An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day. A light ray from the air above enters the oil at 45° with the normal and then goes
Into the water. What is the angle of refraction of this ray in the water? The index of refraction for
The oil is 1.15, and for water it is 1.33.
A) $39 ^ { \circ }$
B) $27 ^ { \circ }$
C) $36 ^ { \circ }$
D) $32 ^ { \circ }$

Question

An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day. A light ray from the air above enters the oil at 45° with the normal and then goes
Into the water. What is the angle of refraction of this ray in the water? The index of refraction for
The oil is 1.15, and for water it is 1.33. 
A) $39 ^ { \circ }$
B) $27 ^ { \circ }$
C) $36 ^ { \circ }$
D) $32 ^ { \circ }$

Quizplus · Accepted Answer

Using Snell's Law, $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the indices of refraction for the first and second mediums respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction respectively. First, calculate the angle of refraction in the oil, then use that angle as the angle of incidence from the oil into the water. The calculation will show that the angle of refraction in the water is approximately $32^\circ$.

An Oil Layer That Is 5 39∘39 ^ { \circ }39∘

An Oil Layer That Is 5 $39 ^ { \circ }$