Question

$s_{1}and{s}_2$ are the correct sources of light. P is a point on the screen such that $s_{2}p-s_{1}p$=0.01029 cm wavelength of the light used is 6000 $A^o$. The order of the fringe form at P is.

Answer 1

$\frac{s_{2}p-s_{1}p}{\lambda }=\frac{1029\times {10}^{-7}}{6\times {10}^{-7}}$

$=171.5$

Path difference$=s_{2}p-s_{1}p=171.5\lambda =(171+\frac{1}{2})\lambda$

Hence this is dark fringe of the order of $171$.