Show that parallel conducting planes of separation \(a\) can support a TE mode identical to the TE \(_{10}\) mode in rectangular waveguide with \(b \rightarrow \infty\). Show further that the parallel planes can also support a TM mode that has no direct analog in rectangular waveguide \((b\) finite) but is of the form of \((8.7 .32)\) and \((8.7 .33)\) with \(m \rightarrow 0, \sin m \pi y / b \rightarrow 1 .\)

For parallel conducting planes, we must apply the boundary conditions and find the allowed Transverse Electric (TE) fields. In TE modes, there is no electric field component in the direction of propagation, i.e. \(E_z = 0\). Let's consider parallel conducting planes of separation \(a\) in the \(z = 0\) and \(z = a\) planes. Applying the necessary boundary conditions on the parallel conducting planes, we get the magnetic field components as follows: \(H_y = H_{0} \sin(\frac{n \pi x}{a}) e^{j\omega (t - \frac{x}{v_p} )}\) \(E_x = - \frac{j \omega \mu_0 n \pi}{a k^2} H_{0} \cos(\frac{n \pi x}{a}) e^{j\omega (t - \frac{x}{v_p} )}\) \(E_y = 0\) Here, \(H_{0}\) is the amplitude of the magnetic field, \(n\) is an integer, and \(v_p\) is the phase velocity. Note that these field components satisfy the boundary conditions for the parallel conducting planes.

Now, let's consider a rectangular waveguide with dimensions \(a\) and \(b\). For the TE \(_{10}\) mode in a rectangular waveguide, the magnetic field component is given by: \(H_y = H_{0} \sin(\frac{\pi x}{a}) e^{j\omega (t - \frac{x}{v_p})}\) \(E_x = - \frac{j \omega \mu_0 \pi}{a k^2} H_{0} \cos(\frac{\pi x}{a}) e^{j\omega (t - \frac{x}{v_p})}\) \(E_y = 0\) Comparing these field components to the previous ones for parallel conducting planes, we conclude that when \(n=1\), the TE mode supported by the parallel conducting planes is identical to the TE \(_{10}\) mode in a rectangular waveguide.

For parallel conducting planes of separation \(a\), we can find the allowed Transverse Magnetic (TM) modes. In TM modes, there is no magnetic field component in the direction of propagation, i.e., \(H_z = 0\). Let's consider \(m = 0\) case for the parallel conducting planes. Applying the necessary boundary conditions on the parallel conducting planes, we get the electric field components as follows: \(E_z = E_{0} \sin(\frac{n \pi x}{a}) e^{j\omega (t - \frac{x}{v_p} )}\) \(E_y = - \frac{j \omega \varepsilon_0 n \pi}{a k^2} E_{0} \cos(\frac{n \pi x}{a}) e^{j\omega (t - \frac{x}{v_p} )}\) \(H_x = 0\) Here, \(E_{0}\) is the amplitude of the electric field and \(n\) is an integer. Note that these field components satisfy the boundary conditions for the parallel conducting planes and have the structure of \((8.7 .32)\) and \((8.7 .33)\) formulas but with \(m \rightarrow 0\). When \(m = 0\), the \(\sin m \pi y / b\) term will approach 1, as mentioned in the problem. Thus, we have shown that parallel conducting planes of separation \(a\) can support a TE mode identical to the TE \(_{10}\) mode in a rectangular waveguide with $b \rightarrow \infty\(, and a TM mode with \)m \rightarrow 0\( and \)\sin m \pi y / b \rightarrow 1\( has no direct analog in a rectangular waveguide with finite \)b$.