Let, us consider the diagram, the ray $(\mathrm{P})$ is incident at an angle $\theta$ and gets reflected in the direction $\mathrm{P}^{\prime}$ and refracted in the direction $\mathrm{P}^{\prime}$ through $\mathrm{O}^{\prime}$. Due to reflection from the glass medium there is a phase change of $\pi$. The time difference between two refracted ray $\mathrm{OP}^{\prime}$ and $\mathrm{O}^{\prime} \mathrm{P}$ " is equal to the time taken by ray to travel along $\mathrm{OO}^{\prime}$.
From Snell's law, $\mathrm{n}=\frac{\sin \theta}{\sin \mathrm{r}}$
$$
\sin r=\frac{\sin \theta}{n}
$$
As we know that,
$$
\cos r=\sqrt{1-\sin ^2 r},
$$
so by putting $\sin r$ value in that relation.
So,
$$
\begin{aligned}
&\cos r=\sqrt{1-\sin ^2 r} \\
&\cos r=\sqrt{1-\frac{\sin ^2 \theta}{n^2}}
\end{aligned}
$$


$$
\begin{aligned}
\Delta \Delta &=\frac{\mathrm{nd}}{\mathrm{c}\left(1-\frac{\sin ^2 \theta}{\mathrm{n}^2}\right)^{1 / 2}} \\
&=\frac{\mathrm{nd}}{\mathrm{c}}\left(1-\frac{\sin ^2 \theta}{\mathrm{n}^2}\right)^{-1 / 2} \\
\text { Phase difference } &=\Delta \phi=\frac{2 \pi}{\mathrm{T}} \times \Delta \mathrm{t} \\
&=\frac{2 \pi \mathrm{d}}{\frac{1}{v} \cdot v \lambda}\left(1-\frac{\sin ^2 \theta}{\mathrm{n}^2}\right)^{-1 / 2} \\
\Delta \phi &=\frac{2 \pi \mathrm{d}}{\lambda}\left[1-\frac{\sin ^2 \theta}{\mathrm{n}^2}\right]^{-1 / 2}
\end{aligned}
$$
$\therefore \quad$ Hence the net phase difference $=\Delta \phi+\pi$
$$
=\frac{2 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{-1 / 2}+\pi
$$