91Ó°ÊÓ

The radio waves of wavelength, $\mathrm{λ}$reach detector D either in straight line or by reflecting from atmosphere layer.

Find the path difference between the direct and reflected wave. Then using the conditions for constructive and destructive interference, we can find two equations. Solving them will give us the wavelength in terms of d, h, and H.

Formulae:

• i)For constructive interference,$\frac{â–³x}{\mathrm{λ}}=n$
• ii) For destructive interference,$\frac{â–³x}{\mathrm{λ}}=(n+\frac{1}{2})$

Consider the waves are reflected from the layer in the atmosphere, which is at height H, then using Pythagorean Theorem, the path difference between direct and reflected wave is given as:
$$\begin{gathered} â–³x=\sqrt{H^2+{\left(\frac{d}{2}\right)}^2}+\sqrt{H^2+{\left(\frac{d}{2}\right)}^2}-\left(\frac{d}{2}+\frac{d}{2}\right) \\ =2\sqrt{H^2+{\left(\frac{d}{2}\right)}^2}-d \end{gathered}$$

For constructive interference, from equation (i), we get
role="math" localid="1661964754919" $$\begin{gathered} 2\sqrt[]{H^2+{\left(\frac{d}{2}\right)}^2}-d=n\mathrm{λ}...............................\left(1\right) \\ Where,n=1,2,3,4,......... \end{gathered}$$

For destructive interference, when the waves are reflected from the layer in the atmosphere, which is at height H +h, the path difference between direct and reflected wave using equation (ii) is given as:

role="math" localid="1661964716776" $$\begin{gathered} 2\sqrt[]{{(H+h)}^2+{\left(\frac{d}{2}\right)}^2}-d=\left(n+\frac{1}{2}\right)\mathrm{λ}.....................................\left(2\right) \\ Where,n=1,2,3,4.......... \end{gathered}$$

Subtracting equation (2) from (1), we get

Hence, the expression value of wavelength is$2(\sqrt{4{(H+h)}^2+d^2}-\sqrt{4H^2+d^2})$

$$\begin{gathered} 2\sqrt{{(H+h)}^2+{\left(\frac{d}{2}\right)}^2}-2\sqrt{H^2+{\left(\frac{d}{2}\right)}^2}=\frac{1}{2}\mathrm{λ} \\ \sqrt{4{(H+h)}^2+d^2}-\sqrt{4H^2+d^2}=\frac{1}{2}\mathrm{λ} \\ \mathrm{λ}=2(\sqrt{4{(H+h)}^2+d^2}-\sqrt{4H^2+d^2}) \end{gathered}$$