Question

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?

Answer 1

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:

$I = I_0 cos^2(\theta)$

$I_0 =$ Indicates the intensity of the light before passing through the Polarizer,

I = The resulting intensity, and

$\theta$ = 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,

$I_2 = I_0cos^2(45)$

For the third polarizer then we have,

$I_3 = I_2 cos^{2}(45)$

Replacing with the first equation,

$I_3 = I_0cos^2(45)cos^{2}(45)$

$I_3 = I_0 (\frac{1}{2})(\frac{1}{2})$

$I_3 = I_0 \frac{1}{4}$

Therefore the transmitted intensity now is $\frac{1}{4}$ of the initial intensity.