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Unpolarised light of intensity I0 falls on a system of three polarising filters. The first and the third filters are kept crossed (their transmission axes at 90 deg to each other). A third (middle) filter is placed between them, initially aligned with the first filter, and is then rotated by an angle theta. Which expression gives the intensity of light emerging from the final (third) filter as a function of theta?

1. I = (I0/8) * sin²(2*theta)
2. I = (I0/2) * cos²(theta)
3. I = (I0/4) * sin²(theta)
4. I = (I0/8) * cos²(2*theta)

Correct answer: I = (I0/8) * sin²(2*theta)

Solution

After the first polariser, unpolarised light becomes polarised with intensity I0/2. The middle filter makes angle theta with this, so by Malus's law it transmits (I0/2)*cos²(theta), now polarised along the middle filter's axis. The final filter is crossed with the first (90 deg), so it makes angle (90 deg - theta) with the middle filter; it transmits a further factor cos²(90 deg - theta) = sin²(theta). The product (I0/2)*cos²(theta)*sin²(theta) simplifies using sin(theta)*cos(theta) = (1/2) sin(2*theta) to (I0/8)*sin²(2*theta). The intensity is zero at theta = 0 and 90 deg (filters crossed) and maximum I0/8 at theta = 45 deg.

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