A layer of oil $(n=1.45)$ floats on water $(n=1.33)$. A ray of light shines onto the oil with an incidence angle of $40.0^{\circ}$. Find the angle the ray makes in the water. (See Fig. $37-5 .$ )
At the air-oil interface, Snell's Law gives
$$
n_{\text {air }} \sin 40^{\circ}=n_{\text {oil }} \sin \theta_{\text {oil }}
$$
At the oil-water interface, we have (using the equality of alternate angles)
$$
n_{\text {oil }} \sin \theta_{\text {oil }}=n_{\text {water }} \sin \theta_{\text {water }}
$$
Thus, $n_{\text {air }} \sin 40.0^{\circ}=n_{\text {water }} \sin \theta_{\text {water }}$; the overall refraction occurs just as though the oil layer were absent. Solving gives
$$
\sin \theta_{\text {water }}=\frac{n_{\text {air }} \sin 40.0^{\circ}}{n_{\text {water }}}=\frac{(1)(0.643)}{1.33} \quad \text { or } \quad \theta_{\text {water }}=28.9^{\circ}
$$