Question

Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room or a car. A library atrium has an overhead skylight that lets in too much light during the day which heats up the interior of the library far too much. The building engineer installs new double paned polarizing sky lights to reduce the intensity. If sunlight, which is unpolarized, has an average intensity of 1340 W/m² what angle should the polarizing axis of the second pane of the window make with the polarizing axis of the first pane of the window in order to reduce the intensity of the sunlight to 29% of the original value?

Answer 1

we know that

I = Io x cos^2(A)

where I = Io x 29/100 = 0.29 x Io

or cos^2(A) = 0.29

or cosA = (0.29)^1/2

or A = cos^-1((0.29)^1/2) = 57.4o

Answer 2

If original intensity is Io,

then the intensity I after polarizer is given by

I = Io * [cos(theta)]^2

According to question, I = 0.29*Io

0.29*Io = Io * [cos(theta)]^2

cos(theta) = 0.5385

theta = 57.42 degrees