Q127 / 200 · NEET 2008 (Screening) · Physics
Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
- A$I_1+I_2$
- B$\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
- C$\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
- Correct answerD$2\left(I_1+I_2\right)$
Solution
Resultant intensity of two periodic waves is given by
$I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \delta$
where $\delta$ is the phase difference between the waves.
For maximum intensity, $\delta=2 n \pi ; n=0,1,2 \ldots$ etc.
Therefore, for zero order maxima, $\cos \delta=1$
$I_{\text {max }}=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
For minimum intensity, $\delta=(2 n-1) \pi$;
$n=1,2 \ldots \text { etc }$
Therefore, for Ist order minima, $\cos \delta=-1$
$\begin{aligned}
I_{\min } & =I_1+I_2-2 \sqrt{I_1 I_2} \\
& =\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\end{aligned}$
Therefore,
$\begin{aligned}
I_{\max }+I_{\min } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2+\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\
& =2\left(I_1+I_2\right)
\end{aligned}$
Hence, the correct answer is option: 'D'.