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

(9/4) fold

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

∴ Total phase difference = $π - π = 0$

Thus the two waves superimpose in phase.

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

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