**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.

## Expert Answer

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 $I_1$:

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

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

\[\cos^2{\theta}\ =\ \frac{2}{10}\ =\ 0.2\]

\[\theta\ \ =\ 63°\]