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=Dλ2d, the intensity measured at P is I. Find the intensity when z is equal to
(a) Dλd, (b) 3Dλ2d and (c) 2Dλd.
Figure

Answer 1

Explanation:

Four $\mathrm{S}_{1}, \mathrm{S}_{2}, \mathrm{S}_{3} \text { and } \mathrm{S}_{4}$ slits.

The distinction between $\mathrm{S}_{3} \text { and } \mathrm{S}_{4}$ slits may be altered.

The common perpendicular bisector of  $\mathrm{S}_{1} \mathrm{S}_{2} \text { and } \mathrm{S}_{3} \mathrm{S}_{4}$ is Point P.  

(a)

$\text { For } z=\frac{\lambda D}{d}$

The position of the slits from the first screen's central point is given by

$\mathrm{y}=\mathrm{OS}_{3}=\mathrm{OS}_{4}=\mathrm{z}_{2}=\frac{\lambda \mathrm{D}}{2 \mathrm{d}}$

The corresponding difference in path in wave fronts that enter S3 is given by

$\Delta x=y d / D=\left[\frac{\lambda D}{2 d}\right] \times d D=\lambda 2$

Similarly at S4, path difference,

$\Delta x=\frac{\lambda}{D}=\frac{\lambda D}{2 d} \times d D=\lambda 2$

i.e. At S3 and S4, dark fringes form.

So, the light intensity at S3 and S4 is nil. Therefore the pressure at P is zero, too.