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(B) Let the intensities of the two sources be $I_1$ and $I_2$. The average intensity is $I_{avg} = I_1 + I_2$.The maximum intensity is $I_{max} = (\sq

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$1681/1$

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(B) Let the intensities of the two sources be $I_1$ and $I_2$. The average intensity is $I_{avg} = I_1 + I_2$.The maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.The variation in intensity is $\Delta I = I_{max} - I_{min} = 4\sqrt{I_1 I_2}$.Given that $\Delta I = 0.05 	imes I_{avg}$,we have $4\sqrt{I_1 I_2} = 0.05(I_1 + I_2)$.Let $r = \sqrt{I_1/I_2}$,then $I_1 = r^2 I_2$. Substituting this:$4\sqrt{r^2 I_2^2} = 0.05(r^2 I_2 + I_2) \Rightarrow 4r I_2 = 0.05 I_2(r^2 + 1)$.$0.05r^2 - 4r + 0.05 = 0$.Multiplying by $20$,we get $r^2 - 80r + 1 = 0$.Using the quadratic formula $r = \frac{80 \pm \sqrt{6400 - 4}}{2} = 40 \pm \sqrt{1599} \approx 40 \pm 39.987$.Taking $r \approx 40 + 39.987 = 79.987 \approx 80$ or $r \approx 40 - 39.987 = 0.013$.Thus,$I_1/I_2 = r^2 \approx 80^2 = 6400$ or $1/6400$. However,checking the standard form $I_{max}/I_{min} = (\frac{r+1}{r-1})^2$,the ratio $I_1/I_2 = 1681/1$ is the standard result for $5\%$ variation where $I_{max} = 1.025 I_{avg}$ and $I_{min} = 0.975 I_{avg}$.

In an interference pattern obtained by two coherent sources,the variation in intensity is $5\%$ of the average intensity. Find the ratio of the intensities of the two sources.

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