Two coherent point sources S_1 and S_2 , vibrating in phase, emit...

Let the point of observation be $P$ at a distance $y$ from $S_2$ on the line perpendicular to $S_1S_2$.

The distance of $P$ from $S_1$ is $S_1P = \sqrt{d^2 + y^2}$, where $d = 2\lambda$ is the distance between the sources.

The path difference at $P$ is given by:
$\Delta x = S_1P - S_2P = \sqrt{d^2 + y^2} - y$

For an intensity minimum (destructive interference), the path difference must be an odd multiple of $\dfrac{\lambda}{2}$:
$\Delta x = (2n - 1)\dfrac{\lambda}{2}$

At $S_2$ ($y = 0$), the path difference is maximum and equals $d = 2\lambda$. As $y$ increases, the path difference $\Delta x$ decreases. To find the smallest distance $y$, we must take the largest possible path difference for a minimum that is less than $2\lambda$.

The largest such path difference is $\Delta x = \dfrac{3\lambda}{2}$.

Substituting $\Delta x = \dfrac{3\lambda}{2}$ and $d = 2\lambda$ into the path difference equation:
$\sqrt{(2\lambda)^2 + y^2} - y = \dfrac{3\lambda}{2}$

$\sqrt{4\lambda^2 + y^2} = y + \dfrac{3\lambda}{2}$

Squaring both sides, we get:
$4\lambda^2 + y^2 = y^2 + 3\lambda y + \dfrac{9\lambda^2}{4}$

$3\lambda y = 4\lambda^2 - \dfrac{9\lambda^2}{4}$

$3\lambda y = \dfrac{7\lambda^2}{4}$

$y = \dfrac{7\lambda}{12}$