A beam of unpolarised light is transmitted through 4 polarising sheets. The angle between each successive sheet is 37.1 degrees. What percentage of the original beam intensity is transmitted through the final sheet
Question
A beam of unpolarised light is transmitted through 4 polarising sheets. The angle between each successive sheet is 37.1 degrees. What percentage of the original beam intensity is transmitted through the final sheet
Solution
To solve this problem, we need to use Malus's law, which states that when completely polarized light is passed through a polarizer, the intensity I of the light that comes out is given by I = I0 * cos^2(θ), where I0 is the initial intensity and θ is the angle between the light's initial polarization direction and the axis of the polarizer.
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The first polarizer will polarize the light, but it will also cut the intensity in half. So after the first polarizer, the intensity is 0.5 * I0.
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The second polarizer is at an angle of 37.1 degrees to the first. So the intensity after the second polarizer is 0.5 * I0 * cos^2(37.1 degrees).
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The third polarizer is at an angle of 37.1 degrees to the second (and 74.2 degrees to the first). So the intensity after the third polarizer is 0.5 * I0 * cos^2(37.1 degrees) * cos^2(37.1 degrees).
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The fourth polarizer is at an angle of 37.1 degrees to the third (and 111.3 degrees to the first). So the intensity after the fourth polarizer is 0.5 * I0 * cos^2(37.1 degrees) * cos^2(37.1 degrees) * cos^2(37.1 degrees).
Now we just need to calculate these values. The cosine of 37.1 degrees is approximately 0.7986, so cos^2(37.1 degrees) is approximately 0.6376.
So the final intensity is approximately 0.5 * I0 * 0.6376 * 0.6376 * 0.6376 = 0.102 * I0.
So the percentage of the original beam intensity that is transmitted through the final sheet is approximately 10.2%.
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