Derive n = 1/sin C for refractive index in terms of critical angle C. Include all steps and assumptions.

Solution & explanation

1. Assume light travels from a transparent material of index n₁ to air of index n₂ ≈ 1, with n₁ > n₂.

2. Write Snell’s law; both angles are measured from the normal.

   $$n_1\times \sin \left(i\right)=n_2\times \sin \left(r\right)$$

3. At the critical angle i = C, the refracted ray grazes the boundary: r = 90°.

   $$n_1\times \sin \left(C\right)=n_2\times \sin \left(\frac{\pi }{2}\right)$$

4. Use sin 90° = 1.

   $$n_1\times \sin \left(C\right)=n_2$$

5. Divide both sides by n₁.

   $$\sin \left(C\right)=\frac{n_2}{n_1}$$

6. Take reciprocals to obtain the relative index.

   $$\frac{n_1}{n_2}=\frac{1}{\sin \left(C\right)}$$

7. For air n₂ ≈ 1, so the material’s index is approximately this value.

   $$n=\frac{1}{\sin \left(C\right)}$$

Start with n₁ sin i = n₂ sin r. At critical incidence i = C and r = 90°, so n₁ sin C = n₂×1. Divide by n₁: sin C = n₂/n₁. Therefore the relative index n₁/n₂ = 1/sin C. For escape to air take n₂ ≈ 1, giving the material’s n ≈ 1/sin C. The air assumption must not be omitted.