Two towers on top of two hills are 40 , km ap | vedclass

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(C) The distance between the two towers is $D = 40 \, km = 40,000 \, m$. The hill is located at the midpoint,so the distance from either tower to the

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$12.5$

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(C) The distance between the two towers is $D = 40 \, km = 40,000 \, m$. The hill is located at the midpoint,so the distance from either tower to the hill is $Z_F = D/2 = 20,000 \, m$.For radio waves to travel between the towers without significant diffraction,the Fresnel distance $Z_F$ must be at least equal to the distance to the obstacle. The condition for the Fresnel distance is given by $Z_F = \frac{a^2}{\lambda}$,where $a$ is the size of the aperture (or the clearance height) and $\lambda$ is the wavelength.Here,the clearance height is $a = 50 \, m$.Rearranging the formula to solve for the wavelength $\lambda$:$\lambda = \frac{a^2}{Z_F}$Substituting the values:$\lambda = \frac{(50 \, m)^2}{20,000 \, m} = \frac{2500}{20,000} \, m = 0.125 \, m$.Converting the wavelength to centimeters:$\lambda = 0.125 \, m 	imes 100 \, cm/m = 12.5 \, cm$.Thus,the longest wavelength is $12.5 \, cm$.

Two towers on top of two hills are $40 \, km$ apart. The line joining them passes $50 \, m$ above a hill halfway between the towers. What is the longest wavelength of radio waves (in $cm$),which can be sent between the towers without appreciable diffraction effects?

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