Please show all work on how to do this. I am lost on this question. Full points

Question

Please show all work on how to do this. I am lost on this question. Full points awarded to first with problem step by step. Thanks

A tank holds a 1.44-m thick layer of oil that floats on a 0.96-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 0.60 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?

Explanation / Answer

let the angle of incidence on the brine-oil interface is alpha =>

tan(alpha) = 0.6/0.96 = 5/8

sin(alpha) = 5/(89^0.5)

applying snells law we get

Ub(sin(alpha) = Uosin(B)

where B is the angle of emergence into oil

By applying snells law at the oil-air interface

we get Uosin(B) = Ua(sin(C))

sinC *1 = Ub *sin(alpha) = 1.52*(5/(89^0.5)) = 0.8

C = 53.13 degrees

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