Additional Snell's Law Example: Refraction by a rectangle

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 2: Continue example 1 and show that light exits the block in a direction parallel to the direction of the incoming ray. That is, when the ray inside the block strikes the right edge, it will refract at an angle of 25 degrees.

Begin with a sketch. Note that the problem of the previous example is shown in red, and the rays for this problem are shown in green.

The first problem is to determine the incident angle, the angle labeled q 1, where the ray strikes the right side of the block of glass. The answer lies in a geometry theorem- "opposite interior angles of parallel lines are equal". In this case, the two normal to the left and right edges of the blocks are parallel lines. The two angles, 15 degrees and q 1, are "opposite interior angles", so q 1 must equal 15 degrees.

Now we can use Snell's law, since we know the angle of incidence (15 degrees), the index of refraction on the incident side (1.63, from Example 1), and the index of refraction on the refracting side (1.0, since the block is surrounded by air).

n1sinq1=n2sinq2

Plugging in the values given for this problem,

(1.63) sin 15o = (1.0) sin q 2

To solve the equation, first evaluate the sine of 15 degrees using a calculator. (If you're not sure how to do this, see the trigonometry tutorial.)

(1.63) (0.2588) = (1.0) sin q 2

Solve first for sin q 2:

sin q 2= 0.4218

When the sine is known and you want the angle, use the "second" sine function (see the tutorial if you don't know what this is)

q 2 = sin-1 (0.4218) = 25 degrees

So, the ray of light that exits the glass is parallel to the ray that entered.

If this weren't the case (and it isn't in some old windows), the view of the outside world would be distorted.