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In an interference experiment, phase difference for points where the intensity is minimum is $(n=1,2,3 \ldots)$
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Verified Answer
The correct answer is:
$(2 n-1) \pi$
The intensity $I$ is minimum for phase difference,
$$
\begin{aligned}
& \delta=(2 n-1) \pi \\
\text { where } & n=1,2,3 \ldots
\end{aligned}
$$
$$
\begin{aligned}
& \delta=(2 n-1) \pi \\
\text { where } & n=1,2,3 \ldots
\end{aligned}
$$
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