Question

Consider the arrangement shown in figure (17-E7). By some mechanism, the separation between the slits $S_3$ and $S_4$ can be changed . the intensity is measured at the point P which is at the common perpendicular bisector

of $S_1{S}_2$ and$S_3{S}_4$ When z = $\frac{d\lambda }{2d},$ the intensity measured at P is I. Find this intensity when z is equal to
(a) $\frac{D\lambda }{d}$, (b) $\frac{3D\lambda }{2d}$ and (c) $\frac{2D\lambda }{d}.$

Answer 1

(a) When , $z=\frac{\lambda D}{d}So,O{S}_3=O{S}_4=\frac{lambdaD}{d}$
$\therefore$ Dark fringe at $S_3$ and $S_4$
$\Rightarrow$ At $S_3$ intensity at $S_3$ = 0
$\Rightarrow l_1=0$
At $S_4$ intensity at $S_4=0$
$\Rightarrow l_2=0$
At P, path difference = 0
$\Rightarrow$ phase difference = 0
$\Rightarrow l=l_1+L_2+2\sqrt{l_1{l}_2}\cos {\theta }^{\circ }=0+0+0=0$
$\Rightarrow$ Intensity at 'p' = 0