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(D) The path difference between the direct wave and the reflected wave is given by $\Delta x = (	ext{Path of reflected wave}) - (	ext{Path of direct

Vedclass · Accepted Answer

$\frac{\lambda}{4 \cos \alpha}$

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(D) The path difference between the direct wave and the reflected wave is given by $\Delta x = (	ext{Path of reflected wave}) - (	ext{Path of direct wave}) + \frac{\lambda}{2}$.From the geometry,the path difference is $\Delta x = (AB + BN) + \frac{\lambda}{2}$,where the $\frac{\lambda}{2}$ term accounts for the phase change upon reflection from a denser medium (water).In $\Delta ACB$,we have $\cos \alpha = \frac{h}{AB}$,so $AB = \frac{h}{\cos \alpha}$.In $\Delta ANB$,the path difference component is $BN = AB \cos(2\alpha)$.Thus,$\Delta x = AB(1 + \cos 2\alpha) + \frac{\lambda}{2} = AB(2 \cos^2 \alpha) + \frac{\lambda}{2}$.Substituting $AB = \frac{h}{\cos \alpha}$,we get $\Delta x = \frac{h}{\cos \alpha} (2 \cos^2 \alpha) + \frac{\lambda}{2} = 2h \cos \alpha + \frac{\lambda}{2}$.For maximum intensity,the path difference must be an integer multiple of $\lambda$. For the first maximum,we set $\Delta x = \lambda$:$\lambda = 2h \cos \alpha + \frac{\lambda}{2} \implies \frac{\lambda}{2} = 2h \cos \alpha \implies h = \frac{\lambda}{4 \cos \alpha}$.

Radio waves coming at an angle $\alpha$ to the vertical are received by a radar after reflection from a nearby water surface and directly. What should be the height of the antenna from the water surface so that it records a maximum intensity? (wavelength $= \lambda$)

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