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 disc becomes just visible. How far below the top of the bowl is the disc placed ?
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Explanation

Refering to the figure, AM is the direction of incidence ray before liquid is filled. After liquied is filled in, BM is the direction of the incident ray. Refracted ray in both cases is same as the along. AM.
Let the disc is separated by O at a distance d as shown in figure. Also, considering angle

`" " =90^(@),OM=a,CB=NB=a-R,AN=a+R`
Hence , in figure
`" " sint =(a-R)/(sqrt(d^(2)+(a-R)^(2)))`
and `" " sinalpha=cos(90-alpha)=(a+R)/(sqrt(d^(2)+(a+R)^(2)))`
But on applying Snell's law.
`" " (1)/(mu)=(sint)/(sinr)=(sint)/(sinalpha)`
On substtituting the values, we have the sepration
`" " d=(mu(a^(2)-b^(2)))/(sqrt((a+r)^(2)-mu(a-r)^(2)))`
This is required expression