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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S1 and S2), i.e., D » d.
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Answer

The condition for constructive interference at $P$ is,
$\Delta l = y _{ n } \frac{d}{D}= n \lambda$
$y_n$ being the position (y-coordinate) of $n$th bright fringe ( $\left.n=0, \pm 1, \pm 2, \ldots\right)$.
$\therefore y _{ n }= n \lambda \frac{D}{d} \ldots . .(2)$
Similarly, the position of $m$ th $(m=+1, \pm 2, \ldots)$ dark fringe (destructive interference) is given by,
$ \Delta 1=y_m \frac{d}{D}=(2 m -1) \lambda \text { giving }$
$y_m=(2 m -1) \lambda \frac{D}{d} \ldots(3) $
The distance between any two consecutive bright or dark fringes, i.e., the fringe width $= W =\Delta y=y_{n+1}-y_n=\lambda \frac{D}{d}$
Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources $S_1$ and $S_2$ are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.

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