Help with polarizers and analyzers: Malus' law?

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So I know the formula is:
Average intensity of the light leaving the analyzer = (Average intensity of the light entering the analyzer)cos^2 theta


Unpolarized light whose intensity is 1.20 W/m2 is incident on the polarizer in the figure.

What is the intensity of the light leaving the polarizer?
...W/m^2

If the analyzer is set at an angle of ? = 70° with respect to the polarizer, what is the intensity of the light that reaches the photocell?
...W/m^2
 
Not much of a diagram, but I'll assume there are two polarizers. Unpolarized light at intensity I0 = 1.2 W/m^2 enters the 1st polarizer. In this situation the polarized light has 1/2 the ave. intensity of the incident light (see ref.).
Ave. intensity of light leaving the 1st polarizer I1 = I0/2 = 0.6 W/m^2
Intensity of light leaving the 2nd polarizer is I1*cos^2(70 deg) = 0.6*0.1170 = 0.07019 W/m^2
 
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