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Chapter 33: Problem 41

A beam of polarized light is sent into a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles \(\theta\) for the first sheet and \(90^{\circ}\) for the second sheet. If \(0.10\) of the incident intensity is transmitted by the two sheets, what is \(\theta\) ?

Short Answer

Expert verified
\( \theta \approx 18.43^{\circ} \).

Step by step solution

01

Understanding Malus's Law

Malus's Law states that the intensity of polarized light after passing through a polarizing filter is given by the equation \( I = I_0 \cos^2\theta \), where \( I_0 \) is the initial intensity and \( \theta \) is the angle between the light's initial polarization direction and the axis of the filter.
02

Apply Malus's Law to First Sheet

For the first sheet, the intensity after passing through is \( I_1 = I_0 \cos^2 \theta \). Here, \( I_0 \) is the initial intensity and \( \theta \) is the angle given in the problem.
03

Transmission Through Second Sheet

The second sheet is oriented at an angle of \(90^{\circ}\) relative to the first. However, the light reaching the second sheet is already polarized along the first sheet's direction. Applying Malus's Law for the second sheet: \[ I_2 = I_1 \cos^2(90^{\circ} - \theta) = I_1 \sin^2 \theta \] Substituting for \( I_1 \), we get \[ I_2 = I_0 \cos^2 \theta \sin^2 \theta \].
04

Express Final Intensity and Solve for \(\theta\)

Given that \(0.10\) of the incident intensity is transmitted, we have: \[ I_2 = 0.10 I_0 \] Substitute the expression from Step 3: \[ I_0 \cos^2 \theta \sin^2 \theta = 0.10 I_0 \] Simplifying, we obtain: \[ \cos^2 \theta \sin^2 \theta = 0.10 \] This can be rewritten using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to form: \[ \sin^2 2\theta = 0.40 \].
05

Solve for \(\theta\)

The equation from the previous step can be solved for \( \theta \): \[ \sin 2\theta = \sqrt{0.40} \] \[ 2\theta = \arcsin(\sqrt{0.40}) \] Calculating, we find: \[ 2\theta \approx 0.6435 \] radians or \(36.87^{\circ}\). Thus, \( \theta = \frac{36.87^{\circ}}{2} \approx 18.43^{\circ} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Malus's Law
Malus's Law offers a fascinating insight into how light behaves when it passes through a polarizing filter. This law describes how the intensity of polarized light changes based on the angle of the light's wavefront with respect to the filter's transmission axis. The mathematical representation, given by \( I = I_0 \cos^2\theta \), neatly quantifies this relationship. Here, \( I \) is the intensity of light after passing through the filter, \( I_0 \) is the initial intensity, and \( \theta \) is the crucial angle. This angle is between the light's initial direction of polarization and the polarizing axis of the filter. When this angle is 0掳, the intensity retained is full, meaning all polarized light passes through unchanged. Conversely, at 90掳, the intensity drops to zero, blocking all the polarized light.
Polarizing Filters
Polarizing filters are simple yet powerful tools that can control light waves. They primarily allow only light waves aligned with their orientation to pass through. When light encounters a polarizing filter, only the component of the light wave that aligns with the filter's axis continues its journey.
  • A polarizing filter has a specific polarization axis.
  • The filter absorbs components of light that are perpendicular to this axis.
  • Natural light, which is unpolarized, will lose exactly half of its intensity when it passes through a polarizing filter.
Polarizers are instrumental in glare reduction, enhancing contrast, and in applications like photography, to selectively control light and improve image clarity.
Intensity of Light
The intensity of light, especially in the context of polarization, refers to the power per unit area carried by light waves. When discussing a series of filters, the intensity of light decreases successively. The initial intensity \( I_0 \) represents the starting strength of the light before any interaction with polarizing filters. Each subsequent filter modifies this intensity according to Malus's Law.
Consider a light beam passing through two filters, with an end intensity of just 10% of the original. This significant reduction happens because each filter processes the incoming wave, aligned or filtered against its axis, thereby progressively decreasing the beam's effective intensity.
Angle of Transmission
The angle of transmission is a vital parameter in determining how much light passes through polarizing filters. It is closely tied to the functionality of Malus's Law, where the angle \( \theta \) directly influences how much intensity is retained. This angle is formed between the light's initial polarization direction and the polarizing filter's axis.
When solving practical problems, identifying this angle accurately is crucial. For example, if given that only 10% of light is transmitted through a set of polarizers, one can use Malus's Law and trigonometric identities to work backwards and find the angle \( \theta \). Through calculations that involve equations like \[ \sin^2 2\theta = 0.40 \], the angle of transmission becomes a critical factor in determining light behavior through polarizers.

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