Q130 / 180 · JIPMER 2017 · Physics
A source emits electromagnetic waves of wavelength 3 m . One beam reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and with intensity reduced to $1 / 4$ as compared to intensity due to the direct beam alone. The resultant intensity will be
- A(1/4) fold
- B(3/4) fold
- C(5/4) fold
- Correct answerD(9/4) fold
Solution
We know that a phase change of $\pi$ occurs when the reflection takes place at the boundary of denser medium. This is equivalent to a path difference of $\lambda / 2$.
$\therefore$ Total phase difference $=\pi-\pi=0$
Thus, the two waves superimpose in phase.
$\begin{aligned}
& \text { Resultant amplitude }=\sqrt{I}+\sqrt{\left(\frac{I}{4}\right)}=\frac{3}{2} \sqrt{I} \\
& \text { Resultant intensity }=\left(\frac{3}{2} \sqrt{I}\right)^2=\frac{9}{4} I=\frac{9}{4} \text { fold. }
\end{aligned}$
Hence, the correct answer is option: 'D'.