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Work out the theory of TM modes for a rectangular wave guide. In particular, find the longitudinal electric field, the cutoff frequencies, and the wave and group velocities. Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency, for a given wave guide. [Caution: What is the lowest TM mode?]

Short Answer

Expert verified

The longitudinal electric field is Ez=E0sinmπxasinnπya, the cut-off frequency is

ωmn=cπma2+nb2, the wave velocity isV=C1-ωmnω2, and the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency for a given waveguide is ω11ω10=1+ab2.

Step by step solution

01

Expression for the components of an electric and magnetic field along the z-axis in a rectangular wave:

Write the expression for the components of electric and magnetic fields along the z-axis in a rectangular wave.

[2x2+2y2+ωc2=k2]Ez=0[2x2+2y2+ωc2-k2]Bz=0

Here,Ez is the longitudinal component of electric field and Bzis the longitudinal component of a magnetic field, ωis the frequency of a wave, c is the speed of light, and k is the wavenumber.

02

Determine the longitudinal electric field:

For the TM wave, the value of the longitudinal component of the magnetic field is zero.

Write the boundary conditions at the wall.

E=0,B=0

Let,Ez(x,y)=X(x)Y(y)

Here, X(x)=Asin(kxx)+Bcos(Kxx)

At walls Ez=0, then at x=0, the value of X and B will be,

X=0B=0

Hence, it is known that:

kx=mπa;m=1,2,3ky=nπa;n=1,2,3.

So, the longitudinal electric field will be,

Ez=E0sinmπxasinnπya

03

Determine the cut off frequency and wave velocity:

Write the expression for wave number.

k=ωc2-π2ma2+nb2

Hence, the cut off frequency will be,

ωmn=cπma2+nb2

Write the expression for the wave velocity,

v=ωk …… (1)

Write the expression for the lowest cut-off frequency (ω11)for modeTM11.

ω11=cπ1a2+1b2 …… (2)

Hence, the wavenumber in terms of the cut-off frequency will be,

k=1cω2-ωmn2

Substitute ω11=cπ1a2+1b2andk=1cω2-ωmn2in equation (1).

v=cπ1a2+1b21cω2-ωmn2

V=C1-ωmnω2

04

Determine the group velocity:

Write the expression for the group velocity .

Vg=1dkdω

Substitute k=1cω2-ωmn2in the above expression.

vg=1ddω1cω2-ωmn2Vg=c1-ωmnω2

Write the expression for the lowest cut-off frequency for Transverse electric mode (TE).

ω10=cπa …… (3)

Take the ratio of equations (2) and (3).

ω11ω10=cπ1a2+1b2cπaω11ω10=1+ab2

Therefore, the longitudinal electric field is Ez=E0sinmπxasinnπya,

the cut-off frequency is ωmn=cπma2+nb2, the wave velocity is

V=C1-ωmnω2, the group velocity is Vg=c11-ωmnω2, and the ratio of the

lowest TM cutoff frequency to the lowest TE cutoff frequency for a given wave guide isω11ω10=1+ab2 .

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Light of (angular) frequency w passes from medium , through a slab (thickness d) of medium 2, and into medium 3(for instance, from water through glass into air, as in Fig. 9.27). Show that the transmission coefficient for normal incidence is given by

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[The naive explanation for the pressure of light offered in section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here's another account, due originally to Planck.] A plane wave travelling through vaccum in the z direction encounters a perfect conductor occupying the region z0, and reflects back:

E(z,t)=E0[coskz-ωt-coskz+ωt]x^,(z>0)

  1. Find the accompanying magnetic field (in the region (z>0))
  2. Assuming B=0inside the conductor find the current K on the surface z=0, by invoking the appropriate boundary condition.
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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.]

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