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The maximum intensity of fringes in Young's experiment is I. If one of the slit is closed, then the intensity at that place becomes $I_0$. Which of the following relation is true?
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The correct answer is:
$I=4 I_0$
Suppose slit width's are equal, so they produces waves of equal intensity say $I^{\prime}$. Resultant intensity at any point $I_R=4 / \cos ^2 \phi$ where $\phi$ is the phase difference between the waves at the point of observation.
For maximum intensity $\phi=0^{\circ} \Rightarrow I_{\max }=4 I^{\prime}=I \ldots$ (i)
If one of slit is closed, Resultant intensity at the same point will be $I^{\prime}$ only i.e. $I^{\prime}=I_0$ (ii)
Comparing equation (i) and (ii) we get
$I=4 I_0$
For maximum intensity $\phi=0^{\circ} \Rightarrow I_{\max }=4 I^{\prime}=I \ldots$ (i)
If one of slit is closed, Resultant intensity at the same point will be $I^{\prime}$ only i.e. $I^{\prime}=I_0$ (ii)
Comparing equation (i) and (ii) we get
$I=4 I_0$
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