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

The intensity at any point on the screen in a Young's double slit experiment is given by:

$I = I_0 \cos^2 \left(\dfrac{\phi}{2}\right)<br><br>where$I_0 is the maximum intensity and $\phi$ is the phase difference.

Given that the intensity falls to one fourth of the maximum:

\dfrac{I_0}{4} =$I_0 \cos^2 \left(\dfrac{\phi}{2}\right)$

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

$\dfrac{\phi}{2} = \dfrac{\pi}{3} \Rightarrow \phi = \dfrac{2\pi}{3}$

The phase difference $\phi$ is related to the path difference $\Delta x$ by:

$\phi = \dfrac{2\pi}{\lambda} \Delta x$

$\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x \Rightarrow \Delta x = \dfrac{\lambda}{3}$

The path difference is also given by $\Delta x = \dfrac{yd}{D}$, where $y$ is the distance from the central maximum.

$\dfrac{yd}{D} = \dfrac{\lambda}{3}$

$y = \dfrac{D\lambda}{3d}$