According to the given figure, $\mathrm{AM}$ is the direction of incidence ray before liquid is filled. After liquid is filled in, BM is the direction of the incident ray. Refracted ray in both cases is same as that along AM. MN is tangent at M, so $\mathrm{MN} \perp \mathrm{AB}$ i.e., $\angle \mathrm{N}=90^{\circ}$
Consider the disc is separated by $\mathrm{O}$ at a distance $\mathrm{d}$ as shown in figure given below,


Now let us consider an angle at
$$
\begin{aligned}
(\mathrm{N}) &=90^{\circ}, \\
\mathrm{OM} &=\mathrm{a} \\
\mathrm{NB} &=\mathrm{a}-\mathrm{R}, \\
\mathrm{MB} &=\sqrt{\mathrm{d}^2+(\mathrm{a}-\mathrm{R})^2}
\end{aligned}
$$
From, above figure,
$$
\begin{aligned}
&\sin \mathrm{i}=\left[\frac{\mathrm{NB}}{\mathrm{MB}}\right]=\frac{a-R}{\sqrt{\mathrm{d}^2+(a-R)^2}} \\
&\because \quad \mathrm{AN}=\mathrm{AB}+\mathrm{BN}=2 \mathrm{R}+\mathrm{a}-\mathrm{R}=(\mathrm{a}+\mathrm{R}) \\
&A M=\sqrt{d^2+(a+R)^2} \\
&\text { And, } \quad \sin \mathrm{r}=\cos (90-\alpha) \\
&=\left[\frac{A N}{A M}\right]=\frac{a+R}{\sqrt{d^2+(a+R)^2}} \\
&
\end{aligned}
$$
So, $\quad{ }_1 \mu_0=\frac{\mu_0}{\mu_1}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$
$\left(\mu_0\right.$ for air = 1) \& ( $\mu_1$ for liquid)
So, $\quad \frac{1}{\mu}=\frac{\sin i}{\sin r}$
By putting the values of $\sin$ i $\& \sin \mathrm{r}$ we get the distance
(d) is
$$
d=\frac{\mu\left(a^2-b^2\right)}{\sqrt{(a+r)^2-\mu(a-r)^2}}
$$
This is the required expression.