Question

Consider a rectangular waveguide of internal dimensions 8 cm × 4 cm. assuming an H₁₀ mode of propagation the critical wavelength would be
1. 8 cm
2. 16 cm
3. 4 cm
4. 32 cm

Answer 1

Correct Answer - Option 2 : 16 cm

Concept:

The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)

'm' and 'n' represents the possible modes.

The cut-off/critical wavelength is given as:

\({{\rm{λ }}_{\rm{C}}} = \frac{2}{{\sqrt {{{\left( {\frac{{\rm{m}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{\rm{n}}}{{\rm{b}}}} \right)}^2}} }}\)

Calculation:

Given: a = 8 cm, b = 4 cm

For H₁₀ mode

i.e. m = 1 and n = 0, we get

\({{\rm{λ }}_{\rm{C}}} = \frac{2}{{\sqrt {{{\left( {\frac{{\rm{1}}}{{\rm{8}}}} \right)}^2} + {{\left( {\frac{{\rm{0}}}{{\rm{4}}}} \right)}^2}} }}\)

λ_C = 2 × 8 cm = 16 cm