Question

Three polarizing filters are stacked with the polarizing axes of the second and third at 45.0${}^{\circ }$ and 90.0${}^{\circ }$, respectively, with that of the first. (a) If unpolarized light of intensity $I_0$ is incident on the stack, find the intensity and state of polarization of light emerging from each filter. (b) If the second filter is removed, what is the intensity of the light emerging from each remaining filter?

Short answer

(a) With all filters: Final intensity is $\frac{I_0}{8}$; light is polarized at 90.0°. (b) Without the second filter: Final intensity is 0.

Step-by-step solution

Understanding Polarized Light

To solve this problem, we need to understand how light intensity changes as it passes through polarizing filters. When unpolarized light with intensity $I_0$ passes through a polarizer, the intensity of the emerging polarized light is reduced to half, or $\frac{I_0}{2}$. This is due to one of the properties of polarized light.

Using Malus's Law

When polarized light then passes through another polarizer, the intensity changes according to Malus's Law, which is given by: $I=I^{\prime }{\cos }^2\theta$, where $I^{\prime }$ is the intensity before the filter and $\theta$ is the angle between the light's polarization direction and the filter's axis.

Intensity after First Filter

The light is initially unpolarized with intensity $I_0$. Passing through the first polarizer, the intensity becomes $I_1=\frac{I_0}{2}$. The polarized light's axis aligns with the first filter.

Intensity after Second Filter

For the second filter, which is at 45.0° to the first: $I_2=I_1{\cos }^2(45.0°)=\frac{I_0}{2}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{4}$. The emerging light is polarized at 45.0°.

Intensity after Third Filter

The third filter is at 90.0° to the first filter, but we consider it at 45.0° to the polarized light from the second filter: $I_3=I_2{\cos }^2(45.0°)=\frac{I_0}{4}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{8}$. The light is polarized at 90.0° with respect to the first filter.

Removing the Second Filter

Without the second filter, the light after the first filter is $I_1=\frac{I_0}{2}$, as calculated earlier. Now, this light passes directly to the third filter, which is at 90.0° to the first filter: $I_3=I_1{\cos }^2(90.0°)=\frac{I_0}{2}\cdot 0=0$. No light emerges.

Key concepts

Malus's Law

Malus's Law is key in understanding how light behaves as it travels through multiple polarizing filters. It states that when polarized light encounters another polarizer, the resulting light intensity can be calculated based on the angle between the polarizer's axis and the light's polarization direction. The formula is given by:

• $I=I^{\prime }{\cos }^2\theta$

where $I$ is the intensity of light after passing through the filter, $I^{\prime }$ is the intensity before the filter, and $\theta$ is the angle between the filter's axis and the light's direction.
This relationship explains the light intensity reduction as it encounters successive polarizers. For example, if the light passes through two polarizers that are oriented at $45.0°$ relative to each other, the intensity of light after the second polarizer would be half of the intensity before it, as ${\cos }^2(45.0°)=\frac{1}{2}$.
Understanding Malus's Law helps us predict how light will be affected by changes in the angle of polarizing filters.

Polarizing Filters

Polarizing filters play a significant role in controlling the light that passes through them. They only allow light waves oscillating in a specific direction to pass through. The rest are filtered out.
At the microscopic level, a polarizing filter consists of multiple long-chain molecules aligned in such a way that they block light waves oscillating perpendicular to these molecules.
When unpolarized light, which consists of waves oscillating in multiple directions, hits a polarizing filter, the outcome is polarized light. This new light only consists of waves aligned with the filter's axis.
Here is how it works:

• Upon passing through the first polarizer, the intensity of unpolarized light reduces to half ($\frac{I_0}{2}$).
• Subsequent polarizers further refine the light wave’s direction and reduce intensity as per Malus's Law.

The fascinating aspect of using multiple polarizers is the ability to dramatically change light's behavior depending on the angle settings of these filters.

Intensity of Light

Intensity of light refers to the amount of energy carried by light waves in a particular direction per unit area. When exploring problems involving polarizers, the intensity of light is a crucial factor as it continuously changes.
Initially, unpolarized light has a specific intensity, let's denote it as $I_0$. As this light passes through a polarizing filter, its intensity shifts according to the properties of the polarizers and the angles involved.
Here's a quick breakdown of intensity changes:

• First polarizer halves the intensity: $I_1=\frac{I_0}{2}$.
• Second polarizer aligned at an angle to the first uses Malus’s Law: $I_2=I_1{\cos }^2\theta$.
• Further polarizers continue to reduce intensity based on their specific alignment angles.

By understanding how intensity transforms through each filter, you ensure proper interpretation of light's journey and its polarization state.
The concept of intensity highlights the energy aspect of light and helps in practical applications, like controlling light in photography or reducing glare through sunglasses.