theta = √3/2
then similarly,
I2 = 3/4*16 = 12 W/m^2

Hence the answer here will be 4 W/m^2 since we can't the option for the other possible answer that is 12 W/m^2