Notice that a polarizer modifies the light intensity according to Malus\'s law a

Question

Notice that a polarizer modifies the light intensity according to Malus's law and also reorients the polarization angle of the beam to match its own transmission axis. Hence it is possible for light to pass through a pair of crossed polarizers if a third polarizer is inserted between them with an intermediate transmission axis direction. What is the new intensity I2 of the light emerging from the final polarizer in Part E if a third polarizer (Polarizer A in the figure (Figure 5) ), whose transmission axis is offset 45 degrees from each of the others, is inserted between the original two polarizers?

Express your answer as a decimal number times the symbol I0 . For example, if I2=(1/4)I0 , enter 0.25 * I0

Explanation / Answer

  Malus' Law states:

I = Io*cos^2(Theta); where Io is the intensity of the incoming light, I is the intensity of the light after the polariser and Theta is the angular difference between the polarisation axis of the incoming light and the axis of the polariser.

So as long as the angular difference between the incoming (polarised) light and the polariser axis is not 90 or 270 degrees, some light is able to pass through. In your case lets assume the incoming light is unpolarised. That means:

I = Io*cos^2(all angles from 0 to 360).
and cos^2(all angles from 0 to 360) = 0.5
thus the amount of light going through the first polariser is 0.5I or 50% of the initial light.

The second polariser has been placed at 45 degrees to the incident light so:
I = Io*cos^2(45)
I = 0.5Io*0.5
I = 0.25Io or 25% of the initial light

The third polariser has been placed another 45 degrees with respect to the incident light so (same as above):
I = Io*cos^2(45)
I = 0.25Io*0.5
I = 0.125Io or 12.5% of the initial light

So the final answer is 12.5% of the original light passes through the final polariser, even though the polariser array contains two polarisers with their axes perpendicular to one another.

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