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Chapter 15: Problem 4

How thick should a half-wave plate of mica be in an application where laser light of \(632.8 \mathrm{nm}\) is being used? Appropriate refractive indices for mica are 1.599 and 1.594.

Short Answer

Expert verified
The mica half-wave plate should be approximately 100 micrometers thick.

Step by step solution

01

Understanding the Concept of a Half-Wave Plate

A half-wave plate is an optical device that shifts the phase of polarized light by half a wavelength, which equals 180 degrees. This means that it inverts the phase of the light, effectively changing its polarization direction.
02

Calculate the Wavelength of Light in Mica

To calculate the wavelength of the laser light in mica, use the formula \( \lambda' = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the wavelength in vacuum (632.8 nm) and \( n \) is the refractive index. Compute for both refractive indices.\[ \lambda'_o = \frac{632.8}{1.599} \approx 395.8 \, \text{nm} \] \[ \lambda'_e = \frac{632.8}{1.594} \approx 396.8 \, \text{nm} \]
03

Determine the Optical Path Difference

The optical path difference (OPD) in a half-wave plate is equal to half the wavelength in the material. Therefore, calculate the OPD using the formula \( OPD = \frac{\lambda'_{o} - \lambda'_{e}}{2} \).\[ OPD = \frac{395.8 - 396.8}{2} \approx 0.5 \text{ nm} \]
04

Calculate the Thickness of the Half-Wave Plate

To find the thickness \( t \) of the half-wave plate, use the OPD calculated in the previous step and the formula \( t = \frac{\text{OPD}}{n_o - n_e} \) where \( n_o \) and \( n_e \) are the refractive indices for the ordinary and extraordinary rays, respectively.\[ t = \frac{0.5}{1.599 - 1.594} \approx 100 \, \text{micrometers} \]

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Key Concepts

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

Optical Path Difference
Optical path difference (OPD) is a crucial concept in optics, referring to the difference in the path traveled by two light waves as they pass through different mediums. This variance can cause shifts in the phase of the waves, leading to interference patterns. In terms of a half-wave plate, the OPD is used to create a phase shift, transforming the polarized light's orientation. To determine the OPD in a half-wave plate, you consider how much one wave is slowed down compared to another due to passing through the plate material. This is calculated by the difference in wavelengths of light inside the material, divided by two.
Wavelength in Material
When laser light enters a material like mica, its speed changes due to the material's properties, which alters its wavelength. This new wavelength is termed as the wavelength in the material. Calculating this wavelength involves using the formula \( \lambda' = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the original wavelength in a vacuum, and \( n \) is the refractive index of the material. For a half-wave plate, understanding these modified wavelengths is key to achieving desired optical effects, such as phase changes and interference patterns.
Phase Shift
Phase shift in the context of a half-wave plate is the change in the phase of polarized light as it passes through the device. The plate is designed to shift the light's phase by exactly half a wavelength (180 degrees). This significant shift effectively alters the polarization orientation of the light. In practical terms, this means the electric field of the light wave is inverted. Phase shifts are vital in optical applications where precise control over light polarization is required, such as in imaging and laser systems.
Refractive Indices
Refractive indices are fundamental in understanding how light propagates through different materials. These indices quantify the extent to which light is slowed down in a medium compared to a vacuum. In the context of a half-wave plate, you often deal with two refractive indices: one for the ordinary ray (\( n_o \)) and one for the extraordinary ray (\( n_e \)). These indices determine the amount of optical path difference created in the plate, which is essential for calculating the necessary thickness of the half-wave plate to achieve the desired phase shift.
Polarized Light
Polarized light refers to light waves in which the vibrations occur in a single plane. Many optical devices, including half-wave plates, rely on manipulating polarized light to perform their functions. In a half-wave plate, the polarized light's plane is rotated by 90 degrees due to the phase shift introduced by the device. This manipulation is key in applications where controlling the orientation of light is crucial, such as reducing glare in photography or enhancing contrast in optical instruments. Understanding polarized light helps in mastering concepts related to optics and wave mechanics.

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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.

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.

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.

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.

The Fresnel equations show that the fraction \(r\) of the incident field that is reflected from a dielectric plane surface for the TE polarization mode has the form, $$r=\frac{\cos \theta-\sqrt{n^{2}-\sin ^{2} \theta}}{\cos \theta+\sqrt{n^{2}-\sin ^{2} \theta}}$$ Thus, the reflectance \(R=r^{2}\) has the form, R=\left(\frac{\cos \theta-\sqrt{n^{2}-\sin ^{2} \theta}}{\cos \theta+\sqrt{n^{2}-\sin ^{2} \theta}}\right)^{2} where \(\theta\) is the angle of incidence and \(n\) is the ratio \(n_{2} / n_{1}\) a. Calculate the reflectance \(R\) for the TE mode when the light is incident from air onto glass of \(n=1.50\) at the polarizing angle. b. The reflectance calculated in part (a) is also valid for an internal reflection as light leaves the glass going into air. This being the case, calculate the net fraction of the TE mode transmitted through a stack of 10 such plates relative to the incident irradiance \(I_{0}\). Assume that the plates do not absorb light and that there are no multiple reflections within the plates. c. Calculate the degree of polarization \(P\) of the transmitted beam, given by $$P=\frac{I_{T M}-I_{T E}}{I_{T M}+I_{T E}}$$ where \(I\) stands for the irradiance of either polarization mode.
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