Total Internal Reflection Question - Optics. Difficult to Understand!?

Question

Question: A layer of liquid B floats on liquid A. A ray of light begins in liquid A and undergoes total internal reflection at the interface between the angle of incidence exceeds 36.5 degrees. When liquid B is replaced with liquid C, total internal reflection occurs for angles of icidence greater than 47.0 degrees. Find the ratio nB/nC of the refractive indices of liquids B and C.

My Approach: I usually have one for every question I post, but this time I am completely STUMPED! I know they provide both angles of incidence, but I cannot understand the part where liquid B is replaced with liquid C. How am I supposed to figure the B in order to put in the nB/nC ratio???

I would love to see how to work this problem!
Answer from the Book: 0.813

Answer 1

The angle for total internal reflection is given by

sinø = n1/n2, where n1 = index of less dense medium

In your case, the mediums B and C both float on medium A, so they are both less dense than A. Therefore

sinø1 = nB/nA

sinø2 = nC/nA

nB/nA divided by nC/nA is nB/nC = sinø1/sinø2 = sin(36.5º)/sin(45º) = 0.813

Answer 2

whilst a delicate ray passes from a denser medium to a rarer medium the sunshine ray is bend faraway from the conventional. So if we pass on increasing the attitude of prevalence then there will be a factor whilst gentle ray will trip on the boundary of the two media. this attitude is talked approximately as serious perspective. Now if we improve the attitude better than serious perspective then gentle ray travels interior the comparable medium and subsequently this is thoroughly internally pondered. This actually relies upon on the refractive indices of the ingredients.