Consider the arrangement shown in figure. By some mechanism, the separation between the slits S₃ and S₄ can be changed. The intensity is measured at the point 𝑃 which is at the common perpendicular bisector of ${\mathrm{S}}_1{\mathrm{S}}_2\text{ and }{\mathrm{S}}_3{\mathrm{S}}_4$. When 𝑧 = 𝐷𝜆/2𝑑, the intensity measured at 𝑃 is 𝐼. Find this intensity when z is equal to 
1) 𝐷𝜆/𝑑,      2) 3𝐷𝜆/2𝑑,     3) 2𝐷𝜆/𝑑

Path difference at S₃ or S₄ corresponding to a given value of z is given by
$\begin{array}{l}\mathrm{\Delta x}-\sqrt{{\mathrm{D}}^2+{\left(\frac{\mathrm{d}}{2}+\frac{\mathrm{z}}{2}\right)}^2}-\sqrt{{\mathrm{D}}^2+{\left(\frac{\mathrm{d}}{2}-\frac{\mathrm{z}}{2}\right)}^2}\\ \mathrm{\Delta x}=\mathrm{D}\left(1+\frac{1}{2{\mathrm{D}}^2}{\left(\frac{\mathrm{d}}{2}+\frac{\mathrm{z}}{2}\right)}^2\right)-\mathrm{D}\left(1+\frac{1}{2{\mathrm{D}}^2}{\left(\frac{\mathrm{d}}{2}-\frac{\mathrm{z}}{2}\right)}^2\right)\\ \mathrm{\Delta x}=\frac{1}{2\mathrm{D}}\left[{\left(\frac{\mathrm{d}}{2}+\frac{\mathrm{z}}{2}\right)}^2-{\left(\frac{\mathrm{d}}{2}-\frac{\mathrm{z}}{2}\right)}^2\right]=\frac{1}{2\mathrm{D}}\mathrm{X}4\mathrm{X}\frac{\mathrm{d}}{2}\mathrm{X}\frac{\mathrm{Z}}{2}=\frac{\mathrm{dz}}{2\mathrm{D}}\end{array}$
If intensity at P is I_p = 4I’ ; I is the intensity at S₃ and S₄ If intensity at S₁ and S₂ is I⁰, then I¹ can be obtained from path difference calculations at S₃ or S₄
1) When $\mathrm{z}=\mathrm{D\lambda }/2\mathrm{d}$
$\mathrm{\Delta x}=\frac{\mathrm{d}}{2\mathrm{D}}\mathrm{X}\frac{\mathrm{D\lambda }}{2\mathrm{d}}=\frac{\mathrm{\lambda }}{4}=>\mathrm{\Delta \varphi }=\frac{\mathrm{\pi }}{2}=>{\mathrm{I}}^1=4{\mathrm{I}}_0{\cos }^2\frac{\mathrm{\pi }}{4}=2{\mathrm{I}}_0\text{ and }{\mathrm{I}}_{\mathrm{P}}=4{\mathrm{I}}^{\mathrm{I}}=8{\mathrm{I}}_0=\mathrm{I}$
2) When $\mathrm{z}=\mathrm{D\lambda }/\mathrm{d}$
$\mathrm{\Delta x}=\frac{\mathrm{d}}{2\mathrm{D}}\mathrm{X}\frac{\mathrm{D\lambda }}{\mathrm{d}}=\frac{\mathrm{\lambda }}{2}=>\mathrm{\Delta \varphi }=\mathrm{\pi }={\mathrm{I}}^{\mathrm{I}}=0=>{\mathrm{I}}_{\mathrm{p}}=4{\mathrm{I}}^{\mathrm{I}}=0$
3) When $\mathrm{z}=3\mathrm{D\lambda }/2\mathrm{d}$
$\mathrm{\Delta x}=\frac{2}{2\mathrm{D}}\mathrm{X}\frac{3\mathrm{D\lambda }}{2\mathrm{d}}=\frac{3\mathrm{\lambda }}{4}=>\mathrm{\Delta \varphi }=\frac{3}{4}\mathrm{X}2\mathrm{\pi }=>{\mathrm{I}}^{\mathrm{I}}=4{\mathrm{I}}_{\mathrm{o}}{\cos }^2\frac{\mathrm{\Delta \varphi }}{2}=4{\mathrm{I}}_0{\cos }^2\frac{3\mathrm{\pi }}{4}=2{\mathrm{I}}_0\text{ and }{\mathrm{I}}_{\mathrm{p}}=4{\mathrm{I}}^{\mathrm{I}}=8{\mathrm{I}}_0=\mathrm{I}$
4) When $\mathrm{z}=2\mathrm{D\lambda }/\mathrm{d}$
$\mathrm{\Delta x}=\frac{\mathrm{d}}{2\mathrm{D}}\mathrm{X}\frac{2\mathrm{D\lambda }}{\mathrm{d}}=\mathrm{\lambda }=>2\mathrm{\pi }=>{\mathrm{I}}^{\mathrm{I}}=4{\mathrm{I}}_0{\cos }^2\frac{\mathrm{\Delta \varphi }}{2}=4{\mathrm{I}}_0{\mathrm{andI}}_{\mathrm{p}}=4{\mathrm{I}}^{\mathrm{I}}=16{\mathrm{I}}_0=2\mathrm{I}$