An electromagnetic wave emitted by source travels 21 km to arrive at a receiver. The wave while travelling in another path is reflected from a surface at 19 km away and further travels 12 km to reach the same receiver. If destructive interference occurs at the receiving end, the maximum wavelength of the wave is

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5 km

0.5 km

20 km

10 km

answer is B.

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Detailed Solution

The path difference $= (19 + 12) - 21 = 10km\$

As the reflected wave produces the destructive interference, the phase difference is nπ.where n = 1,3,5,.. waves = $\large \frac{{2\pi }}{\lambda } \times \,path\,\,difference$

$\large \therefore \,\frac{{2\pi }}{\lambda } \times 10\, = \,n\pi \,\left( {n = 1,3,5,...} \right)$

$\large \Rightarrow \,\lambda \, = \,\frac{{20}}{n}km$

So $\large {\lambda _{\max }}\, = \,\frac{{20}}{1}km\, = \,20km$

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