Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ∘ from vertical is inserted between the first two. What is the transmitted intensity now?

To solve this problem it is necessary to apply the concepts related to Malus' law. Malus' law indicates that the intensity of a linearly polarized ray of light that passes through a perfect analyzer with a vertical optical axis is equivalent to:

Indicates the intensity of the light before passing through the Polarizer,

I = The resulting intensity, and

= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

There is 3 polarizer, then

For the exit of the first polarizer we have that the intensity is,

For the third polarizer then we have,

Replacing with the first equation,

Therefore the transmitted intensity now is of the initial intensity.

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