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10.18 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, which can be sent between the towers without appreciable diffraction effects?

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Payal Gupta | Contributor-Level 10

5 months ago

10.18 Distance between the towers, d = 40 km

Height of the line joining the hills, h = 50 m

Thus, the radial spread of the radio wave should not exceed 40 km

Since the hill is located halfway between the towers,

Fresnel's distance $Z_{P}$ = $\frac{40}{2}$ = 20 km = 2 $* 10^{4} m$

Aperture can be taken as a = h = 50 m

Fresnel's distance is given by the relation,

$Z_{P}$ = $\frac{a^{2}}{?}$ or

$? = \frac{a^{2}}{Z_{P}}$ = $\frac{50^{2}}{2 * 10^{4}}$ = 0.125 m = 12.5 cm

Therefore, the wavelength of the radio wave is 12.5 cm

10.18 Distance between the towers, d = 40 kmHeight of the line joining the hills, h = 50 mThus, the radial spread of the radio wave should not exceed 40 kmSince the hill is located halfway between the towers, Fresnel's distance <math> <msub> <mrow> <mrow> <mi>Z</mi> </mrow> </mrow> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> </msub> </math>  = <math> <mfrac> <mrow> <mrow> <mn>40</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> = 20 km = 2 <math> <mo>*</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mi> </mi> <mi>m</mi> </math> Aperture can be taken as a = h = 50 mFresnel's distance is given by the relation, <math> <msub> <mrow> <mrow> <mi>Z</mi> </mrow> </mrow> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>? </mi> </mrow> </mrow> </mfrac> </math> or <math> <mi>? </mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>Z</mi> </mrow> </mrow> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> =  <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mn>50</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi mathvariant="normal"> </mi> <mo>*</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 0.125 m = 12.5 cmTherefore, the wavelength of the radio wave is 12.5 cm

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10.18 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, which can be sent between the towers without appreciable diffraction effects?

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