Two ideal polarizing filters are oriented so that they transmit the maximum amou

Question

Two ideal polarizing filters are oriented so that they transmit the maximum amount of light when unpolarized light is shone on them.

a. To what fraction of its maximum value is the intensity of the transmitted light reduced when the second filter is rotated through 24.0 degrees?

b.To what fraction of its maximum value is the intensity of the transmitted light reduced when the second filter is rotated through 43.0 degrees?

c.To what fraction of its maximum value is the intensity of the transmitted light reduced when the second filter is rotated through 64.5 degrees?

Explanation / Answer

When dealing with polarizers, just do one at a time, in the order that light will hit it. When you put down the first polarizer, the intensity of unpolarized light is cut by 1/2. When you have two polarizers in a row, the light becomes polarized after leaving the first one, so you have to use that equation of I = I_0*cos^2(theta) where theta is the angle between the polarization directions, and I_0 is the initial intensity. ASCII art time: I_0 ---> | I_0/2 ---> / I_0/2*cos^2(theta) Unpolarized light is incident on a vertically polarized sheet. Its intensity is halved. Then it's incident on a polarized sheet that is tilted by some angle theta. Then it's intensity changes again by a factor of cos^2(theta). Note that if you have a horizontal and vertical polarizer next to each other, the intensity of light is zero, because cos(90 degrees) is zero. Also note that cos^2(theta) = 1 is a maximum, which occurs when theta = 0. In other words, if the two polarizers are oriented in the same direction, the second one won't affect the intensity at all!

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