Consider three incident rays of light encountering an interface between two media. In this example, the second medium is the slower medium and the rays are refracted towards the normal - note that angle A is greater than angle B in the diagram.

Since all rays are perpendicular to their respective wavefronts,m angle A + m angle 1 = 90m angle B + m angle 2 = 90

Since all normals are perpendicular to their respective interfaces,m angle C + m angle 1 = 90m angle D + m angle 2 = 90

Therefore, m angle C = m angle A and m angle D = m angle B so we can now examine the following new relationships:sin A = d1/Lsin B = d2/L

where L is the distance along the interface between points P1 and P2 as shown in the diagram below.

Solving each equation for L yields:L = d1/sin AL = d2/sin B

Therefore, d1/sin A = d2/sin B If d1 and d2 represent the distances traveled in the respective mediums during the same amount of time, then we can replace them with the expressionsd1 = v1td2 = v2t

But v1 and v2 represent the speed of the waves in each medium and can be replaced with the expressions n1 = c/v1 ---> v1 = c/n1 n2 = c/v2 ---> v2 = c/n2

where n1 and n2 are the respective indices of refraction and c is the speed of light. At this junction, we can now writed1/sin A = d2/sin Bv1t/sin A = v2t/sin B

(c/n1)(t/sin A) = (c/n2)(t/sin B) Canceling the common terms (c and t) yields(1/n1)(1/sin A) = (1/n2)(1/sin B)n1sin A = n2sinB

Or, as Snell's Law is more commonly expressed: n1sin theta1 = n2sin theta2

Notice that Snell's Law shows that the index of refraction and the sine of the angle of refraction are inversely proportional - that is, as the refractive index gets larger [n2 > n1] the sine of the refracted angle gets smaller [sinθ2 < sinθ1], since the product of the two terms must remain a constant.