A tank holds a layer of oil, 1.58 m thick, which floats on a layer of syrup that

Question

A tank holds a layer of oil, 1.58 m thick, which floats on a layer of syrup that is 0.66 m thick. Both liquids are clear and do not intermix. A ray, which originates at the bottom of the tank on a vertical axis, crosses the oil-syrup interface at a point 0.90 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m from the axis and at the critical angle. In the figure, the index of refraction of the syrup is closest to:

A) 1.37 B) 1.24 C) 1.17 D) 1.52 E) 1.66

A tank holds a layer of oil, 1.58 m thick, which floats on a layer of syrup that is 0.66 m thick. Both liquids are clear and do not intermix. A ray, which originates at the bottom of the tank on a vertical axis, crosses the oil-syrup interface at a point 0.90 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m from the axis and at the critical angle. In the figure, the index of refraction of the syrup is closest to: A) 1.37 B) 1.24 C) 1.17 D) 1.52 E) 1.66

Explanation / Answer

First we can calculate the angle at which the ray strikes the oil-syrup interface.

From the geometry of the problem, tan = .66/.90

= 36.25o

Then, from the geometry of the triangle in the oil, we know the light will move be 2.0 - .9 or 1.1 m more away from the axis.

Thus tan = 1.58/1.1

= 55.15o

Then, for the oil-air interface, we can use the critical angle formula

sin c = 1/noil

Substituting, sin 55.15 = 1/noil

noil = 1.219

Finally, using Snells Law

n1sin1 = n2sin2

(1.219)(sin 55.15) = (n2)(sin 36.25)

n2 = 1.69   (with rounding, the closest answer is choice E)

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