In Young's double slit interference experiment the wavelength of light used is 6000A. If the path difference between waves reaching a ponit P on the screen is 1.5 microns, then at that point P:

The correct option is C third dark band occurs

Given $\lambda =6000A=6000\times {10}^{-10}m=6\times {10}^{-7}mand△=1.5microns=1.5\times {10}^{-6}m.$ For bright fringes: $△=n\lambda ;$ where n is an integer.
$n=\frac{△}{\lambda }=\frac{1.5\times {10}^{-6}}{6\times {10}^{-7}}=\frac{5}{2},$which is not an integer.
hence, path difference of $1.5\times {10}^{-6}m$ does not correspond to a bright fringe. For dark fringes, we have
$△=(n-\frac{1}{2})\lambda or1.5\times {10}^{-6}=(n-\frac{1}{2})\times (6\times {10}^{-7})$
which gives $n-\frac{1}{2}=\frac{5}{2}orn=3.$, Hence a path difference of $1.5\times {10}^{-6}m$ corresponds to the third dark fringe.