Question

A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (Figure). 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 µ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Answer 1

In figure AM and BM are the rays from the ends of disc AB reaching at one end of bowl at M.MN is tangent at M, so MN ⊥ AB i.e., ∠N = 90°

Taking incidence ray BM and refracted ray MD

BN = CN – CB = OM – CB = a – R

MB = `sqrt(d^2 + (a - R^2))`

∴ `sin i = (BN)/(BM) = ((a - R))/sqrt(d^2 + (a - R)^2)`

 ∠r = ∠a = ∠AMN

`sin r = cos(90° - α) = (AN)/(AM) = (a + R)/sqrt(d^2 + (a + R)^2`

For incidence ray MN to the horizontal level of liquid MP, MN will be normal at M. ∠i and ∠r will be incidence and refracted angles when ray BM passes from liquid (μ) to air. By Snell's law, as ray passes from liquid to air

`1mu_0 = (sin  i)/(sin  r)` ⇒ `mu_0/mu_1 = (sin i )/(sin  r)`  .....[μ₀ for air = 1] [μ₁ for liquid]

`1/mu_0 = (sin i)/(sin r)`

`1/mu_1 = (sqrt(a - R)/sqrt(d^2 + (a - R)^2))/((a + R)/sqrt(d^2 + (a + R)^2)) = ((a - R) sqrt(d^2 + (a + R)^2))/((a + R)sqrt(a^2 + (a - R)^2)`

`d = (mu(a^2 - d^2))/(sqrt((a + r)^2 - mu(a - r)^2)`

 It is required expression.