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(A) In an interference pattern,the intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum and minimum intensity,$\cos \phi$ is

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$4$

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(A) In an interference pattern,the intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum and minimum intensity,$\cos \phi$ is $1$ and $-1$ respectively.$I_{	ext{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{	ext{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$.Given $\frac{I_{	ext{max}}}{I_{	ext{min}}} = \frac{9}{1}$,we have $\frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 9$.Taking the square root on both sides: $\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 3$.Using componendo and dividendo: $\frac{\sqrt{I_1}}{\sqrt{I_2}} = \frac{3+1}{3-1} = \frac{4}{2} = 2$.Therefore,the ratio of intensities of the sources is $\frac{I_1}{I_2} = (2)^2 = 4$.

In Young's double-slit experiment,the ratio of intensities of bright and dark fringes is $9$. What is the ratio of the intensities of the sources?

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