Answer to Question #162791 in Optics for Odu Ephraim

Question

The difference between the indices of carbon bi sulphide for blue light and red light is 0.48 while the critical angle for the red light at carbon bi sulphide air interface is 49°. Calculate the critical Ray of the carbon bi sulphide for the blue light. Thank you

Answer 1

Instead of always having to measure the critical angles of different materials, it is possible to calculate the critical angle at the surface between two media using Snell's Law. To recap, Snell's Law states:

$n_1sinθ_1=n_2sinθ_2$

where $n_1$ is the refractive index of material 1, $n_2$  is the refractive index of material 2,

$θ_1$  is the angle of incidence and $θ_2$ is the angle of refraction. For total internal reflection, we know that the angle of incidence is the critical angle. So,

$θ_1=θ _c.$

However, we also know that the angle of refraction at the critical angle is 90°. So we have:

$θ_2=90^∘.$

We can then write Snell's Law as:

$n_1sinθ_c=n_2sin90^∘$

Solving for $θ_c$  gives:

$n_1sinθ _c = n_2sin90°$

$sinθ_c = \large\frac{n_2}{n_1}$

$∴θ_c=sin^{−1}(\large\frac{n_2}{n_1})$

We have $n_2 = 1$ for air $\theta_c = 48^o$

$sin48^o = \large\frac{1}{n_1} \to 0.74n_1 = 1 \to n_1 = 1.35$

$n_2 = n_1 + 0.48 = 1.81$

Critical ray for carbon bi sulphide for the blue light

$sin\theta _c = \frac{n_2}{n_1} = \frac{1}{1.85} = 0.54 \to \theta_c = 33.5^o$