In a YDSE setup,the intensity of two coherent | vedclass

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(D) Let the intensities of the two coherent beams be $I_1 = I$ and $I_2 = I + \Delta I$.Given that the intensities differ by $1\%$,we have $\frac{\Del

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

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(D) Let the intensities of the two coherent beams be $I_1 = I$ and $I_2 = I + \Delta I$.Given that the intensities differ by $1\%$,we have $\frac{\Delta I}{I} = 1\% = 10^{-2}$.The intensity of the minima in a $YDSE$ setup 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 $x$,where $x = \frac{\Delta I}{I} = 10^{-2}$:$I_{	ext{min}} \approx I \left( (1 + \frac{1}{2} \cdot \frac{\Delta I}{I}) - 1 ight)^2 = I \left( \frac{1}{2} \cdot \frac{\Delta I}{I} ight)^2$.$I_{	ext{min}} = I \cdot \frac{1}{4} \cdot \left( \frac{\Delta I}{I} ight)^2 = \frac{I}{4} (10^{-2})^2 = \frac{I}{4} (10^{-4})$.

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

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