Question

A coin of radius 'r' is kept at the bottom of a hemispherical vessel of radius 'a'. The coin is not visible when viewed from the edge of the vessel but just visible when a liquid of refractive index 'mu' is poured into the vessel. Prove that r=a( ((mu^2) - 1)/(mu^2 +1))

Answer 1

r;
a;
μ.
Prove.
r=a (μ^2 - 1)/(μ^2+1).
Solution.
First of all, let's analyze the result

r=a (μ^2 - 1)/(μ^2+1)

If μ=1 (without a liquid) then r=0! but according to the condition of the problem the radius of the coin is given. Hence, what is r in this formula? Can the radius of the coin depend on the μ? Probably no.

Answer. It is necessary to clarify the condition of the problem.