MHT CET · Physics · Waves and Sound
If the ratio of the intensities of two waves producing interference is \(49: 16\), then the ratio of the resultant maximum intensity to minimum intensity will be
- A 11:3
- B 121:9
- C \(7: 4\)
- D 49:16
Answer & Solution
Correct Answer
(B) 121:9
Step-by-step Solution
Detailed explanation
Let the intensities of the two waves be \(I_1\) and \(I_2\).
Given: \(I_1: I_2=49: 16\)
Therefore, the ratio of maximum and minimum intensities is given by,
\(\frac{I_{\max }}{I_{\min }}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{7+4}{7-4}\right)^2=\frac{121}{9}\)
Given: \(I_1: I_2=49: 16\)
Therefore, the ratio of maximum and minimum intensities is given by,
\(\frac{I_{\max }}{I_{\min }}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{7+4}{7-4}\right)^2=\frac{121}{9}\)
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