**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:**

_{3}) 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

_{ce}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 θ_{1} that gives you 90 degrees under a refractive index of 1:

n_{1}*sin(θ_{1}) = n_{2}*sin(θ_{2})

1.4864 * sin(θ_{1}) = 1 * sin(90)

1.4864 * sin(θ_{1}) = 1

sin(θ_{1}) = 0.672766

θ_{1} = 42.28°

42.28°

Part B

_{co}for o-waves in calcite.

Just use the same formula as in Part A:

n_{1}*sin(θ_{1}) = n_{2}*sin(θ_{2})

1.6584 * sin(θ_{1}) = 1 * sin(90)

1.6584 * sin(θ_{1}) = 1

sin(θ_{1}) = 0.60829908

θ_{1} = 37.08°

37.08°

Part C

_{i}= 40°, which of the following pictures correctly shows what will happen?>

The correct answer should look like this picture:

See above for the answer

Part D

parallel

Part E

The correct answer should look like this picture:

See above for the answer