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