# Optics Brewster's Angle Reflection Intensity

An incident unpolarised light beam of intensity $I_{0}$ strikes glass plate B at Brewster's Angle. The reflected light travels vertically and strikes a second glass plate A, again at Brewster's Angle. (We ignore the light transmitted by the glass plates.) Plate A is then rotated about the z-axis as shown. Briefly explain how the intensity of the light reflected by the apparatus varies with the angle of plate A. Illustrate with a qualitative sketch of intensity vs. angle of plate A.

We have an unpolarised beam incident on plate B at Brewster's Angle which upon reflection should become s-polarised or polarised perpendicular to the incident plane. My understanding of Brewster's Angle then says that upon the second reflection at plate A if the plate is oriented such that the incident plane is the same as for plate B there will be no reflection. Otherwise as it rotates there will be light reflected, is this correct?

Also does the intensity of the light reflected off of A increase gradually to a maximum when the beam is parallel to the incident plane at A? Any help would be much appreciated.

Simply, at Brewster's angle, the $\pi$ - component of incident electromagnetic wave always transmits 100%.
What is left is the $\sigma$ - component. The $\sigma$ - component of EM wave (which is already lying in xz - plane) striking on the second plate with respect to that plate is equal its amplitude times cosine of the angle the glass plate B will rotate.
The average reflected intensity is that thing to the square time $\epsilon_{0} \cdot 0.5$.
The reflected light off of the second Brewster window will be maximum in the configuration shown. It will be a minimum if you rotate it by $90^\circ$ around the $z$ axis. For angles in between it will act just like any other pair of polarizers.
• So when graphing it will appear like a cos squared graph? Since the intensity varies with $E^{2}$ hence the contribution that is reflected is $E_{0}cos(\theta)$? Feb 13, 2014 at 18:32