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

Question

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?

Please show work for 5*

Explanation / Answer

Let angle of incidence at brine-oil interface be a.

tana = 0.6/0.96

a = 32 degrees.

Let angle of refraction be r1

n1sina = n2sinr1

sinr1 = n1sina/n2

r1 = sin-1(n1sina/n2)

r1 = 35.13 degrees

Angle of incidence between air and oil = angle of refraction between brine and oil

Angle of refraction = r2

n3sinr2 = n2sinr1

r2 = sin-1(n2sinr1/n3)

n3 = 1

so, r2 = 53.67 degrees.

Angle that ray makes with vertical = 53.67 degrees

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