A circular disc of radius 'R' is placed co-axially and horizontally inside and opaque hemispherical bowl of radius 'a', Fig. The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index `mu` and the near edge of the dise becomes just visible. How far below the top of the bowl is the disc placed ?
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Explanation

In Fig., we have shown an opaque hemisperical bowl of radius `a` with centre `O`.
A circular disc of radius `R` with centre `C` is placed co-axially and horizontally inside the bowl.We have to calculate `OC = d. AMA'` is the direction of incident index `mu` is filled in the bowl, the near edge `B` of the disc just becomes visible.
Here, `BM` is the incident ray, which is refracted along `MA'`
`NN'` is normal to the horizontal surface of liquid in the bowl. `i` is the angle of incidence and `r = prop` is the angle of refraction.
As refraction occurs at `M` in going from denser to rarer medium, therefore, according to Snell's law
`(1)/(mu) = (sin i)/(sin r) = (sin i)/(sin prop)` ....(i)
Now, `sin i = (BN')/(BM) = (a - R)/(sqrt(d^2 + (a - R)^2))` and `sin prop = cos (90 - prop) = (AN')/(AM) = (a + R)/(sqrt(d^2 + (a - R)^2))`
Putting in (i), we get
`(1)/(mu) = (a - R)/(sqrt(d^2 + (a - R)^2)) xx (sqrt(d^2 + (a + R)^2))/(a + R) ` or `mu xx (a - R) sqrt(d^2 + (a + R)^2) = (a + R) sqrt(d^2 + (a - R)^2)`
On simplifying, we get `d = (mu (a^2 - r^2))/(sqrt((1 + R)^2 - mu (a - R)^2))`.