# A rectangular waveguide has dimensions of 5.1 cm × 2.4 cm. For the dominant mode TE10 , the cut-off frequency is

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A rectangular waveguide has dimensions of 5.1 cm × 2.4 cm. For the dominant mode TE10 , the cut-off frequency is
1. 2.94 GHz
2. 5.88 GHz
3. 6.25 GHz
4. 68.99 GHz

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Correct Answer - Option 1 : 2.94 GHz

Concept:

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

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.

c = speed of light = 3 × 1010 cm/s

Calculation:

For TE10 mode, m = 1, and n = 0.

The cut-off frequency will be:

${{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{0}{b} \right)}^{2}}}$

${{f}_{c\left( mn \right)}}=\frac{c}{2a}$

With a = 5.1 cm, the cut-off frequency will be:

${{f}_{c\left( mn \right)}}_{TE_{(1,0)}}=\frac{3\times 10^{10}}{2\times 5.1}$

fc(1,0) = 2.94 GHz