Question

The cut off frequency of a waveguide depends upon

Answer 1

Ans : (a & b)

$f_c=\frac{v_0}{2\pi }\sqrt{{\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2}$

where, $V_0=\frac{C}{\sqrt{{\mu }_r{ϵ}_r}}$

$f_c=\frac{c}{2\sqrt{{\mu }_r{ϵ}_r}}\sqrt{{\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2}$

So, $f_c$ depends on 'a' and 'b' (dimensions of waveguide)

$f_c\alpha \frac{1}{\sqrt{{\mu }_r{ϵ}_r}}$

So, $f_c$ depends upon the dielectric property of the medium in the waveguide.