Physics
posted by Brian on .
We want to rotate the direction of polarization of a beam of polarized light through 90 degrees by sending the beam through one or more polarizing sheets.(a) What is the minimum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than 60% of the original intensity?

Since when can polarizing sheets rotate a linearly polarized light beam?

I don't really know. I did ln(1/100)=4.605 and ln(cos(90)^2)=1.605 and 4.605/1.605=2.869. The answer for part (a) should be 2 but 2.869 is closer to 3 so am I doing anything wrong?

(a)If we place one sheet with its polarization direction 90⁰ to the polarized light then the transmitted intensity will be zero (since I=I₀•cos²α = I₀•cos²90⁰=0).
If we use 2 sheets, the 1st sheet rotates by angle α (0< α<90⁰) and the 2nd sheet rotate by the angle (90⁰α)
I=I₀•cos²α•cos²(90⁰α)= I₀•cos²α•sin²α.
Thus, we need two sheets.
(b) I=I₀•cos²ⁿ(90⁰/n)
n=2 I₂=I₀•cos⁴(90⁰/2)=0.25•I₀
n=3 I₃=I₀•cos⁶(90⁰/3)=0.422•I₀
n=4 I₄=I₀•cos⁸(90⁰/4)=0.53•I₀
n=5 I₅=I₀•cos¹⁰(90⁰/5)=0.6054•I₀
I ո ≥0.6•I₀
Thus, the minimum value n=5 sheets. 
I still don't understand part (a). I know that cos^2(90)=0 but how did you get to the answer of 2 sheets?

2 because....that's the minimum no. of angles of that can ad up to 90