/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 A plane plate of beryl is cut wi... [FREE SOLUTION] | 91Ó°ÊÓ

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A plane plate of beryl is cut with the optic axis in the plane of the surfaces. Linearly polarized light is incident on the plate such that the \(\overrightarrow{\mathbf{E}}\) -field vibrations are at \(45^{\circ}\) to the optic axis. Determine the smallest thickness of the plate such that the emergent light is (a) linearly and (b) circularly polarized.

Short Answer

Expert verified
(a) \(d = \frac{\lambda}{n_e - n_o}\); (b) \(d = \frac{\lambda}{4(n_e - n_o)}\).

Step by step solution

01

Understanding Birefringence

Beryl is a birefringent material, meaning it has two different refractive indices for light polarized along different axes. We denote these indices as \( n_o \) (ordinary) and \( n_e \) (extraordinary). When light enters at \(45^{\circ}\) to the optic axis, it splits into two components, each traveling with different speeds due to these indices.
02

Linearly Polarized Light Condition

For emergent light to be linearly polarized, the phase difference between the ordinary and extraordinary rays must be an integer multiple of \(2\pi\). The phase difference \(\Delta \phi = \frac{2\pi}{\lambda} (n_e - n_o) d\), where \(d\) is the thickness of the plate, \(\lambda\) is the wavelength of light. Set \(\Delta \phi = 2m\pi\) for integer \(m\).
03

Calculating Thickness for Linear Polarization

From the condition \(\frac{2\pi}{\lambda} (n_e - n_o) d = 2m\pi\), we simplify to:\[ d = m\frac{\lambda}{n_e - n_o}. \] For the smallest thickness, use \(m = 1\):\[ d = \frac{\lambda}{n_e - n_o}. \]
04

Circularly Polarized Light Condition

For light to be circularly polarized, the phase difference between the two components must be an odd multiple of \(\frac{\pi}{2}\). The requirement is \(\Delta \phi = (2m + 1)\frac{\pi}{2}\) where \(m\) is an integer.
05

Calculating Thickness for Circular Polarization

Substitute \(\Delta \phi = (2m + 1)\frac{\pi}{2}\) into \(\frac{2\pi}{\lambda} (n_e - n_o) d\):\[\frac{2\pi}{\lambda} (n_e - n_o) d = (2m + 1)\frac{\pi}{2}. \] Solving for \(d\) gives:\[ d = (2m + 1)\frac{\lambda}{4(n_e - n_o)}. \] For the smallest thickness, use \(m = 0\):\[ d = \frac{\lambda}{4(n_e - n_o)}. \]

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Key Concepts

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

Refractive Indices
Refractive indices are crucial when it comes to understanding how light behaves as it passes through different materials. In birefringent materials like beryl, there are usually two distinct refractive indices: the ordinary index ( _o") and the extraordinary index ( _e"). These indices determine how light travels through the material, depending on its polarization state relative to the optic axis.
  • The "ordinary" refractive index refers to light polarized perpendicular to the optic axis.
  • The "extraordinary" refractive index refers to light polarized parallel to the optic axis.
When light enters such a material at an angle, it generally splits into two rays, each experiencing different speeds and paths due to these two indices. This splitting causes a phase shift between the two components, which is the basis for many optical phenomena including polarization changes.
Understanding the difference between these refractive indices is key to calculating how thick the beryl plate should be for achieving the desired polarization of emergent light.
Linear Polarization
Linear polarization involves aligning the electrical field of a light wave in a single direction. When light encounters a birefringent material like beryl with its E ield vibrations at 45° to the optic axis, it transforms into two components. These components travel at different velocities due to different refractive indices.
  • For light to remain linearly polarized after passing through, the phase difference between its components should be a multiple of 2m ight).
  • The condition is expressed mathematically as: \(\Delta \phi = \frac{2\pi}{\lambda} (n_e - n_o) d = 2m\pi\).
This equation helps in determining the thickness of the material required to keep the outgoing light linearly polarized. By ensuring the components are synchronized in phase, they combine into a single linear polarization state. For the smallest thickness, set \(m = 1\), reflecting the desire to achieve the minimal required phase alignment.
Circular Polarization
Circular polarization differs from linear in that the electric field rotates in a circle as the light travels. Achieving circular polarization after passing through a birefringent material like beryl requires a specific phase condition: the components must differ by an odd multiple of \(\frac{\pi}{2}\).
  • The mathematical requirement is \(\Delta \phi = (2m + 1)\frac{\pi}{2}\).
  • The fundamental phase difference ensures that each component of light is out of phase just enough to create a circular pattern.
This requires careful calculation of the plate's thickness. Calculating this involves the formula: \( d = (2m + 1)\frac{\lambda}{4(n_e - n_o)} \).
By choosing \(m = 0\), you calculate the smallest thickness necessary to convert the entering linear polarized light into circularly polarized. Thus, the components match the phase and amplitude criteria crucial for circular polarization.
Optic Axis
The optic axis of a birefringent crystal is a crucial concept in understanding light propagation in such materials. It is the direction within the crystal along which light does not experience birefringence or splitting into two polarized rays.
  • Light traveling along this direction will encounter no difference between the ordinary and extraordinary refractive indices.
  • Light approaching from other angles splits into rays with speeds determined by \(n_o\) and \(n_e\).
In the given exercise, the optic axis is within the plane of the beryl plate surfaces. When light enters at 45° to this axis, the maximal phase separation occurs between the ordinary and extraordinary components. This setup is meticulously chosen to harness differences in refractive indices, enabling control over the emerging light's polarization state.
Understanding the optic axis helps in visualizing and comprehending how changes in the light path through birefringent materials are possible, making it integral to mastering the topics of refraction and polarization.

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Most popular questions from this chapter

Light from a source immersed in oil of refractive index 1.62 is incident on the plane face of a diamond \((n=2.42),\) also immersed in the oil. Determine (a) the angle of incidence at which maximum polarization occurs and (b) the angle of refraction into the diamond.

The indices of refraction for the fast and slow axes of quartz with \(546 \mathrm{nm}\) light are 1.5462 and 1.5553 , respectively. a. By what fraction of a wavelength is the e-ray retarded, relative to the o-ray, for every wavelength of travel in the quartz? b. What is the thickness of a zeroth-order QWP? c. If a multiple-order quartz plate \(0.735 \mathrm{mm}\) thick functions as a QWP, what is its order \(m ?\) d. Two quartz plates are optically contacted so that they produce opposing retardations. Sketch the orientation of the OA of the two plates. What should their difference in thickness be such that they function together like a zeroth-order QWP?

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.

Initially unpolarized light passes in turn through three linear polarizers with transmission axes at \(0^{\circ}, 30^{\circ},\) and \(60^{\circ}\) respectively, relative to the horizontal. What is the irradiance of the product light, expressed as a percentage of the unpolarized light irradiance?

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 \(90^{\circ}\)

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