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Confirm that the energy in theTEmnmode travels at the group velocity. [Hint: Find the time-averaged Poynting vector <S>and the energy density <u>(use Prob. 9.12 if you wish). Integrate over the cross-section of the waveguide to get the energy per unit time and per unit length carried by the wave, and take their ratio.]

Short Answer

Expert verified

Answer

The velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.

Step by step solution

01

Expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide:

Write the expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide.

V=<S>dA<u>dA …… (1)

Here, v is the velocity of electromagnetic wave, Sis the time-averaged pointing vector for the electromagnetic wave, uis the energy per unit volume of the electromagnetic wave, and Ais the cross-section of the waveguide.

Write the expression for the time-averaged pointing vector for the electromagnetic wave.

S=12μ0ReE×B …… (2)

Write the expression for the energy per unit volume of the electromagnetic wave.

u=14Reε0EE+1μ0BB …… (3)

Here, *represents the complex conjugate of the quantity.

02

Determine the time-averaged pointing vector for the electromagnetic wave:

Write the value of E×B.

(B×B)=(Bz*Ey)x^-(Bz*Ex)x^+(Bz*Ex-Bz*Ey)z^

As (Bz*Ey)and (Bz*Ex)are imaginary components, the value of localid="1658406246663" Re(E×B)will be,

localid="1658406255214" Re(E×B)=(Bz*Ex-Bz*Ey)z^

Using equation 9.180 and 9.186, write the value of By*,Ex,Bx*and Ey.

By*=-ik(ωc)2-k2(-nπb)B0cos(mπxa)sin(nπyb)Ex=-(ωc)2-k2(-b)B0cos(mπxa)sin(nπyb)Bx*=-ik(ωc)2-k2(a)B0sin(mπxa)cos(nπyb)Ey=-ik(ωc)2-k2(-a)B0sin(mπxa)cos(nπyb)

Substitute all the above values in equation (2).

<S>=12μ0(By*Ex-Bx*Ey)z^<S>=ωkπ2B02(ωc)2-k2[nb2cos2mπxasin2nπyb+ma2sin2mπxacos2nπybz^]<S>·da=18μ0ωkπ2B02(ωc)2-k2ab[ma2+nb2]<S>·da=18μ0ωkc2B02ab(ωmn2)

03

Determine the energy per unit volume of the electromagnetic wave:

Using the equation9.176, write the value of Eand B.

E=E0ei(kz-ωt)B*=B*0ei(kz-ωt)

Similarly, using equation 9.180 and 9.186, write the value of Bz*and Ez.

Bz*=B0cos(mπxa)cos(nπyb)Ez=0

Substitute the value of localid="1658406358582" Eand localid="1658406367803" Bin equation (3).

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04

Determine the group velocity:


Substitute <S>·da=18μ0ωkc2B02(ωmn2)and <u>·da=B02abω28μ0ωmn2in equation (1).

V18μ0ωkc2B02ab(ωmn3)B02abω28μ0ωmn2V=18μ0ωkc2B02ab(ωmn2)×8μ0ωmn2B02abωV=kc2ωV=cωω2-ωmn2

Therefore, the velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.

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Most popular questions from this chapter

The "inversion theorem" for Fourier transforms states that

ϕ(Z)=-ϕ(k)eikzdkϕ(k)=12π-ϕ(z)e-ikzdz

Use this to determine A(k), in Eq. 9.20, in terms of f(z,0)andf*(z,0)

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

SupposeAeiax+Beibx=Ceicx, for some nonzero constants A, B, C, a, b, c, and for all x. Prove that a = b = cand A + B = C.

a) Derive Eqs. 9.179, and from these obtain Eqs. 9.180.

(b) Put Eq. 9.180 into Maxwell's equations (i) and (ii) to obtain Eq. 9.181. Check that you get the same results using (i) and (iv) of Eq. 9.179.

Question:The index of refraction of diamond is 2.42. Construct the graph analogous to Fig. 9.16 for the air/diamond interface. (Assume .) In particular, calculate (a) the amplitudes at normal incidence, (b) Brewster's angle, and (c) the "crossover" angle, at which the reflected and transmitted amplitudes are equal.

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