Question

Calcite (CaCO_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.A.) Find the critical angle theta_ce for e-waves in calcite. This was 42.28 degrees.B.) Find the critical angle theta_co for o-waves in calcite. I got 38.68 then 35.8 and am still wrong. Please help me find part B.

Answer 1

Answer:

(a) 42.28°

(b) 37.08°

Explanation:

From the principle of refraction of light, when light wave travels from one medium to another medium, we have:

$\frac{n_{b} }{n_{a} }$ = sinθ$_{a}$/sinθ$_{b}$

In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.

For critical angle θ$_{a}$ = θ$_{c}$, θ$_{b}$ = 90°; $n_{b} = 1$

(a) $n_{a} = 1.4864$

$\frac{1 }{1.4864 }$ = sinθ$_{c}$/sin90°

0.6728 = sinθ$_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584$

$\frac{1 }{1.6584}$ = sinθ$_{c}$/sin90°

0.60299 = sinθ[tex]_{c}

θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°