Part A A tank containing methanol has walls 2.50 cm thick made of glass of refra

Question

Part A A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3 angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. Methanol has a the refractive index of 1.329. Part B The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2 from the normal, what is the refractive index of the unknown liquid?

Explanation / Answer

Part A
If X = angle in glass
then sin(41.3) = 1.55*sin(X) by Snell's Law
so X = arcsin(sin(41.3)/1.55) = 25.2 degrees

If Y = required angle in methanol
then 1.55*sin(25.2) = 1.329*sin(Y) which also = sin(41.3) from above
so Y = arcsin(sin(41.3)/1.329) = arcsin(0.4966) = 29.8 degrees

Part B
If n = refractive index of unknown liquid
then 1.55*sin(25.2) = sin(41.3) = n*sin(20.2)
so n = sin(41.3) / sin(20.2)
n = 1.91

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