Final answer:
The cutoff frequencies for the TE₁₀ and TE₂₀ modes of a rectangular waveguide with dimensions a=1 cm and b=0.6 cm are 15 GHz and 30 GHz, respectively, calculated using the formula based on the dimensions of the waveguide and the mode indices.
Step-by-step explanation:
The student has asked to determine the cutoff frequencies for the TE₁₀ and TE₂₀ modes of a rectangular waveguide with dimensions a=1 cm and b=0.6 cm. The cutoff frequency for a given mode in a rectangular waveguide can be calculated using the formula:
f_c = (c/2) * sqrt((n/a)^2 + (m/b)^2),
where f_c is the cutoff frequency, c is the speed of light in a vacuum (approximately 3 x 10^8 m/s), n and m are the mode indices, a is the width of the waveguide, and b is the height of the waveguide.
For the TE₁₀ mode (n=1, m=0), we have:
f_c(TE₁₀) = (3 x 10^8 m/s / 2) * sqrt((1/0.01 m)^2 + (0/0.006 m)^2) = 15 x 10^9 Hz,
and for the TE₂₀ mode (n=2, m=0), we have:
f_c(TE₂₀) = (3 x 10^8 m/s / 2) * sqrt((2/0.01 m)^2 + (0/0.006 m)^2) = 30 x 10^9 Hz.
Therefore, the cutoff frequencies for the TE₁₀ and TE₂₀ modes are 15 GHz and 30 GHz, respectively.