%3Cp%3EIn%20the%20figure%2C%20initially%20unpolarized%20light%20is%20sent%20into
ID: 2296171 • Letter: #
Question
%3Cp%3EIn%20the%20figure%2C%20initially%20unpolarized%20light%20is%20sent%20into%20a%20system%0Aof%20three%20polarizing%20sheets%20whose%20polarizing%20directions%20make%20angles%0Aof%20%3F1%20%3D%20%3F2%20%3D%20%3F3%20%3D%2038%C2%B0%20with%20the%20direction%20of%20the%20y%20axis.%20What%0Apercentage%20of%20the%20initial%20intensity%20is%20transmitted%20by%20the%20system%3F%0A(Hint%3A%20Be%20careful%20with%20the%20angles.)%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E%3Cimg%20src%3D%0A%22http%3A%2F%2Fedugen.wileyplus.com%2Fedugen%2Fcourses%2Fcrs4957%2Fart%2Fqb%2Fqu%2Fch0%2Fimg1337192962815_7418372303131279.gif%22%20%2F%3E%3C%2Fp%3E%0AExplanation / Answer
when unpolarized light hits a polarizer, the inensity on the output is about 1/2 of the intensity on the input. So if the first intensity is 'I', after the first polarizer, the intesity will be I_1 = 0.5*I.
When polarized light hist a polarizer at a different angle the new intensity is given by:
I_new = I_orig*(cos theta)^2 where theta is the angle difference between the between their axes.
Hence after the second and future polarizers, the intesity would be:
I_2 = I_1*(cos ( 38- 38))^2
I_3 = I_2*(cos (38- 38))^2
When you do all the math and multiplication you should end up the I_3 = 0.5 * I
and answer is 50 % .. thank you
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