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In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.
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Verified Answer
The correct answer is:
$9 \frac{\lambda}{2}$
In a YDSE, the path difference for $\mathrm{n}$ th minima is given by
$$
\Delta y=(2 n-1) \frac{\lambda}{2}
$$
For 5 th minima, $\mathrm{n}=5$
$$
\therefore \quad \Delta y=[2(5)-1] \frac{\lambda}{2}=\frac{9 \lambda}{2}
$$
$$
\Delta y=(2 n-1) \frac{\lambda}{2}
$$
For 5 th minima, $\mathrm{n}=5$
$$
\therefore \quad \Delta y=[2(5)-1] \frac{\lambda}{2}=\frac{9 \lambda}{2}
$$
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