A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30$\degree$ relative to the normal.If the index of refraction of water and glass are 1.33 and 1.5,respectively,at what angle (in degrees)does the light leave the glass (relative to its normal)?
A)26
B)35
C)42
D)22
E)48

Question

Quizplus · Accepted Answer

Using Snell's Law, n1 * sin(theta1) = n2 * sin(theta2), where n1 = 1.33 (water), theta1 = 30 degrees, n2 = 1.5 (glass). Solving for theta2 gives sin(theta2) = (1.33/1.5) * sin(30), which results in theta2 ≈ 26.4 degrees. When the light exits the glass into air (n = 1), using Snell's Law again, sin(theta3) = (1.5/1) * sin(26.4), which results in theta3 ≈ 42 degrees.

A Diver Shines Light Up to the Surface of a Flat