## Mastering Physics Solutions: Birefringence in Calcite

Birefringence in Calcite

Part A = 42.28°
Part B = 37.08°
Part C = See below for the answer
Part D = parallel
Part E = See below for the answer

Solution Below:

Calcite (CaCO3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field parallel to the optical axis (called extraordinary waves or e-waves) is 1.4864. The index of refraction for light with electric field perpendicular to the optical axis (called ordinary waves or o-waves) is 1.6584.

Part A

Find the critical angle θce for e-waves in calcite.

Just use Snell’s law. The critical angle is (by definition) 90 degrees, so just solve for the value of θ1 that gives you 90 degrees under a refractive index of 1:

n1*sin(θ1) = n2*sin(θ2)
1.4864 * sin(θ1) = 1 * sin(90)
1.4864 * sin(θ1) = 1
sin(θ1) = 0.672766
θ1 = 42.28°

42.28°

Part B

Find the critical angle θco for o-waves in calcite.

Just use the same formula as in Part A:

n1*sin(θ1) = n2*sin(θ2)
1.6584 * sin(θ1) = 1 * sin(90)
1.6584 * sin(θ1) = 1
sin(θ1) = 0.60829908
θ1 = 37.08°

37.08°

Part C

Assume that the optical axis of the crystal is oriented horizontally (perpendicular to the direction of the incident ray). If unpolarized light shines into the calcite crystal shown and strikes the far side with θi = 40°, which of the following pictures correctly shows what will happen?>

The correct answer should look like this picture:

Part D

Is the transmitted ray (the one that passes into the air) polarized parallel to or perpendicular to the optical axis?

parallel

Part E

Now assume that the optical axis of the crystal is vertical (parallel to the direction of the incident ray). (The same figure still applies; just assume now that the optical axis is vertical.) Which of the following figures shows what would happen to the incident ray from Part C?

The correct answer should look like this picture: