Analysis of Interference Patterns

The two sources $S_1$ and $S_2$ are incoherent, meaning they do not maintain a constant phase relationship with each other. Therefore, their electric fields do not interfere. Instead, their intensities add up on Screen B.

However, each individual source acts as a coherent source for the double slit on Screen A, creating its own Young’s Double Slit interference pattern on Screen B.

• Source $S_1$: Located at $y = +l/2$. Due to the vertical shift of the source, its central maximum (where path difference is zero) will shift downwards on Screen B.
  Using the standard result for source shift: $\Delta y_{center} = – \frac{D}{D_{source}} y_{source}$.
  Here, $D_{source}=D$ and $D_{screen}=D$.
  Shift for $S_1$: $y_1 = – \frac{D}{D} (\frac{l}{2}) = -\frac{l}{2}$.
• Source $S_2$: Located at $y = -l/2$.
  Shift for $S_2$: $y_2 = – \frac{D}{D} (-\frac{l}{2}) = +\frac{l}{2}$.

Condition for Disappearance of Pattern

We have two separate interference patterns superimposed on Screen B. One is centered at $-l/2$ and the other at $+l/2$. The distance between the centers of these two patterns is $l$.

A “stationary interference pattern” will not be observed if the maxima of one pattern overlap exactly with the minima of the other pattern. This results in uniform illumination (washout).

The condition for this is that the shift between the patterns must be an odd multiple of half the fringe width ($\beta/2$).

$$ \text{Shift} = (2n + 1) \frac{\beta}{2} \quad \text{where } n = 0, 1, 2… $$

The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$ (Note: the distance from slits to screen is $D$).

Substituting the shift $l$:

$$ l = (2n + 1) \frac{\lambda D}{2d} $$