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Additional Exercises · 10.18

Q.Two towers on top of two hills are 40 km40\ \text{km} apart. The line joining them passes 50 m50\ \text{m} above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

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Treating the 50 m50\ \text{m} clearance over the hill as the effective aperture, the Fresnel-distance condition zF=a2/λ≥xz_F = a^2/\lambda \geq x (with xx the tower-to-hill distance) gives the longest wavelength that avoids appreciable diffraction spread: λmax=0.125 m\lambda_{max} = 0.125\ \text{m}.

Step 1: Identify the geometry

The two towers are 40 km40\ \text{km} apart, with a hill exactly halfway between them, so the distance from each tower to the hill is:

x=40 km2=20 km=2×104 mx = \frac{40\ \text{km}}{2} = 20\ \text{km} = 2\times10^{4}\ \text{m}

The line joining the towers clears the hill by a=50 ma = 50\ \text{m} — this clearance acts as the effective aperture through which the radio beam must pass without being significantly diffracted by the hill.

Step 2: Condition for negligible diffraction

For the transmitted beam to travel to the far tower without spreading out appreciably (i.e. for line-of-sight/ray propagation to remain a good approximation over the path length xx), the Fresnel distance for the aperture aa must be at least as large as the actual path length: …

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