Unpolarized light of intensity i_0 is incident on a series of three polarizing f

Question

Unpolarized light of intensity i_0 is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45 degree to that of the first filter, while the axis of the third filter is oriented at 90 degree to that of the first filter. What is the intensity of the light transmitted through the third filter? Vertically polarized light with an intensity of 500W/m^2 passes through a polarizer oriented at an angle to the vertical. Find the transmitted intensity of the light for (a) 15 degree (b) 25 degree, (c) 90 degree

Explanation / Answer

1. As unpolarized light passes through polarizer, intensity get halved and light becomes polarizer.

I1 = Io / 2

after that If = I ( cos(theta))^2

after second, I2 = I1 (cos45)^2 = 0.5 Io (1/2)

I2 = 0.25 Io

after third, If = I2 (cos(90 - 45))^2 = (0.25 Io) (cos45)^2

If = 0.125 Io .........ANs

2. I = Io (cos15)^2 (cos(25 - 15))^2 cos(90 - 25) ^2

I = (500) (cos15 cos10 cos65)^2

I = 80.8 W/m^2

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