Question

Two towers on top of two hills are 40km apart. The line joining them passes 50m 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 ?

Answer 1

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

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\orange{Solution:}}}}}}$

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$\hookrightarrow$ Distance between the towers, $\tt d\:=\:40km$

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$\hookrightarrow$ Height of the line joining the hills, $\tt d_1\:=\:50m$

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

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$\hookrightarrow$ Since the hill is located halfaway between the towers, Fresnel's distance can be obtained as

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$\tt Z_p\:=\:20\,km\:=\:2\times 10^4$

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Aperture can be taken as,

$\tt a\:=\:d_1\:=50m$

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Fresnel's distance is given by the relation,

$\tt Z_p\:=\: \dfrac{a^2}{\lambda}$

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Where, $\lambda$ = Wavelength of radio waves

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$\tt \therefore\:\lambda\:=\: \dfrac{a^2}{z_p}$

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$\tt \lambda\:=\: \dfrac{(50)^2}{2\times 10^4}$

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$\tt \lambda\:=\:1250\times 10^{-4}\:=\:0.1250\,m\:=\:12.5\,cm$

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$\hookrightarrow$ Therefore, the wavelength of the radio waves is $\tt{\pink{12.5\,cm}}$