A layer of water (n = 1.333)floats on carbon tetrachloride (n = 1.461)contained in an aquarium.To the nearest degree,what is the critical angle at the interface between the two liquids?
A)88$\degree$
B)78$\degree$
C)66$\degree$
D)58$\degree$
E)43$\degree$

Question

Quizplus · Accepted Answer

The critical angle can be calculated using Snell's Law: sin(θc) = n2/n1, where n1 is the refractive index of water and n2 is the refractive index of carbon tetrachloride. Thus, sin(θc) = 1.461/1.333. Solving for θc gives approximately 66 degrees.

A Layer of Water (N = 1 °\degree° B)78 °\degree° C)66 °\degree° D)58 °\degree°

A Layer of Water (N = 1 $\degree$ B)78 $\degree$ C)66 $\degree$ D)58 $\degree$