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MHT CET · Physics · Waves and Sound

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

  1. A \(I_{1}+I_{2}\)
  2. B \(\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\)
  3. C \(\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}\)
  4. D \(2\left(I_{1}+I_{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2\left(I_{1}+I_{2}\right)\)

Step-by-step Solution

Detailed explanation

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,
\(
\delta=2 n \pi ; \quad n=0,1,2, \ldots \text { etc. }
\)
Therefore, for zero order maxima, \(\cos \delta=1\)
\(
I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}
\)
For minimum intensity,
\(
\delta=(2 n-1) \pi ; n=1,2, \ldots \text { etc }
\)
Therefore, for Ist order minima, \(\cos \delta=-1\)
\(
\begin{array}{c}
I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}} \\
=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2} \\
\text { Therefore, } I_{\max }+I_{\min }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2} \\
\quad+\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2} \\
=2\left(I_{1}+I_{2}\right)
\end{array}
\)