What is the intensity of the transmitted light?

Question

Vertically polarized light with an intensity of 0.60 W/m² passes through a polarizer whose transmission axis is at an angle of 45 degrees  with the vertical.

What is the intensity of the transmitted light?

Answer in W/m²

Answer 1

Use Malus's Law  I = I_o cos²θ where θ is the angle between the polarization direction of the light and the transmission axis of the polarizer

so I = 0.06•cos²45° (this will yield an answer in the required units since I₀ is in the same units.)

Answer 2

There are two rules that govern all questions of this sort. The cases deal in whether 1) the incident light is unpolarized or 2) the incident light is polarized.

1) Unpolarized light that is incident upon a polarizing sheet will leave the sheet with half its initial intensity, and the polarization angle of the sheet. In other words, a single sheet turns unpolarized into weaker, polarized light with intensity reduced by half. e.g. I = (1/2) Io.

2) If instead the light incident on a polarizing sheet is initially polarized, it will still obtain the polarization angle of the final sheet, but reduce its intensity reduced by a factor equal to the square of the cosine of the difference between the angles available to you - that would be the angle ∆Θ between the polarization of the incoming light and the sheet: it strikes e.g., I = cos^2(∆Θ).

So - in our case, the first sheet reduces the intensity by 1/2: e.g. the intensity leaving polarizer 1 is I = 1/2 Io = .3 W/m^2. This is because of rule 1: unpolarized light always leaves a polarizer with half the incoming intensity. Its a rule, just memorize it.

The question doesn't mention any subsequent polarizing sheets, so we are done. The transmitted intensity is I=(1/2)*Io = 0.3 W/m^2.

Note: Another way to summarizer the simple rules that govern this sort of problem are:

1') light entering a polarizer will obtain the polarization angle of the polarizer -> light leaves with new polarization determined by sheet and

2') The intensity of light incident on a polarizer is reduced by the factor

a) 1/2 if the indicent light is not polarized and

b) the square of the cosine of the angular difference Θ2-Θ1 between the incident and transmited polarization angles.

TO further simplify, if you can even consider two sheets, the intensity drops due to the angle between the sheets. If you have no such angle available to you, the intensity is just reduced to half.