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Chapter 15: Problem 24

A half-wave plate is placed between crossed polarizer and analyzer such that the angle between the polarizer TA and the FA of the HWP is \(\theta\). How does the emergent light vary as a function of \(\theta ?\)

Short Answer

Expert verified
The emergent light intensity varies as \(I = I_0 \sin^2(2\theta)\).

Step by step solution

01

Understanding the System

A half-wave plate (HWP) is placed between a crossed polarizer (TA) and analyzer. The angle between the transmission axis (TA) of the polarizer and the fast axis (FA) of the half-wave plate is given as \(\theta\). The purpose is to find how the emergent light varies with \(\theta\).
02

Recognizing Malus's Law

When dealing with polarizers, Malus's Law is important, which states that the intensity of light passing through a polarizer is given by \(I = I_0 \cos^2(\phi)\), where \(I_0\) is the initial intensity and \(\phi\) is the angle between the light's polarization direction and the polarizer's axis.
03

Analyzing Light through the Half-Wave Plate

A half-wave plate shifts the polarization direction of incident light, rotating it by \(2\theta\), effectively mirroring the angle across the fast axis of the plate. Thus, light's polarization direction exiting the HWP is dependent on \(\theta\) and given by \(2\theta\) relative to its input.
04

Light through the Analyzer

After exiting the HWP, the light encounters the analyzer (crossed with the initial polarizer). Since the angle between the polarization of the light after the HWP and the analyzer is \(90^\circ - 2\theta\), according to Malus's Law, the intensity of the emergent light is \(I = I_0 \cos^2(90^\circ - 2\theta)\).
05

Simplifying the Expression

Utilizing the trigonometric identity \(\cos(90^\circ - x) = \sin(x)\), we simplify \(I = I_0 \sin^2(2\theta)\). This expression explains how the emerging light intensity varies as a function of \(\theta\).

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Key Concepts

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

Half-wave plate
A half-wave plate (HWP) is a fascinating optical device that plays a key role in the manipulation of light polarization. Its main function is to alter the polarization direction of the incident light. This is achieved by slowing down one component of the polarized light more than the other, due to the birefringent nature of the plate.
The plate is designed such that it introduces a phase difference of half a wavelength (\(\lambda/2\)) between the fast and slow axis. As a result, when linearly polarized light passes through, its polarization direction rotates by a precise angle.
In this exercise, the angle is given by \(2\theta\), which means the polarization direction is effectively flipped around the fast axis of the plate by twice the angle \(\theta\). Therefore, the HWP's function is critical in applications needing precise control of light polarization.
Malus's Law
When dealing with polarized light, Malus's Law is a fundamental principle that governs light intensity. It states that the intensity of polarized light passing through another polarizer, known as the analyzer, is proportional to the square of the cosine of the angle between the light's initial polarization direction and the axis of the analyzer.
Mathematically, it is given by:
  • \(I = I_0 \cos^2(\phi)\)
Where \(I\) is the emergent intensity, \(I_0\) is the initial intensity, and \(\phi\) is the angle of interest.
This concept is crucial when light is manipulated through multiple optical elements, as in this scenario. After passing through the half-wave plate, the light encounters the analyzer, and Malus's Law helps to determine the final light intensity.
Light intensity
Light intensity, in the context of polarization, refers to the amount of power of the optical field that passes through an optical system, like the one comprising a polarizer and half-wave plate.
In this exercise, by utilizing both the characteristics of the half-wave plate and applying Malus's Law, we can derive how light intensity varies with the angle \(\theta\).
  • The HWP rotates the polarization direction by \(2\theta\)
  • The analyzer, placed after the HWP, crosses at an angle \(90^\circ\) relative to the initial polarizer
According to Malus's Law, the light intensity after the analyzer is given as:
  • \(I = I_0 \cos^2(90^\circ - 2\theta)\)
Utilizing the trigonometric identity \(\cos(90^\circ - x) = \sin(x)\), we simplify it to:
  • \(I = I_0 \sin^2(2\theta)\)
This relationship shows how the emergent light intensity is a function of the angle \(\theta\), dynamically changing along with it.

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Most popular questions from this chapter

since a sheet of Polaroid is not an ideal polarizer, not all the energy of the \(\overrightarrow{\mathbf{E}}\) -vibrations parallel to the TA are transmitted, nor are all \(\overrightarrow{\mathbf{E}}\) -vibrations perpendicular to the TA absorbed. Suppose an energy fraction \(\alpha\) is transmitted in the first case and a fraction \(\beta\) is transmitted in the second. a. Extend Malus' law by calculating the irradiance transmitted by a pair of such polarizers with angle \(\theta\) between their TAs. Assume initially unpolarized light of irradiance \(I_{0} .\) Show that Malus' law follows in the ideal case. b. Let \(\alpha=0.95\) and \(\beta=0.05\) for a given sheet of Polaroid. Compare the irradiance with that of an ideal polarizer when unpolarized light is passed through two such sheets having a relative angle between TAs of \(0^{\circ}\) \(30^{\circ}, 45^{\circ},\) and \(90^{\circ}\)

Light is incident on a water surface at such an angle that the reflected light is completely linearly polarized. a. What is the angle of incidence? b. The light refracted into the water is intercepted by the top flat surface of a block of glass with index of 1.50 . The light reflected from the glass is completely linearly polarized. What is the angle between the glass and water surfaces?

In each of the following cases, deduce the nature of the light that is consistent with the analysis performed. Assume a \(100 \%\) efficient polarizer. a. When a polarizer is rotated in the path of the light, there is no intensity variation. With a QWP in front of the rotating polarizer (coming first), one finds a variation in intensity but no angular position of the polarizer that gives zero intensity. b. When a polarizer is rotated in the path of the light, there is some intensity variation but no position of the polarizer giving zero intensity. The polarizer is set to give maximum intensity. A QWP is allowed to intercept the beam first with its OA parallel to the TA of the polarizer. Rotation of the polarizer now can produce zero intensity.

Vertically polarized light of irradiance \(I_{0}\) is incident on a series of \(N\) successive linear polarizers, each with its transmission axis offset from the previous one by a small angle \(\theta\) With the help of the Law of Malus, determine the value of \(N\) such that the final transmitted irradiance is \(I_{N}=0.9 I_{0}\) when the small angle offsets sum to \(90^{\circ},\) that is when the initial vertical polarization is rotated to a horizontal polarization.

When a plastic triangle is viewed between crossed polarizers and with monochromatic light of \(500 \mathrm{nm},\) a series of alternating transmission and extinction bands is observed. How much does \(\left(n_{\perp}-n_{\|}\right)\) vary between transmission bands to satisfy successive conditions for HWP retardation? The plastic triangle is \(\frac{1}{16}\) in. thick.
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