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(B) The propagation of radio waves into a tunnel depends on the phenomenon of diffraction. Diffraction is significant when the wavelength $(\lambda)$

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Radio signal of frequency $f_2$

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(B) The propagation of radio waves into a tunnel depends on the phenomenon of diffraction. Diffraction is significant when the wavelength $(\lambda)$ of the wave is comparable to the size of the aperture (diameter $D = 10 \, m$).The wavelength is given by $\lambda = c/f$,where $c \approx 3 	imes 10^8 \, m/s$.For $f_1 = 3 	imes 10^6 \, Hz$,$\lambda_1 = (3 	imes 10^8) / (3 	imes 10^6) = 100 \, m$.For $f_2 = 30 	imes 10^6 \, Hz$,$\lambda_2 = (3 	imes 10^8) / (30 	imes 10^6) = 10 \, m$.For $f_3 = 3 	imes 10^9 \, Hz$,$\lambda_3 = (3 	imes 10^8) / (3 	imes 10^9) = 0.1 \, m$.Since the diameter of the tunnel is $10 \, m$,the signal with $\lambda_2 = 10 \, m$ matches the aperture size,leading to effective diffraction and better penetration into the tunnel compared to the others.

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