AIIMS 2018 Physics Young's Double Slit Experiment MCQ Question

In Young's Double Slit Experiment (YDSE), the path difference $\Delta x$ at a point $P$ on the screen can be calculated using the formula $\Delta x = \frac{d \cdot y}{D}$, where $d$ is the distance between the slits, $y$ is the distance from the central maximum to point $P$, and $D$ is the distance from the slits to the screen. Substituting the values $d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}$, $y = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}$, and $D = 1 \, \text{m}$, we find $\Delta x = \frac{(1 \times 10^{-3}) \cdot (2 \times 10^{-3})}{1} = 2 \times 10^{-6} \, \text{m}$. Other options are incorrect as they do not match the calculated path difference based on the given parameters.