Dashed line $MN$ and $PQ$ are incident wave front.

Just before entering the slab $PQ$ is maxima.

$P_1$ and $Q_1$ are the points on other face of slab.

So, $P_1$ will be at a lead of $n_{2}d$ w.r.t. $P$

and $Q\text{'}$ will be at lead of $n_{1}d$ w.r.t. $Q$

As the medium on the right side is homogeneous, so to have same path lead as $P_1, Q\text{'}$ must travel to $Q_1$, such that $n_{2}d=n_{1}d+h \sin \theta$.

$$\begin{gathered} \sin \theta =\frac{\left(n_2-n_1\right)d}{h} \\ \Rightarrow \theta ={\sin }^{-1}\left[\frac{\left(n_2-n_1\right)d}{h}\right] \end{gathered}$$