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 0.5 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?

The answer is 44.6. I need an explanation as to why this is the answer. Thanks!

Answer 1

firs consider refraction at brine-oil interface.

let incident angle be theta.

(this is the angle the incident ray makes with vertical at the interface)

then with horizontal, it will make an angle of (90-theta) degree.

let the point where the incident ray strikes is named P and the point where vertical and horizontal axis (here horizontal axis mean the line made by brine-oil interface) be named Q.

given that PQ=0.5 m

OQ=0.96 m

then 90-theta=arctan(OQ/PQ)=62.488 degrees

==>theta=90-62.488=27.512 degrees

now let the refraction angle (angle made by the refracted ray with vertical at the interface) is theta1.

then using snell's law:

1.52*sin(27.512)=1.4*sin(theta1)

==>sin(theta1)=0.50153

==>theta1=30.101 degrees

now, consider the oil-air interface.

here the previous refracted ray will act

like incident ray.

incident angle=30.101 degrees

let refraction angle be theta2.

refractive index of air=1

using snell's law:

1.4*sin(30.101)=1*sin(theta2)

==>theta2=44.599 degrees=44.6 degrees