Question

A tank holds a 1.44-m thick layer of oil that floats on a 0.98-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 0.7 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 axis

Answer 1

Answer:

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

$tan\theta_i = \frac{0.7}{0.98}$

$\theta_i = 35.5^0$

now by Snell'a law at that interface we have

$\mu_1 sin\theta_i = \mu_2 sin \theta_r$

now we will have

$1.52 sin35.5 = 1.40 sin\theta_r$

$\theta_r = 39.12^0$

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

$\mu_2 sin\theta_i' = \mu_{air} sin\theta$

$1.40 sin39.12 = 1 sin\theta$

$\theta = 62.1^0$