Question

The figure drawn here, shows a modified Young's double slit experimental set up in which $s{s}_2-s{s}_1=\lambda /4$,
(i) State the condition for constructive and destructive interference.
(ii) Obtain the expression for fringe width.
(iii) Locate position of the central fringe.

Answer 1

Let the initial path different between $S_1$ and $S_2$ be ${\Delta }_0=S{S}_2-S{S}_1=\frac{\lambda }{4}$

path difference between disturbance from $S_1$ and $S_2$, at point P be, $\Delta =\frac{yd}{D}$

Total path difference between the two disturbances at P, be ${\Delta }_T={\Delta }_0+\Delta =\frac{\lambda }{4}+\frac{yd}{D}$

Therefore For constructive interference

${\Delta }_T=\frac{\lambda }{4}+\frac{yd}{D}=n\lambda ;n=0,1,2,3,...$

$\therefore \frac{y_{n}d}{D}=(n-\frac{1}{4})\lambda ........\left(i\right)$

For destructive interference

${\Delta }_T=\frac{\lambda }{4}+\frac{yd}{D}=(2n-1)\frac{\lambda }{2}........................\left(ii\right)\therefore \frac{y_n^{\prime }d}{D}=(2n-1-\frac{1}{2})\frac{\lambda }{2}or,\frac{y_n^{\prime }d}{D}=(2n-\frac{3}{2})\frac{\lambda }{2}$

Finge width, $\beta =y_{n+1}-y_n=\frac{\lambda D}{d}$

The position $Y_0$ of central fringe is obtained by putting $n=0$ in Eqn (i). Therefore

$y_0=-\frac{\lambda D}{4d}$

[Negative sign shows that the central fringe is obtained at a point below the (central) point O.]