Two waves of equal frequencies have their amplitudes in the ratio of 3 : 5. They are superimposed on each…
Two waves of equal frequencies have their amplitudes in the ratio of 3 : 5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.
Solution
Sol. Given, $A_1/A_2 = 3/5$
\therefore $\sqrt{I_1}/\sqrt{I_2} = 3/5$ (As, $I \propto A^2$)
Maximum intensity is obtained, where cos $\phi = 1$ and $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$
Minimum intensity is obtained, where
cos $\phi$ = - 1 and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$
Hence,
$\displaystyle \frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2 = \left(\frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1}\right)^2$
$= \left(\frac{3/5 + 1}{3/5 - 1}\right)^2 = \frac{64}{4} = \frac{16}{1}$
Answer: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$