A tank holds a 1.44-m thick layer of oil that floats on a 0.97-m thick layer of

Question

A tank holds a 1.44-m thick layer of oil that floats on a 0.97-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.40 and 1.52, respectively. A ray originating at O crosses the brine-oil interface at a point 0 m from the axis. The ray continues and emerges into the air above the oil. What is the angle that the ray in the air makes with the vertical? (Give the angle to the nearest 0.1°).

Explanation / Answer

A ray originating at O reaches the brine-oil interface at the critical angle."
Which is given by sin(theta) = 1.40/1.52, which tells you what theta is.

theta = arcsin ( 1.40/1.52) = 67.08 degree

You have a right triangle whose height is 0.97 m, and that's the adjacent of theta. The unknown horizontal distance is the opposite of theta.

So tan(theta) = x/0.97

x = tan 67.08 degree * 0.97

= 2.294 m

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