Two coherent narrow slits emitting light of wavelength l in the same phase are placed parallel to each other at a small separation of 2 l. The light is collected on a screen S which is placed at a distance $D\left(\gg \lambda \right)$ from the slit S₁ as shown in figure. Find the finite distance x such that the intensity at P is equal to intensity at O.

                                                         

Path difference at O,                                                                                                                                                

                                                                                $S_{1}O-S_{2}O=2\lambda$                                                                                                                                

i.e. maximum intensity is obtained at O. Next maxima will be obtained at point P where,

                                                                                  $$\begin{gathered} S_{1}P-S_{2}P=\lambda \\ d \cos \theta =\lambda \\ 2\lambda \cos \theta =\lambda \\ \cos \theta =\frac{1}{2} \\ \theta =60° \end{gathered}$$          

Now in $△S_{1}PO$,

                                                                                  $$\begin{gathered} \frac{PO}{S_{1}O}=\tan \theta or \frac{x}{D}=\tan 60°=\sqrt{3} \\ ∆x=\sqrt{3}D \end{gathered}$$

     At point O, path difference is $2\lambda$, i.e. we get second order maxima. At point P, where path difference is $\lambda (i.e., x=\sqrt{3}D$we get first order maxima. The next, i.e. zero order maxima will be obtained where path difference, i.e. $d \cos \theta =0 or \theta =90°$.At $\theta =90°, x=\infty$ .So, our answer, i.e. finite distance of x should be x D = 3 , corresponding to first order maxima.