A layer of ethyl alcohol (n = 1.361)is on top of water (n = 1.333).To the nearest degree,at what angle relative to the normal to the interface of the two liquids is light totally reflected?
A)78$\degree$
B)88$\degree$
C)68$\degree$
D)49$\degree$
E)the critical angle is undefined

Question

Quizplus · Accepted Answer

The critical angle is given by sin(theta_c) = n2/n1, where n1 is the index of refraction of the less dense medium and n2 is the index of refraction of the denser medium. In this case, n1 = 1.333 and n2 = 1.361, so sin(theta_c) = 1.333/1.361 = 0.979. Therefore, theta_c = 78.1 degrees.

A Layer of Ethyl Alcohol (N = 1 °\degree° B)88 °\degree° C)68 °\degree°

A Layer of Ethyl Alcohol (N = 1 $\degree$ B)88 $\degree$ C)68 $\degree$