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Chapter 12: Problem 18

A uniform plane wave is normally incident onto a slab of glass \((n=1.45)\) whose back surface is in contact with a perfect conductor. Determine the reflective phase shift at the front surface of the glass if the glass thickness is (a) \(\lambda / 2 ;(b) \lambda / 4 ;(c) \lambda / 8\).

Short Answer

Expert verified
Question: Determine the reflective phase shift at the front surface of a glass slab for 3 different thickness scenarios when a uniform plane wave is incident on it and the back surface of the glass is in contact with a perfect conductor. The refractive index of the glass is 1.45. Consider the following glass thickness scenarios: (a) \(d=\lambda / 2\), (b) \(d=\lambda / 4\), and (c) \(d=\lambda / 8\). Answer: The reflective phase shift at the front surface of the glass for different thicknesses is: (a) \(\Delta \phi_{a} = -2(1.45) \pi\) (b) \(\Delta \phi_{b} = -1(1.45) \pi\) (c) \(\Delta \phi_{c} = -0.5(1.45) \pi\)

Step by step solution

01

Identifying the given variables and constants

The given variables are: Refractive index of glass, \(n = 1.45\) Glass thickness, \(d\) is given in 3 cases: - (a) \(d = \lambda / 2\) - (b) \(d = \lambda / 4\) - (c) \(d = \lambda / 8\)
02

Calculate the reflective phase shift for each thickness

Reflective phase shift can be calculated using the formula \(\Delta \phi = 2 \phi_i - 2n \pi (\frac{2d}{\lambda})\). But since the incident angle is normal, \(\phi_i = 0\). So the formula simplifies to: \(\Delta \phi = -2n \pi (\frac{2d}{\lambda})\) Calculate the reflective phase shift for each case: (a) When \(d = \lambda / 2\), \(\Delta \phi_{a} = -2(1.45) \pi (\frac{2 (\lambda/2)}{\lambda}) = -2(1.45) \pi\) (b) When \(d = \lambda / 4\), \(\Delta \phi_{b} = -2(1.45) \pi (\frac{2 (\lambda/4)}{\lambda}) = -1(1.45) \pi\) (c) When \(d = \lambda / 8\), \(\Delta \phi_{c} = -2(1.45) \pi (\frac{2 (\lambda/8)}{\lambda}) = -0.5(1.45) \pi\)
03

Presenting the results

The reflective phase shift at the front surface of the glass for different thicknesses is: (a) \(\Delta \phi_{a} = -2(1.45) \pi\) (b) \(\Delta \phi_{b} = -1(1.45) \pi\) (c) \(\Delta \phi_{c} = -0.5(1.45) \pi\)

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Key Concepts

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

Uniform Plane Wave
Imagine light as a collection of waves spreading out from a source. A uniform plane wave is a simplification used to describe these waves when they travel through space without any disturbance 鈥� meaning they have a constant amplitude and frequency and their wavefronts (the peaks and valleys of the waves) are infinite, flat planes.

Picture a perfectly calm ocean with waves that are all the same size and move together side by side without changing shape. This is similar to how uniform plane waves behave. They are a useful concept in physics because they make it easier to predict how light and other forms of electromagnetic radiation will behave under certain conditions, like when they strike a surface.
Glass Refractive Index
When light enters a different material, such as glass, it changes speed and direction. This bending of light is known as refraction. The glass refractive index, denoted as n, is a number that tells us how much the light will bend.

The refractive index of glass (around 1.45 for common glass) is a way of saying that light travels about 1.45 times slower in glass than it does in a vacuum. A higher refractive index indicates that the light is significantly slower, causing a larger bend. This property is fundamental in optics, affecting how lenses focus light and how prisms create spectrums.
Perfect Conductor Boundary Condition
Conductors are materials that allow electrons to move freely. A perfect conductor is an idealized material where the electrons move with no resistance at all; in other words, it has an infinite electrical conductivity. When an electromagnetic wave encounters a perfect conductor, it can't penetrate the surface and is completely reflected. This is known as the perfect conductor boundary condition.

In practice, no material is a perfect conductor, but some materials like superconductors come very close under certain conditions. The boundary condition is vital for understanding how waves behave when encountering different materials and is used in designing antennas and other radio frequency devices.
Wave Interference
When two or more waves meet while traveling through the same medium, they affect each other's displacement. This phenomenon is called wave interference. There are two main types of interference: constructive, where waves add up to make a larger wave, and destructive, where they cancel each other out to make a smaller wave or no wave at all.

An everyday example of wave interference is the patterns seen when two sets of ripples cross each other in a pond. When applied to light, interference is responsible for the colorful patterns in soap bubbles and oil on water. It's also the principle behind technologies such as noise-cancelling headphones.

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Most popular questions from this chapter

A wave starts at point \(a\), propagates \(1 \mathrm{~m}\) through a lossy dielectric rated at \(0.1 \mathrm{~dB} / \mathrm{cm}\), reflects at normal incidence at a boundary at which \(\Gamma=0.3+j 0.4\), and then returns to point \(a .\) Calculate the ratio of the final power to the incident power after this round trip, and specify the overall loss in decibels.

A uniform plane wave in region 1 is normally incident on the planar boundary separating regions 1 and 2. If \(\epsilon_{1}^{\prime \prime}=\epsilon_{2}^{\prime \prime}=0\), while \(\epsilon_{r 1}^{\prime}=\mu_{r 1}^{3}\) and \(\epsilon_{r 2}^{\prime}=\mu_{r 2}^{3}\), find the ratio \(\epsilon_{r 2}^{\prime} / \epsilon_{r 1}^{\prime}\) if \(20 \%\) of the energy in the incident wave is reflected at the boundary. There are two possible answers.

The plane \(z=0\) defines the boundary between two dielectrics. For \(z<0\), \(\epsilon_{r 1}=9, \epsilon_{r 1}^{\prime \prime}=0\), and \(\mu_{1}=\mu_{0} .\) For \(z>0, \epsilon_{r 2}^{\prime}=3, \epsilon_{r 2}^{\prime \prime}=0\), and \(\mu_{2}=\mu_{0}\) Let \(E_{x 1}^{+}=10 \cos (\omega t-15 z) \mathrm{V} / \mathrm{m}\) and find \((a) \omega ;(b)\left\langle\mathbf{S}_{1}^{+}\right\rangle ;(c)\left\langle\mathbf{S}_{1}^{-}\right\rangle\); (d) \(\left\langle\mathbf{S}_{2}^{+}\right\rangle\).

You are given four slabs of lossless dielectric, all with the same intrinsic impedance, \(\eta\), known to be different from that of free space. The thickness of each slab is \(\lambda / 4\), where \(\lambda\) is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally incident. The slabs are to be arranged such that the airspaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and airspaces such that \((a)\) the wave is totally transmitted through the stack, and \((b)\) the stack presents the highest reflectivity to the incident wave. Several answers may exist.

A left-circularly polarized plane wave is normally incident onto the surface of a perfect conductor. (a) Construct the superposition of the incident and reflected waves in phasor form. (b) Determine the real instantaneous form of the result of part ( \(a\) ). \((c)\) Describe the wave that is formed.
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