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Q. In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength $6000 \,\mathring{A}$ coming from the coherent sources $S_{1}$ and $S_{2}$. At certain point $P$ on the screen third dark fringe is formed. Then the path difference $S_{1} P-S_{2} P$ in microns is

Wave Optics

Solution:

For dark fringe at $P$
$S_{1} P-S_{2} P=\Delta=(2 n-1) \lambda / 2$
Here $n=3$ and $\lambda=6000 \,\mathring{A}$
So, $\Delta=\frac{5 \lambda}{2}=5 \times \frac{6000}{2}$
$=15000 \,\mathring{A}=1.5$ micron