Problem 32 - University Physics with Modern Physics

Question

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at 23.0\(^\circ\) and 62.0\(^\circ\), respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 55.0 \(\mathrm {W/cm}^2\) after it passes through the stack. If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack?

Step-by-Step Solution

Answer

The intensity of the light after the stack, without the second polarizer, is less than with all three polarizers.

Step 1: Understand Polarization Concept

When unpolarized light passes through a polarizer, its intensity is reduced by half. If any additional polarizers are used, the intensity further decreases according to Malus's Law, which is given by \( I = I_0 \cos^2(\theta) \), where \( I_0 \) is the initial light intensity, \( \theta \) is the angle between the light’s polarization direction and the axis of the polarizer, and \( I \) is the resulting intensity.

Step 2: Initial Setup with Three Polarizers

The angles between the polarizers are as follows: the first and second polarizer differ by 23.0° and the second and third differ by 39.0° (62.0° - 23.0°). The given output intensity after all three polarizers is 55.0 W/cm².

Step 3: Effect of the First Polarizer

An unpolarized light initially has its intensity reduced by half after passing through the first polarizer. So, the intensity after the first polarizer, \( I_1 \), is \( \frac{I_0}{2} \).

Step 4: Effect of the Second Polarizer

Using Malus's Law for the second polarizer, we calculate the intensity as \( I_2 = I_1 \cos^2(23.0°) \).

Step 5: Effect of the Third Polarizer

The intensity after the third polarizer using Malus's Law is \( I_3 = I_2 \cos^2(39.0°) \). Given \( I_3 = 55.0 \ \mathrm{W/cm}^2 \), substitute \( I_2 \) to solve for \( I_0 \), the original incident intensity of unpolarized light.

Step 6: Calculate Original Intensity \( I_0 \)

We have \( 55.0 = \left(\frac{I_0}{2}\right) \cos^2(23.0°) \cos^2(39.0°) \). Solving for \( I_0 \), we determine the original incident intensity of light before any polarizer.

Step 7: Remove Second Polarizer

Without the second polarizer, the light after the first polarizer is \( \frac{I_0}{2} \). The intensity after the third polarizer is then \( I = \left(\frac{I_0}{2}\right) \cos^2(62.0°) \). Substitute the value of \( I_0 \) calculated previously to find the intensity after the light passes through the remaining two polarizers.

Step 8: Calculate New Intensity

Substitute the value of \( I_0 \) acquired from step 6 into \( I = \left(\frac{I_0}{2}\right) \cos^2(62.0°) \) to find the final intensity after removing the second polarizer.

Key Concepts

Malus's Law

When light passes through polarizing filters, its intensity is altered based on a fundamental principle known as Malus's Law. The law provides a mathematical way to determine the intensity of polarized light after it travels through a polarizer.The equation for Malus's Law is simple: \[ I = I_0 \cos^2(\theta) \]where:

• \( I \) is the resulting intensity of light after passing through the polarizer.
• \( I_0 \) represents the initial intensity of the polarized light.
• \( \theta \) denotes the angle between the light's initial polarization direction and the axis of the polarizer.

It is important to know that this law only applies when light is already polarized when it enters the polarizer. If the light is unpolarized, like sunlight or light from a regular bulb, Malus's Law is not directly applied until after the first polarizer, because unpolarized light becomes half as intense once it passes through a single polarizer.This concept is crucial for understanding how filters control light intensity, especially in various practical applications like photography and optics.

Unpolarized Light

Unpolarized light consists of waves vibrating in multiple planes perpendicular to the direction of propagation. Unlike polarized light, which has waves vibrating predominantly in one plane, unpolarized light's electric field vectors change direction randomly. When unpolarized light hits a polarizer, its intensity is reduced by half. This is because the polarizer only allows one plane of the electric field to pass through it, blocking others. Therefore, while polarized light has a definite direction of electric field oscillations, unpolarized light has no such direction until it encounters a polarizer. Here's what happens to unpolarized light once it passes through different optical configurations:

• **First Polarizer:** The intensity drops to exactly half of its original value.
• **Subsequent Polarizers:** Further intensity reduction occurs according to Malus's Law as the angles between the polarizers' axes affect the intensity.

You can think of unpolarized light as an equal mix of horizontal and vertical waves. When initially encountering a polarizer, the waves of just one orientation are allowed through, simplifying the remaining light to be manageable using Malus's Law.

Light Intensity

The concept of light intensity refers to the amount of energy a light wave carries per unit area, and it is typically measured in watts per square centimeter (W/cm²). In polarization, light intensity is a crucial factor as it determines how bright the light appears after passing through polarizing filters. For instance, when unpolarized light with a certain intensity naturally encounters multiple polarizers:

• The first polarizer cuts the intensity in half because it filters out one plane of vibration.
• Each additional polarizer further modifies the light intensity following Malus's Law, reducing it depending on the angle between consecutive polarizers.

Light intensity is a key aspect when considering how different layers of filters affect light passing through them. In practical terms, this principle is applied in sunglasses, photography filters, and even LCD screens to control brightness and reduce glare, enhancing the user's visual experience.