Polarization of light through a sugar solution

[KaiserBrandon]

Homework Statement

A glass vessel is placed between a pair of crossed HN-50 linear polarizers, and 50% of the natural light incident on the first polarizer is transmitted through the second polarizer. By how much did the sugar solution in the cell rotate the light passed by the first polarizer.

Homework Equations

I(\vartheta)=I(0)cos^{2}(\vartheta)
where \vartheta is the angle between the transmission axes of the polarizers, and I(0) is the irradience incident on the analyzer.

The Attempt at a Solution

I'm having a very hard time understanding how HN polarizers work, but here is my stab at the question:
An HN-50 polarizer transmits 50% of natural light incident on it. So If the natural light incident through the first polarizer is I, then I(0)=0.5I. So:
I(\vartheta)=0.5I*cos^{2}(90+\varphi)=0.5I
the 90 is in the cos because the polarizers are crossed, and I'm assuming that means their transmission angles are perpindicular. and \varphi is the angle that the sugar rotates the light. So, \varphi=-90.

 

[hikaru1221]

Perhaps you misunderstood the question. The way I understand it, 50% is the ratio between the intensity of incident light on 1st polarizer & the intensity of light coming out of 2nd polarizer. Both polarizers have the same polarization axis. That is, if there were no sugar solution in between, the light coming out would have the same intensity as incident light. But thanks to the sugar solution which rotates the polarization plane of the beam by some angle, the light coming out has different intensity.
From here, you can work it out, can't you?

 

[KaiserBrandon]

too late, already turned in the assignment. However, I'm pretty sure the light incident on the first polaraizer would be 50% transmitted (since the incident light is natural light with no specific polarization). Without the sugar, no light would go through the polarizer since their axes are crossed (at right angles to each other). So the sugar rotates it 90 degrees so that the light incident on the second polarizer has a polarization parallel to it's axis, and so gets transmitted through completely (so 50% the intensity of the natural light ends up at the end).