"Correct Answer - Option 1 : 0.9a
The cutoff wavelength of TMmn mode is given as,
\({{\lambda }_{C}}~=~\frac{2}{\sqrt{{{\left( \frac{m}{a} \right)}^{2}}~+~{{\left( \frac{n}{b} \right)}^{2}}}}\) ---(1)
And, the aspect ratio is \(\frac{a}{b}\)
Calculation:
Given:
Aspect ratio \(=~\frac{2}{1}\), TM11 mode,
\(\frac{a}{b}~=~\frac{2}{1}\)
a = 2b
From equation (1)
\({{\lambda }_{C}}~=~\frac{2}{\sqrt{\frac{1}{{{a}^{2}}}~+~\frac{1}{{{b}^{2}}}}}\) ∵ (a = 2b)
\({{\lambda }_{C}}~=~\frac{2a}{\sqrt{5}}~=~0.89~a\approx 0.9~a\)
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