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## Homework Statement

Describe how to rotate the plane of polarization of a plane-polarized beam of light by $90^\circ$ and produce only a 10 percent loss in intensity using "perfect" polarizers.

The solution says to use 24 polarizers, with each at an angle of $3.75^\circ$ with the next. How is this allowed by the problem statement which says you must

*rotate*the plane of polarization?

## Homework Equations

Polarization equation - [itex]\cos^2(\theta)[/itex].

## The Attempt at a Solution

My solution was let the initial angle of polarization be $\theta$ so we have

[itex]\frac{\cos^2(90+\theta)}{\cos^2(\theta)}=\frac{9}{10}[/itex]

[itex]\frac{\cos(90-(-\theta))}{\cos(\theta)}=\frac{3}{\sqrt{10}}[/itex]

[itex]\frac{\sin(-\theta)}{\cos(\theta)}=-\frac{\sin(\theta)}{\cos(\theta)}=\frac{3}{\sqrt{10}}[/itex]

[itex]\tan(\theta)=\frac{\sqrt{3}}{10}[/itex]

[itex]\theta\approx \boxed{43^\circ}[/itex]

The solution says to use 24 polarizers, with each at an angle of [itex]3.75^\circ[/itex] with the next. How is this allowed by the problem statement which says you must

*rotate*the plane of polarization? How should I interpret the problem?