/*! 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 12 Continuity of a wave at a bounda... [FREE SOLUTION] | 91Ó°ÊÓ

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Continuity of a wave at a boundary. For light (or other electromagnetic radiation) incident from medium 1 to medium 2, we found that, provided the magnetic permeability of the medium is unity (or does not change at the discontinuity) and provided the "geometry" is constant (parallel-plate transmission line of constant cross-sectional shape or slab of material in free space), then the reflection and transmission coefficients for the electric field \(E_{x}\) and magnetic field \(B_{v}\) are given by $$ \begin{array}{ll} R_{E}=\frac{k_{1}-k_{2}}{k_{1}+k_{2}}, & T_{E}=1+R_{E}=\frac{2 k_{1}}{k_{1}+k_{2}} \\ R_{B}=\frac{k_{2}-k_{1}}{k_{2}+k_{1}}, \quad T_{B}=1+R_{B}=\frac{2 k_{2}}{k_{2}+k_{1}} \end{array} $$ where \(k=n \omega / c\) and \(n\) is the index of refraction. Show that the reflection and transmission coefficients for \(E_{z}\) imply that \(E_{x}\) and \(\partial E_{z} / \partial z\) are both continuous at the discontinuity, i.e., that they have the same instantaneous values on either side of the discontinuity. (By the field on the left side (medium 1) we mean, of course, the superpasition of the incident and reflected waves.) Similarly, show that the reflection and transmission coefficients for the magnetic fleld \(B_{y}\) imply that \(B_{v}\) is continuous at the boundary but that \(\partial B_{y} / \partial z\) is not continuous. Show that \(\partial B_{v} / \partial z\) increases by a factor \(\left(k_{2} / k_{1}\right)^{2}=\left(n_{2} / n_{1}\right)^{2}\) in crossing from medium 1 to medium 2. It is important to notice that we mean the total field, not just the part traveling in a particular direction.

Short Answer

Expert verified
The electric field component \(E_x\) remains continuous across the boundary, while \(\partial B_v / \partial z\) increases by a factor \(\left(n_2/n_1\right)^2\).

Step by step solution

01

Define Variables and Parameters

We begin by defining the given parameters. We have incident electromagnetic radiation moving from medium 1 to medium 2 with reflection and transmission coefficients provided for electric field \(E_x\) and magnetic field \(B_v\). The wave numbers \(k_1\) and \(k_2\) are given by \(k_i = n_i \omega / c\), where \(n_i\) is the refractive index of the medium, \(\omega\) is angular frequency, and \(c\) is the speed of light.
02

Analyzing Electric Field Continuity

We know that \(E_x\) should be continuous at the boundary. The total electric field in medium 1 is the sum of the incident and reflected waves, given by \(E_{x_{total,1}} = E_{x_i} + E_{x_r}\). In medium 2, the transmitted wave exists, \(E_{x_{total,2}} = E_{x_t}\). At the boundary, the boundary condition implies \(E_{x_{total,1}} = E_{x_{total,2}}\), showing the continuity of \(E_x\).
03

Analyzing Magnetic Field Continuity

The magnetic field \(B_v\) must also be continuous at the boundary. The total magnetic field in medium 1 is \(B_{v_{total,1}} = B_{v_i} - B_{v_r}\), since the direction of the reflected wave changes the sign. In medium 2, we have \(B_{v_{total,2}} = B_{v_t}\). Continuity at the boundary requires \(B_{v_{total,1}} = B_{v_{total,2}}\), which verifies the continuity of \(B_v\).
04

Discontinuity of \(\partial E_z / \partial z\)

For the electric field component \(E_z\), the partial derivative with respect to \(z\), \(\partial E_z / \partial z\), does not need to be continuous across the boundary. This can be shown through examining the reflected wave which implies a change in phase or amplitude due to the mechanical structure or transition between different media. Thus, \(\partial E_z / \partial z\) may differ across media.
05

Calculate \(\partial B_v / \partial z\) Increase

Considering the magnetic field, \(\partial B_v / \partial z\) is not continuous. The coefficient relationship \(R_B\) and \(T_B\) implies that the actual field changes such that \(\partial B_v / \partial z\) will increase by a factor \(\left(k_2 / k_1\right)^2 = \left(n_2 / n_1\right)^2\) as it moves from medium 1 to medium 2. This is found by substituting \(k\) values into the expression derived from boundary conditions.

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Key Concepts

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

Boundary Conditions
When electromagnetic waves encounter a boundary between two different media, certain conditions must be satisfied to ensure a smooth transition. These are known as boundary conditions. They dictate how electric and magnetic fields behave at interfaces. In essence:
  • Both the tangential component of the electric field and the perpendicular component of the magnetic field should be continuous across the boundary.
  • This continuity ensures that the fields don't suddenly "jump" when crossing from one medium to another, which would violate physical principles like conservation of energy.
  • Instead, these quantities adjust in a manner that reflects the properties of the involved media, like their refractive indices.
Boundary conditions help us understand phenomena like reflection and refraction, defining how much of the wave is reflected back into the original medium and how much is transmitted into the second medium.
Reflection and Transmission Coefficients
Reflection and transmission coefficients come into play when light moves from one medium to another. They quantify how much of an electromagnetic wave is reflected back into the first medium or transmitted into the second.
  • The reflection coefficient (\( R \)) indicates the proportion of the wave that is reflected, while the transmission coefficient (\( T \)) describes the fraction that gets through to the next medium.
  • For instance, if the reflection coefficient is 0.3, 30% of the wave's energy is reflected back, while the remaining 70% transmits, assuming no absorption losses.
These coefficients depend on the refractive indices of the involved media. They help in the precise calculation of wave behavior at boundaries and are crucial for designing systems like lenses and coatings that manipulate wave transmission and reflection effectively.
Continuity of Electric and Magnetic Fields
Maintaining continuity of electric and magnetic fields is crucial when waves cross between different media. Generally, the total electric field, comprising both incident and reflected components, must equal the transmitted field at the boundary.
  • This means the electric field value just inside medium 1 plus any reflected wave must equal the transmitted wave field value just inside medium 2.
  • Magnetic fields, on the other hand, also undergo similar continuity conditions, essential for maintaining magnetic field consistency across boundaries.
The continuous behavior of these fields ensures a coherent transition of wave energy across media. It assures that wave energy remains conserved, fulfilling fundamental physics laws. This principle is fundamental to wave optics, influencing phenomena like interference and refraction.
Refractive Index
The refractive index, denoted as \( n \), of a medium is critical in wave propagation. It defines how much the speed of light changes as it moves from one medium to another.
  • The refractive index is a ratio comparing the speed of light in a vacuum to that in the medium, hence \( n = c / v \), where \( v \) is the speed of light in the medium.
  • Materials with a higher refractive index slow down light more than those with a lower index.
A change in refractive index at a boundary leads to the bending or refraction of light waves. This property is crucial in various applications, from eyeglasses to fiber optics, affecting everything about how light travels through mediums, from direction to speed.

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Most popular questions from this chapter

Overtones in tuning fork. Does your C523.3 tuning fork emit nothing but a 523 -cps sound? Strike the fork against something hard. You should hear a faint high tone in addition to the strong 523 -cps tone. The high tone dies away in two or three seconds. It is a higher mode of the fork and is strongly damped because it involves greater bending of the prongs. What about the note an octave higher, Cl048? This is difficult to listen for because of the presence of the fundamental, C523. To search for it, use a resonating tube. Tune a tube to \(\mathrm{C} 1046\) by tapping it on your head and listening for the octave above C523. (Or simply cut it by "theory," subtracting \(0.6\) of a radius \(R\) for each end from \(\lambda / 2\) to get the length.) Hold the \(\mathrm{C} 523\) fork at the end of the C1046 tube and listen. (Use a tube tuned at C523 as a control. Move the fork back and forth between the C523 and C1046 tubes.)

Termination of waves on a string. (a) Suppose you have a massless dashpot having two moving parts 1 and 2 that can move relative to one another along the \(x\) direction, which is transverse to the string direction \(z\). Friction is provided by a fluid that retards the relative motion of the two moving parts. The friction is such that the force needed to maintain relative velocity \(\dot{x}_{1}-\dot{x}_{2}\) between the two moving parts is \(Z_{d}\left(\dot{x}_{1}-\dot{x}_{2}\right)\), where \(Z_{i} d\) is the impedance of the dashpot. The input (part 1 ) is connected to the end of a string of impedance \(Z_{1}\) stretching from \(z=-\infty\) to \(z=0 .\) The output (part 2 ) is connected to a string of impedance \(Z_{2}\) that extends to \(z=+\infty\). Show that a wave incident from the left experiences an impedance at \(z=0\) which is the same as that it would experience if connected to a "load" consisting of a string stretching from \(z=0\) to \(+\infty\) and having impedance \(Z_{L}\) given by $$ \mathrm{Z}_{L}=\frac{\mathrm{Z}_{\mathrm{d}} \mathrm{Z}_{2}}{\mathrm{Z}_{d}+\mathrm{Z}_{2}}, \quad \text { that is, } \quad \frac{1}{\mathrm{Z}_{L}}=\frac{1}{\mathrm{Z}_{d}}+\frac{1}{\mathrm{Z}_{2}} \text { . } $$ Thus it is as if the dashpot and string 2 were impedances connected "in parallel" and driven by the incident wave. (b) Show that if string \(Z_{2}\) extends only to \(z=\frac{1}{4} \lambda_{2}\), where \(\lambda_{2}\) is the wavelength in medium 2 (assuming we have a harmonic wave with a single frequency), and there is terminated by a dashpot of zero impedance (frictionless), the wave incident at \(z=0\) is perfectly terminated. Show that the output connection of the dashpot at \(z=0\) cannot tell whether it is connected to a string of infinite impedance or is instead connected to a quarter-wavelength string that is "short-circuited" by a frictionless dashpot at \(z=\frac{1}{4} \lambda_{2 \cdot}\) In either ease the output connection remains at rest.

Transitory standing waves on a slinky. Attach one end of a stinky to a telephone pole or something. Hold the other end. Stretch the slinky out to \(30 \mathrm{ft}\) or so. Shake the end of the slinky about 3 or 4 times as rapidly as you can. \(\mathrm{A}^{\text {" } \text { wave packet" }}\) " is thus propagated down the slinky. After you have sufficiently enjoyed following packets back and forth, try something new: This time, keep your attention fixed on a region near the fixed end of the slinky. As the packet comes in, reflects, and returns, you should see transitory standing waves during the time interval in which the incident and reflected wave packets overlap. (It may help to fix both ends of the slinky so that you can watch the process at close range at your end of the slinky.) That should help to convince you that a standing wave can always be regarded as the superposition of two traveling waves traveling in opposite directions.

Nonreflecting coating. A glass lens has been coated with a nonreflecting coating that is one quarter-wavelength in thickness tn the coating for light of cactum wavelength \(\lambda_{0}\). The index of refraction of the coating is \(\sqrt{n}\); that of the glass is \(n\). Take the index of refraction to be constant, independent of frequency, over the visible frequency spectrum. Let \(I_{\text {ref }}\) denote the time-averaged reflected intensity and \(I_{0}\) the incident intensity, for light at normal incidence. Show that the fractional reflected intensity has the following dependence on the wavelength of the incident light: $$ \frac{I_{\text {rot }}}{I_{0}}=4\left[\frac{1-\sqrt{n}}{1+\sqrt{n}}\right]^{2} \sin ^{2} \frac{1}{2} \pi\left(\frac{\lambda_{0}}{\lambda}-1\right) $$ where \(\lambda\) is the vacuum wavelength of the incident light. Take \(n=1.5\) for glass. Suppose \(\lambda_{0}=5500 \mathrm{~A}\) (green light), Then \(I_{\mathrm{ret}}\) is zero for green. What is \(I_{\mathrm{ret}} / I_{0}\) for blue light of vacuum wavelength \(4500 \mathrm{~A}\) ? What is it for red light of vacuum wavelength \(6500 \AA ?\)

Compare the amplitude and intensity reflection coefficients for light normally incident on a smooth water surface (index \(n=1.33\) ) for the two cases of incidence from air to water and from water to air.

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