Calculating Intensity Reduction by a Polarizing Filter

When the intensity is reduced by $\text{90}\text{.}\mathrm{0\%}$, it is $\text{10}\text{.}\mathrm{0\%}$ or 0.100 times its original value. That is, $I=0\text{.}\text{100}{I}_0$. Using this information, the equation $I=I_0{\text{cos}}^2\theta$ can be used to solve for the needed angle.

Solving the equation $I=I_0{\text{cos}}^2\theta$ for $\text{cos}\theta$ and substituting with the relationship between $I$ and $I_0$ gives

Solving for $\theta$ yields

A fairly large angle between the direction of polarization and the filter axis is needed to reduce the intensity to $\text{10}\text{.}\mathrm{0\%}$ of its original value. This seems reasonable based on experimenting with polarizing films. It is interesting that, at an angle of $\text{45º}$, the intensity is reduced to $\text{50\%}$ of its original value (as you will show in this section’s Problems & Exercises). Note that $\text{71}\text{.}\mathrm{6º}$ is $\text{18}\text{.}\mathrm{4º}$ from reducing the intensity to zero, and that at an angle of $\text{18}\text{.}\mathrm{4º}$ the intensity is reduced to $\text{90}\text{.}\mathrm{0\%}$ of its original value (as you will also show in Problems & Exercises), giving evidence of symmetry.