Two towers on top of two hills are 40 km apart.

Distance between the towers, d = 40 km

Height of the line joining the hills, d = 50 m.

Thus, the radial spread of the radio waves should not exceed 50 km.

Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:

ZP = 20 km = 2 × 104 m

Aperture can be taken as:

a = d = 50 m

Fresnel’s distance is given by the relation,

$Z_{\mathrm{p}}=\frac{a^{2}}{\lambda}$

Where,

λ = Wavelength of radio waves

$\therefore \lambda=\frac{a^{2}}{Z_{\mathrm{P}}}$

$=\frac{(50)^{2}}{2 \times 10^{4}}=1250 \times 10^{-4}=0.1250 \mathrm{~m}=12.5 \mathrm{~cm}$

Therefore, the wavelength of the radio waves is 12.5 cm.