In a Young's double slit interference experiment, the fringe...

The intensity at a point on the screen in Young's double slit experiment is given by I = I_{\text\max} \cos^2 \left(\dfrac{\phi}{2}\right), where\phi is the phase difference.

Given that the intensity falls to half the maximum, $I =$\dfrac{I_{\text\max}}{2}.

\dfrac{I_{\text\max}}{2} =I_{\text\max} \cos^2 \left(\dfrac{\phi}{2}\right)

$\cos^2\left(\dfrac{\phi}{2}\right) = \dfrac{1}{2} \Rightarrow \cos\left(\dfrac{\phi}{2}\right) = \dfrac{1}{\sqrt{2}}$

$\dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{\pi}{2}$

The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \dfrac{2\pi}{\lambda} \Delta x$.

$\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{2} \Rightarrow \Delta x = \dfrac{\lambda}{4}$

For a point at a distance $y$ from the central maximum, the path difference is $\Delta x = \dfrac{yd}{D}$.

$\dfrac{yd}{D} = \dfrac{\lambda}{4} \Rightarrow y = \dfrac{D\lambda}{4d}$