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The light beam in the figure passes through a polarizer whose transmission axis makes an angle \(\phi\) with the vertical. The beam is partially polarized and partially unpolarized, and the average intensity \(\bar{S}_{0}\) of the incident light is the sum of the average intensity \(\bar{S}_{0, \text { polar }}\) of the polarized light and the average intensity \(S_{0, \text { unpolar }}\) of the unpolarized light; \(\bar{S}_{0}=S_{0, \text { polar }}+S_{0, \text { unpolar }}\) The intensity \(\bar{S}\) of the transmitted light is also the sum of two parts: \(\bar{S}=\bar{S}_{\text {pobar }}+\bar{S}_{\text {unpola }}\). As the polarizer is rotated clockwise, the intensity of the transmitted light has a minimum value of \(\bar{S}=2.0 \mathrm{W} / \mathrm{m}^{2}\) when \(\phi=20.0^{\circ}\) and has a maximum value of \(\bar{S}=8.0 \mathrm{W} / \mathrm{m}^{2}\) when the angle is \(\phi=\phi_{\max } .\) Concepts: (i) How is \(\bar{S}_{\text {unpolar }}\) related to \(\bar{S}_{0, \text { unpolur }} ?\) (ii) How is \(\bar{S}_{\text {polur }}\) related to \(\bar{S}_{0, \text { polar }}\) ? (iii) The minimum transmitted intensity is \(2.0 \mathrm{W} / \mathrm{m}^{2} .\) Why isn't it \(0 \mathrm{W} / \mathrm{m}^{2}\) ? Calculations: (a) What is the intensity \(\bar{S}_{0, \text { unpolar }}\) of the incident light that is unpolarized? (b) What is the intensity \(\bar{S}_{0, \text { polur }}\) of the incident light that is polarized?

Step by step solution

01

Understanding Unpolarized Light

For unpolarized light, the transmitted intensity is affected half of its initial intensity value. Therefore, the relationship is given by:\[ \bar{S}_{\text{unpolar}} = \frac{1}{2} \bar{S}_{0, \text{unpolar}} \]

02

Understanding Polarized Light

The intensity of polarized light after passing through a polarizer is given by Malus's Law:\[ \bar{S}_{\text{polar}} = \bar{S}_{0, \text{polar}} \cos^2(\phi) \] where \( \phi \) is the angle between the light's polarization direction and the polarizer's transmission axis.

03

Calculating Total Minimum Intensity

The given condition when the transmission intensity is minimum, \( \bar{S} = 2.0 \, \text{W/m}^2 \), occurs at \( \phi = 20.0^\circ \). Using the formula:\[ \bar{S} = \bar{S}_{\text{polar}} + \bar{S}_{\text{unpolar}} \] we write:\[ 2.0 = \bar{S}_{0, \text{polar}} \cos^2(20^\circ) + \frac{1}{2}\bar{S}_{0, \text{unpolar}} \]

04

Calculating Total Maximum Intensity

The maximum intensity occurs at another angle \( \phi = \phi_{\max} \), where \( \bar{S} = 8.0 \, \text{W/m}^2 \):\[ 8.0 = \bar{S}_{0, \text{polar}} \cos^2(\phi_{\max}) + \frac{1}{2}\bar{S}_{0, \text{unpolar}} \]At maximum, \( \cos^2(\phi_{\max}) = 1 \), so:\[ 8.0 = \bar{S}_{0, \text{polar}} + \frac{1}{2}\bar{S}_{0, \text{unpolar}} \]

05

Solving the System of Equations

We have two equations now:1. \( 2.0 = \bar{S}_{0, \text{polar}} \cos^2(20^\circ) + \frac{1}{2}\bar{S}_{0, \text{unpolar}} \)2. \( 8.0 = \bar{S}_{0, \text{polar}} + \frac{1}{2}\bar{S}_{0, \text{unpolar}} \)Rearrange the first equation to solve for \( \bar{S}_{0, \text{polar}} \), and substitute into the second equation to solve for \( \bar{S}_{0, \text{unpolar}} \).\[\cos^2(20^\circ) \approx 0.9397^2 \approx 0.883 \]Using these, simplify to find:\[ \bar{S}_{0, \text{polar}} \approx \frac{8.0 - 2.0 \tan^2(20^\circ)}{1- \tan^2(20^\circ)} \approx 6.78 \, \text{W/m}^2 \]and,\[ \bar{S}_{0, \text{unpolar}} = 2(8.0 - 6.78) \approx 2.44 \, \text{W/m}^2 \]

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Key Concepts

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

Malus's Law

When light passes through a polarizer, its intensity changes depending on the angle between the light's polarization direction and the polarizer's transmission axis. This relationship is described by **Malus's Law**. According to Malus's Law, the intensity of light after passing through a polarizer is given by the formula:\[\bar{S}_{\text{polar}} = \bar{S}_{0, \text{polar}} \cos^2(\phi)\]Here, \( \bar{S}_{\text{polar}} \) is the transmitted intensity, \( \bar{S}_{0, \text{polar}} \) is the initial intensity of the polarized light, and \( \phi \) is the angle between the light's polarization direction and the polarizer's axis. The **cosine squared component** tells us that the intensity decreases as the angle \( \phi \) increases.

• When \( \phi = 0 \) degrees, the light is perfectly aligned with the polarizer, and the intensity is maximized.
• When \( \phi = 90 \) degrees, the polarized light is orthogonal to the polarizer, resulting in zero transmitted intensity.

Understanding Malus's Law is crucial for predicting how much light comes through a polarizer.

Polarizers in Optics

**Polarizers** are optical filters that allow only light waves with a specific polarization to pass through. These are essential tools in **optics** for controlling and analyzing light waves. When natural or unpolarized light, which contains all possible polarization directions, encounters a polarizer, the polarizer filters out all directions except one. This process results in polarized light. Using multiple polarizers can demonstrate fascinating effects. By rotating one polarizer relative to the other, you can control the amount of light passing through due to the interaction between their respective transmission axes. This technique is even applied in sunglasses and camera lenses to reduce glare.

• **Polarizer 1**: transforms unpolarized light into polarized light.
• **Polarizer 2**: the angle between the transmission axis and the light's polarization determines the amount of light that passes through.

Applications of polarizers illustrate the fundamental principles of polarization, simultaneously enhancing our everyday experiences with optical devices.

Intensity of Polarized Light

The **Intensity of Polarized Light** is a measure of how much light is transmitted after it passes through a polarizer. For unpolarized light, the average transmitted intensity is half of its initial intensity:\[\bar{S}_{\text{unpolar}} = \frac{1}{2} \bar{S}_{0, \text{unpolar}}\]This means that even without applying Malus's Law, unpolarized light is naturally reduced to 50% when passing through a polarizer. This is because only half of the light's polarization components align with the polarizer's axis at any given time. Once light is polarized, its intensity is further adjusted by the angle \( \phi \) following Malus's Law, as previously discussed. This explains why light can never be fully diminished to zero except when polarized perpendicular to the polarizer.

• Polarized light intensity changes with angle \( \phi \) due to the cosine squared relationship.
• Unpolarized light inherently reduces to half intensity upon first encounter with a polarizer.

Understanding these changes in intensity helps us manipulate light effectively in various scientific and practical applications, like in communications and imaging technologies.