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

Before the liquid is filled, far edge of disc i.e. A is just visible. Therefore, the direction of the incident ray is AM. In the presence of liquid, the near edge of the disc i.e., B is just visible. Therefore, the direction of the incident ray is BM. In both the cases, the refracted ray will be along MN.
If `mu` is the refractive index of the transparent liquid in bowl, then
`(1)/(mu)=(sini)/(sinr)=(sini)/(sin alpha)" ....(i)"`
From the figure, in `DeltaBMP`
`sini=(BP)/(BM)=(a-R)/(sqrt(d^(2)+(a-R)^(2)))`
`and sin alpha=cos(90^(@)-alpha)=(AP)/(AM)=(a+R)/(sqrt(d^(2)+(a+R)^(2)))`
Substituting these values in equation (i), we get:
`(1)/(mu)=(a-R)/(sqrt(d^(2)+(a-R)^(2)))xx(sqrt(d^(2)+(a+R)^(2)))/(a+R)`
`rArr muxx(a-R)xxsqrt(d^(2)+(a+R)^(2))=(a+R)xxsqrt(d^(2)+(a+R)^(2))`
`rArr" "d=(mu(a^(2)-R^(2)))/([(a+r)^(2)-mu(a-R)^(2)]^(1//2))`