In Young's double slit experiment using a monochromatic light of...

The intensity at any point in Young's double slit experiment is given by $I = I_0 \cos^2 \left(\dfrac{\phi}{2}\right)<br><br>Given that$ I = $\dfrac{I_0}{4}<br><br>\dfrac{I_0}{4} =$ I_0 \cos^2 \left(\dfrac{\phi}{2}\right)$<br><br>$\cos\left(\dfrac{\phi}{2}\right) = \pm \dfrac{1}{2}$<br><br>$\dfrac{\phi}{2} = n\pi \pm \dfrac{\pi}{3}$<br><br>$\phi = 2n\pi \pm \dfrac{2\pi}{3}$<br><br>The phase difference$\phi$is related to the path difference$\Delta x$by$\phi = \dfrac{2\pi}{\lambda} \Delta x$<br><br>$\dfrac{2\pi}{\lambda} \Delta x = 2\pi \left( n \pm \dfrac{1}{3}\right)<br><br>\Delta x = \left($n$\pm \dfrac{1}{3}\right)\lambda

Answer: \left($n$\pm\dfrac{1}{3}\right)\lambda,$\ n=0,1,2$