$A$ small transparent slab of refractive index $\mu = 1.5$ and thickness $L = d/4$ is placed along the path $AS_2$ (see figure). What will be the distance from $O$ of the principal maxima and the first minima on either side of the principal maxima obtained in the absence of the glass slab? Given $AC = CD = D$ and $S_1C = S_2C = d$ (where $d << D$).

(N/A) The introduction of the glass slab in the path $AS_2$ introduces an additional optical path difference. The optical path difference introduced by the slab is $\Delta x_{slab} = (\mu - 1) L = (1.5 - 1) (d/4) = d/8$.
Since the slab is in the path of the lower ray $AS_2$,the interference pattern shifts downwards. The net path difference at a point $P$ at an angle $\theta$ is $\Delta x = 2d \sin \theta - d/8$ (taking downward direction as positive).
For the principal maxima (central bright fringe),$\Delta x = 0$,so $2d \sin \theta_0 = d/8$,which gives $\sin \theta_0 = 1/16$. Since $\theta$ is small,$\sin \theta \approx \tan \theta = x/D$. Thus,the distance from $O$ is $x_0 = D \sin \theta_0 = D/16$.
For the first minima,the condition is $\Delta x = \pm \lambda/2$.
Case $1$: $2d \sin \theta_1 - d/8 = \lambda/2 \implies 2d \sin \theta_1 = d/8 + \lambda/2 \implies x_1 = D(1/16 + \lambda/4d)$.
Case $2$: $2d \sin \theta_2 - d/8 = -\lambda/2 \implies 2d \sin \theta_2 = d/8 - \lambda/2 \implies x_2 = D(1/16 - \lambda/4d)$.