Question

Consider the arrangement shown in the figure (17-E7). 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 S₁S₂ and S₂S₄. When $z=\frac{D\lambda }{2d}$, the intensity measured at P is I. Find the intensity when z is equal to

$\left(\mathrm{a}\right)\frac{D\lambda }{d},\left(\mathrm{b}\right)\frac{3D\lambda }{2d}\mathrm{and}\left(\mathrm{c}\right)\frac{2D\lambda }{d}.$

Figure

Answer 1

Given:
Fours slits S₁, S₂, S₃ and S₄.
The separation between slits S₃ and S₄ can be changed.
Point P is the common perpendicular bisector of S₁S₂ and S₃S₄.



(a) $\mathrm{For}z=\frac{\lambda D}{d}$:
The position of the slits from the central point of the first screen is given by
$y={\mathrm{OS}}_3={\mathrm{OS}}_4=\frac{z}{2}=\frac{\lambda D}{2d}$
The corresponding path difference in wave fronts reaching S₃ is given by
$∆x=\frac{yd}{D}=\frac{\lambda D}{2d}\times \frac{d}{D}=\frac{\lambda }{2}$
Similarly at S₄, path difference, $∆x=\frac{yd}{D}=\frac{\lambda D}{2d}\times \frac{d}{D}=\frac{\lambda }{2}$
$\mathrm{i}.\mathrm{e}.\mathrm{dark}\mathrm{fringes}\mathrm{are}\mathrm{formed}\mathrm{at}{\mathrm{S}}_3\mathrm{and}{\mathrm{S}}_4$.
So, the intensity of light at S₃ and S₄ is zero. Hence, the intensity at P is also zero.

(b) $\mathrm{For}z=\frac{3\lambda D}{2d}$
The position of the slits from the central point of the first screen is given by
$y={\mathrm{OS}}_3={\mathrm{OS}}_4=\frac{z}{2}=\frac{3\lambda D}{4d}$
The corresponding path difference in wave fronts reaching S₃ is given by
$∆x=\frac{yd}{D}=\frac{3\lambda D}{4d}\times \frac{d}{D}=\frac{3\lambda }{4}$
Similarly at S₄, path difference, $∆x=\frac{yd}{D}=\frac{3\lambda D}{4d}\times \frac{d}{D}=\frac{3\lambda }{4}$
Hence, the intensity at P is I.

(c) Similarly, for $z=\frac{2D\lambda }{d}$,
the intensity at P is 2I.