A cicular disc of radius R is placed co-axially and horizontally inside an opaque hemi spherical bowl of radius .a. (see figure). The faredge of the disc in 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 ?

Explanation

Consider situartion before pouring water in an opaque bowl of radius a. Now consider a circular disc of radius R and centre C which si placed inside the bowl horizontnally and coaxially with the bowl. Here we have to calculate OC=d .
`rArr` Here before filling bowl with water, incident light ray is AMA. . Now when water is paured in the hemispherical bowl completely, the nearer and B of the disc is seen for which inciden ray is `vec (BM)` and refracted ray is `vec(MA)`. Here NN. is perpedicular to water surface, drawn at point M. Hence `anlge (BMN)` = i (angle of incidence) and `angle NMA = alpha` (angle of refraction)
`rArr` Now, applying Snell.s law at point M ,
`u sin i = (1) sin r`
`therefore mu sin i = sin alpha (therefore r = alpha)`
`therefore 1/mu = (sin i)/(sin alpha)` ........(1) `[because r = alpha]`
Now, from , figure
`sin i = (BN.)/(BM)`
but `BN. = CN. - CB = OM -CB`
and `BM = sqrt(d^2 + (a-R)^2)`
`therefore sin i = sqrt(a-R)/(sqrt(d^2 + (a-R)^2))`...... (2)
and `sin alpha = cos(90^@ - alpha) = (AN.)/(AM) = (AC + CN.)/(AM)`
`therefore sin alpha = (AC +OM)/(AM)`
`= (a+R)/(sqrt(d^2(a+R)^2)) ........(3)`
`therefore` From equation (1),(2) and (3),
`1/mu = (a-R)/(sqrt(d^2 +(a-R)^2)) xx (sqrt(d^2 + (a+R)^2))/(a+R)`
`therefore mu(a-R)(sqrt(d^2 + (a+R)^2))=(a+R)(sqrt(d^2+(a-R)^2))`
Taking squares on the both sides,
`therefore mu^2(a-R)^2 {d^2 + (a+R)^2}=(a+R)^2{d^2 + (a-R)^2}`
`therefore {mu^2(a-R)^2d^2} + {mu^2(a-R)^2(a+R)^2} = {d^2(a+R)^2} + {(a+R)^2(a-R)^2}`
`therefore(a-R)^2(a+R)^2(mu^2 -1)=d^2{(a+R)^2-mu^2(a-R)^2}`
`therefore d^2 = ((mu^2-1){(a-R)(a+R)}^2)/((a+R)^2 - mu^2 (a-R)^2)`
`therefore d^2 = ((mu^2-1)(a^2-R^2)^2)/((a+R)^2-mu^2(a-R)^2)`
`therefore d=sqrt((mu^2-1)/((a+R)^2-mu^2(a-R)^2))xx (a^2 xxb^2)`
`rArr` Above equation gives required result.