 # Unpolarized light with intensity I₀ is incident on two polarizing filters. Find intensity of the light after passing through second filter. The first filter is oriented at an angle of $60.0°$ between its axis and vertical whereas the second filter is oriented at the horizontal axis.

The aim of this question is to find the intensity of polarized light after it has passed through two filters which are oriented at a certain angle and axis.

The article uses the concept of Malus Law, which explains that when a plane-polarized light passes through an analyzer oriented at a certain angle, the intensity of that polarized light is directly proportional to the square of the cosine of the angle between the plane at which polarizer is oriented and the axis of the analyzer at which it transmits the polarized light. It is represented as per the following expression:

$I\ =\ I_o\cos^2{\theta}$

Where:

$I\ =$ Intensity of Polarized light

$I_o\ =$ Intensity of Unpolarized light

$\theta\ =$  Angle between initial polarization direction and polarizer axis

When an unpolarized light passes through a polarizer, the intensity of light is reduced to half irrespective of the axis of polarization.

Given that:

The Angle between Filter Axis and Vertical $\phi\ =\ 60.0°$

$I_o\ =$ Intensity of Unpolarized light

So the angle $\theta$ between initial polarization direction and polarizer axis will be:

$\theta\ =\ 90° -\ ϕ$

$\theta\ =\ 90° -\ 60°$

$\theta\ =\ 30°$

When the unpolarized light with Intensity $I_o$ is passed through the first filter, its Intensity $I_1$ after polarization will be reduced to half of its initial value.

Hence Intensity $I_1$ after the  first filter will be:

$I_1\ =\ \frac{I_o}{2}$

In order to find the Intensity of Polarized Light $I_2$ after the  second filter, we will use the concept of Malus Law which is expressed as follows:

$I_2\ =\ I_1\cos^2{\theta}$

Substituting the value of $I_1$ from above equation, we get:

$I_2\ =\ \left(\frac{I_o}{2}\right)\cos^2{\theta}$

Substituting the value of $\theta$, we get:

$I_2\ =\ \left(\frac{I_o}{2}\right)\cos^2(30°)$

As we know that:

$\cos(30°) = \dfrac{\sqrt3}{2}$

$\cos^2(30°) =\ \left(\frac{\sqrt3}{2}\right)^2 = \dfrac{3}{4}$

Substituting the value of $\cos^2(30°) =\dfrac{3}{4}$:

$I_2\ =\ \left(\frac{I_o}{2}\right)\times\left(\frac{3}{4}\right)$

$I_2\ =\ \frac{3}{8}\times I_o$

$I_2\ =\ 0.375I_o$

## Numerical Result

The Intensity $I_2$ of the light after it has passed through the second filter will be:

$I_2\ =\ 0.375I_o$

## Example

Unpolarized light having an intensity $I_o$ is allowed to pass through two polarized filters. If the Intensity of light after passing through the second filter $I_2$ is  $\dfrac{I_o}{10}$, calculate the angle that exists between the axes of the two polarized filters.

Solution

Given that:

The intensity of the light after the second filter $I_2\ =\ \dfrac{I_o}{10}$

When the unpolarized light with Intensity $I_o$ is passed through the first filter, its intensity $I_1$ after polarization will be reduced to half of its initial value.

Intensity $I_1$ after first filter will be:

$I_1\ =\ \frac{I_o}{2}$

As per Malus Law, we know that:

$I_2\ =\ I_1\cos^2{\theta}$

Substituting the values of $I_2$ and $I_1$:

$\frac{I_o}{10}\ =\ \left(\frac{I_o}{2}\right)\cos^2{\theta}$

$\frac{I_o}{10}\ =\ \left(\frac{I_o}{2}\right)\cos^2{\theta}$

$\cos^2{\theta}\ =\ \frac{2}{10}\ =\ 0.2$

$\theta\ \ =\ 63°$