In a YDSE setup,the intensity due to two cohe | vedclass

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(D) Let the intensities of the two beams be $I_1 = I$ and $I_2 = I + \Delta I$. Given that the difference in intensity is $1\%$,we have $\frac{\Delta

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$\frac{I}{4}(10^{-4})$

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(D) Let the intensities of the two beams be $I_1 = I$ and $I_2 = I + \Delta I$. Given that the difference in intensity is $1\%$,we have $\frac{\Delta I}{I} = 1\% = 10^{-2}$. The intensity of the minima in a $YDSE$ is given by $I_{	ext{min}} = (\sqrt{I_2} - \sqrt{I_1})^2$. Substituting the values: $I_{	ext{min}} = (\sqrt{I + \Delta I} - \sqrt{I})^2 = I \left( \sqrt{1 + \frac{\Delta I}{I}} - 1 ight)^2$. Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$: $I_{	ext{min}} \approx I \left( 1 + \frac{1}{2} \frac{\Delta I}{I} - 1 ight)^2 = I \left( \frac{1}{2} \frac{\Delta I}{I} ight)^2$. $I_{	ext{min}} = \frac{I}{4} \left( \frac{\Delta I}{I} ight)^2 = \frac{I}{4} (10^{-2})^2 = \frac{I}{4} (10^{-4})$.

In a $YDSE$ setup,the intensity due to two coherent beams differs from each other by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is:

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