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(A) Given that the intensity of the first slit $I_1 = 2I_2$,where $I_2$ is the intensity of the second slit.The ratio of maximum intensity to minimum

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$34$

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(A) Given that the intensity of the first slit $I_1 = 2I_2$,where $I_2$ is the intensity of the second slit.The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$Substituting $I_1 = 2I_2$ into the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{2I_2} + \sqrt{I_2}}{\sqrt{2I_2} - \sqrt{I_2}} \right)^2 = \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)^2$Rationalizing the denominator:$\frac{I_{\max}}{I_{\min}} = \left( \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \right)^2 = (\sqrt{2} + 1)^4 = (2 + 1 + 2\sqrt{2})^2 = (3 + 2\sqrt{2})^2$Calculating the value:$\frac{I_{\max}}{I_{\min}} = 9 + 8 + 12\sqrt{2} = 17 + 12(1.414) \approx 17 + 16.968 \approx 33.968 \approx 34$.

In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

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