Unpolarized light with intensity I 0 is incident on an ideal polarizing filter.

Question

Unpolarized light with intensity I0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at 42.0 to that of the first.

Determine the intensity of the beam after it has passed through the second polarizer.

Enter the factor only. For example, enter 0.250, that means your answer is I=0.250I0.

Determine its state of polarization.

The light is linearly polarized along the axis of the first filter. The light is linearly polarized along the axis of the second filter. The light is linearly polarized perpendicular to the axis of the second filter. The light is linearly polarized perpendicular to the axis of the first filter.

Explanation / Answer

50% of unpolarised light passes through the first filter because, on average, 50% of the waves are aligned with the fiter's axis. Intensity is reduced by a factor 0.5.

The second filter then reduces the intensity by a factor cos²() (Law of Malus).

cos²(42.0°) = (0.74314)² = 0.5523

So overall, the intensity is reduced by a factor 0.5 x 0.5523 = 0.276

I = 0.276 I

And

it would still be vertically polarized (same as the first filter) because the second filter does not rotate the light, it just blocks some of it

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