MHT CET · Physics · Waves and Sound
Two waves are superimposed whose ratio of intensities is \(9: 1\). The ratio of maximum and minimum intensity is
- A 9:1
- B \(4: 1\)
- C \(3: 1\)
- D \(5: 3\)
Answer & Solution
Correct Answer
(B) \(4: 1\)
Step-by-step Solution
Detailed explanation
Given, the ratio of intensities, we can obtain ratio of amplitudes of the waves:
\(\begin{aligned} & \frac{I_1}{I_1}=\left(\frac{a_1}{a_2}\right)^2=\frac{9}{1} \\ & \Rightarrow \frac{a_1}{a_2}=\frac{3}{1}\end{aligned}\)
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{\left(3 a_2+a_2\right)^2}{\left(3 a_2-a_2\right)^2}=\left(\frac{4 a_2}{2 a_2}\right)^2=\frac{4}{1}\)
\(\begin{aligned} & \frac{I_1}{I_1}=\left(\frac{a_1}{a_2}\right)^2=\frac{9}{1} \\ & \Rightarrow \frac{a_1}{a_2}=\frac{3}{1}\end{aligned}\)
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{\left(3 a_2+a_2\right)^2}{\left(3 a_2-a_2\right)^2}=\left(\frac{4 a_2}{2 a_2}\right)^2=\frac{4}{1}\)
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