Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

(N/A) In Young's double-slit experiment,light from a single source $S$ reaches slits $S_1$ and $S_2$ simultaneously,making them coherent sources.
Let the distance between $S_1$ and $S_2$ be $d$ and the distance to the screen $GG'$ be $D$.
Consider a point $P$ on the screen at a distance $x$ from the central point $O$.
The path difference $\Delta p$ is given by $\Delta p = S_2P - S_1P$.
From the geometry of the setup:
$S_1P^2 = D^2 + (x - d/2)^2$
$S_2P^2 = D^2 + (x + d/2)^2$
Subtracting the two equations:
$S_2P^2 - S_1P^2 = (D^2 + x^2 + dx + d^2/4) - (D^2 + x^2 - dx + d^2/4) = 2xd$
$(S_2P - S_1P)(S_2P + S_1P) = 2xd$
Since $D \gg d$ and $D \gg x$,we can approximate $S_2P + S_1P \approx 2D$.
Therefore,$\Delta p(2D) = 2xd$,which simplifies to $\Delta p = \frac{xd}{D}$.