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In Young's double slit experiment, first slit has width four times the width of the second slit. the ratio of the maximum intensity to the minimum intensity in the interference fringe system is :
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The correct answer is:
$9: 1$
Let intensity of light coming from each slit of a coherent source is $I$.
As first slit has width 4 times the width of the second slit, so
$I_1=4 I \text { and } I_2=I$
$\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}$
$=\frac{(\sqrt{4 I}+\sqrt{I})^2}{(\sqrt{4 I}-\sqrt{I})^2}=\frac{9}{1}$
As first slit has width 4 times the width of the second slit, so
$I_1=4 I \text { and } I_2=I$
$\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}$
$=\frac{(\sqrt{4 I}+\sqrt{I})^2}{(\sqrt{4 I}-\sqrt{I})^2}=\frac{9}{1}$
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