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(B) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.Taking the square

Vedclass · Accepted Answer

$\frac{7}{1}$

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(B) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.Taking the square root on both sides: $\sqrt{\frac{I_{\max}}{I_{\min}}} = \frac{a_1 + a_2}{a_1 - a_2}$.Given $\frac{I_{\max}}{I_{\min}} = \frac{144}{81}$,so $\sqrt{\frac{144}{81}} = \frac{12}{9} = \frac{4}{3}$.Let $r = \frac{a_1}{a_2}$. Then $\frac{r+1}{r-1} = \frac{4}{3}$.$3(r+1) = 4(r-1) \implies 3r + 3 = 4r - 4$.$r = 7$.Thus,the ratio of amplitudes is $\frac{a_1}{a_2} = \frac{7}{1}$.

In the interference of two coherent sources of different intensities,the ratio of maximum and minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{144}{81}$. What is the ratio of their amplitudes?

Similar Questions

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