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In Young' double slit experiment, for the $n^{\text {th }}$ dark fringe ( $n=1,2,3$.....) the phase difference of the interfering waves in radian will be
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The correct answer is:
$(2 n+1) \pi$
The resultant intensity of the interfering waves is given by,
$\mathrm{E}^2 \mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos (\delta)$
where, $\delta$ is the phase difference between the waves.
For a dark fringe, $\mathrm{E}_1=\mathrm{E}_2$ and $\cos (\delta)=-1$.
Therefore, the phase $\delta=(2 n+1) \pi$ is an odd multiple of pi.
$\mathrm{E}^2 \mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos (\delta)$
where, $\delta$ is the phase difference between the waves.
For a dark fringe, $\mathrm{E}_1=\mathrm{E}_2$ and $\cos (\delta)=-1$.
Therefore, the phase $\delta=(2 n+1) \pi$ is an odd multiple of pi.
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