$A$ circular disc of radius $R$ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius $a$ (See figure). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with a transparent liquid of refractive index $\mu$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

(N/A) Let the depth of the disc below the top of the bowl be $d$.
$1$. Before filling the bowl with liquid,the far edge $A$ is just visible from the edge $M$ of the bowl. From the geometry of the triangle formed by the depth $d$ and the distance $(a+R)$,we have $\tan \alpha = \frac{a+R}{d}$,where $\alpha$ is the angle of the ray with the vertical.
$2$. When the bowl is filled with liquid of refractive index $\mu$,the near edge $B$ becomes visible. The ray from $B$ reaches the surface at $M$ and refracts into air. By Snell's Law at point $M$: $\mu \sin i = 1 \sin \alpha$,where $i$ is the angle of incidence from $B$ and $\alpha$ is the angle of refraction.
$3$. From the geometry,$\sin i = \frac{a-R}{\sqrt{d^2 + (a-R)^2}}$ and $\sin \alpha = \frac{a+R}{\sqrt{d^2 + (a+R)^2}}$.
$4$. Substituting these into Snell's Law: $\mu \frac{a-R}{\sqrt{d^2 + (a-R)^2}} = \frac{a+R}{\sqrt{d^2 + (a+R)^2}}$.
$5$. Squaring both sides: $\mu^2 \frac{(a-R)^2}{d^2 + (a-R)^2} = \frac{(a+R)^2}{d^2 + (a+R)^2}$.
$6$. Solving for $d^2$: $d^2 = \frac{(a-R)^2 (a+R)^2 (\mu^2 - 1)}{(a+R)^2 - \mu^2 (a-R)^2}$.
$7$. Therefore,$d = \sqrt{\frac{(a^2-R^2)^2 (\mu^2-1)}{(a+R)^2 - \mu^2(a-R)^2}}$.