Snell's law: 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?

Question

Quizplus · Accepted Answer

Using Snell's law, we can find the angle of refraction at the oil-brine interface:
$n_1\sin	heta_1=n_2\sin	heta_2$
where $n_1$ and $n_2$ are the indices of refraction of the two mediums and $	heta_1$ and $	heta_2$ are the angles of incidence and refraction relative to the normal.
At the brine-oil interface, we have:
$n_{brine}\sin	heta_1=n_{oil}\sin	heta_2$
Plugging in the given values:
$1.52\sin	heta_1=1.40\sin	heta_2$
We also know that the ray crosses this interface 0.60 m from the axis, which means that $	heta_1$ and $	heta_2$ must be related by:
$	heta_2=	heta_1+\alpha$
where $\alpha$ is the angle of incidence at the oil-air interface. Using Snell's law again, we have:
$n_{oil}\sin	heta_2=n_{air}\sin\alpha$
Plugging in the given values:
$1.40\sin(	heta_1+\alpha)=1.00\sin\alpha$
Expanding the sine:
$1.40\sin	heta_1\cos\alpha+1.40\cos	heta_1\sin\alpha=1.00\sin\alpha$
We can use the first equation we derived to eliminate $\sin	heta_1$:
$1.52\frac{1.40}{1.52}\cos\alpha+1.40\cos	heta_1\frac{1.40}{1.52}=1.00$
Simplifying:
$\cos\alpha+0.809\cos	heta_1=0.658$
We know that $	heta_1$ must be acute since the ray is traveling away from the axis. Furthermore, if $	heta_1$ were close to 90°, then $	heta_2$ would be close to 90° as well and the ray would not emerge into the air. Therefore, we can make the approximation $\cos	heta_1\approx 1$ (i.e. $	heta_1\approx 0$) and solve for $\cos\alpha$:
$\cos\alpha+0.809\approx 0.658$
$\cos\alpha\approx -0.151$
Since $\alpha$ is acute as well, we must have:
$\alpha\approx 104^\circ$
Finally, the ray in the air makes an angle of $	heta_2+\alpha$ with the vertical line, so we have:
$	heta_2+\alpha\approx 54^\circ$

Snell's Law: a Tank Holds a 1