Additional Snell's Law Example: Finding Index of Refraction

The first step in solving a problem with Snell's law is to make a sketch, showing surface between the two media, the normal to the surface and the incident and refracted rays. Any quantities that are known should be labeled.

 

Example 1: A student is measuring the index of refraction of a block of glass. The student directs a ray of light from a laser so that it is incident on the surface of the glass at a 25 degree angle. The student determines that the angle of refraction (inside the glass) is 15 degrees. What is the index of refraction of the glass?

 

Begin with a sketch. The angles in the sketch below aren't exact, but they are close enough to be "reasonable", which is important. Use a protractor if you aren't sure.

In the sketch, it is assumed that the material on the left of the glass is air, with in index of refraction of 1. Snell's Law is

n1sinq1=n2sinq2

Plugging in the values given for this problem,

(1) sin 25o = n2 sin 15o

In this equation, n2, the index of refraction of the glass, is the unknown. To solve the equation, first evaluate the sines using a calculator. (If you're not sure how to do this, see the angle and sines tutorial in the LECTURES area.)

(1) (0.4226) = n2 (0.2588)

To solve for n2, divide both sides of the equation by 0.2588:

n2 = 1.63

This is a reasonable value for an index of refraction for glass. To get a better experimental result, the student should repeat the experiment at several different angles of incidence.