Let the surface separating medium 1 and medium 2, as shown. Let 𝑣₁ and 𝑣₂ represent the speed of light in medium 1 and medium 2, respectively. We assume a plane wavefront AB propagating in the direction AC incident on the interface at an angle i. Let Δt be the time taken by the wavefront to travel the distance BC. Thus, BC = v₁Δt. In this time, the secondary wavelet from B travels to D in medium 2, covering a distance BD = v₂Δt. In the right-angled triangle BCD, we have sin i = BC/CD and sin r = BD/CD. Therefore,
sin i / sin r = BC/BD = (v₁Δt)/(v₂Δt) = v₁/v₂
We have n₁ = c/v₁ and n₂ = c/v₂, where n₁, n₂ are refractive indices of respective media and c is the speed of light in vacuum. Hence
n₁ sin i = n₂ sin r