91Ó°ÊÓ

The cutoff frequency \(\omega_c\) in a rectangular waveguide of dimensions \(a\) and \(b\) is given by the following equations for TE and TM modes. \begin{align} \omega_{c,\text{TE}} = \frac{c}{2\pi}\sqrt{\frac{(m\pi/a)^2+(n\pi/b)^2}{\mu\epsilon}}, \end{align} \begin{align} \omega_{c,\text{TM}} = \frac{c}{2\pi}\sqrt{\frac{(m\pi/a)^2+(n\pi/b)^2}{\mu\epsilon}}, \end{align} where \(c\) is the speed of light, \(\mu\) and \(\epsilon\) are the permeability and permittivity of the medium, respectively, and \(m\) and \(n\) are the mode indices.

We are given the frequency \(\omega_{\max}x\). To find the total number of modes that have a cutoff frequency less than this given frequency, we need to find the mode indices \((m, n)\) for both TE and TM modes such that their cutoff frequencies are less than the given frequency. For TE and TM modes, the expression for the cutoff frequencies is the same: $$\omega_c = \frac{c}{2\pi}\sqrt{\frac{(m\pi/a)^2+(n\pi/b)^2}{\mu\epsilon}} \leq \omega_{\max}x.$$ We can solve this inequality to find the values of \(m\) and \(n\). First, let's rearrange the inequality: $$\frac{(m\pi/a)^2+(n\pi/b)^2}{\mu\epsilon} \leq \frac{4\pi^2\omega_{\max}^2 x^2}{c^{-2}}.$$ As we are looking for the number of modes \(N\), we can sum over the mode indices: $$N = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} (m, n) = \frac{\omega_{\max}^2}{2\pi c^2}ab,$$

Next, we need to find the derivative of \(N\) with respect to \(\omega\) to compute the density of modes per unit frequency interval, \(dN/d\omega\). Taking the derivative of \(N\) with respect to \(\omega\), we get: $$\frac{dN}{d\omega} = \frac{d}{d\omega} \left( \frac{\omega_{\max}^2}{2\pi c^2} ab \right) = \frac{2\omega_{\max}}{2\pi c^2} ab,$$ where we can see that the density of modes per unit frequency interval increases linearly with frequency, as required.