A source emits electromagnetic waves of wavelength 3m. One beam reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and with an intensity reduced to 1/4 as compared to intensity due to the direct beam alone. The resultant intensity will be.

(1/4) fold

(3/4) fold

(5/4) fold

(9/4) fold

Solution

(9/4) fold

We know that a phase change of π occurs when the reflection takes place at the boundary of the denser medium. This is equivalent to a path difference of λ/2.

Therefore, Total phase difference = π -π = 0

Thus, the two waves superimpose in phase.

Resultant,

$Resultant amplitude = \sqrt{I} + \sqrt{(\frac{I}{4})} = \frac{3}{2} \sqrt{I} Resultant amplitude = {(\frac{3}{2} \sqrt{I})}^{2} = \frac{9}{4} I = \frac{9}{4} fold$

Some More Questions From Communication Systems Chapter

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Communication Systems

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