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Velocity vector of light changes as it passes from one medium to another. It's a change at magnitude and direction. And it will be the starting point of this solution. (See the figures left and below)
AO is the vector v1 and BO is the vector v2, AB is the vectoral change. (The magnitude of v1=c/n1) |
| Draw perpendicular lines to AO at the point A, and to BO at the point B respectively. They cross at point K. OK is hypotenuse of both triangles AOK and BOK. The points O,A,B,K are on the same circle of which diameter is OK. Then NO is the tangent line to that circle.
The angles of AKO and AON have the same arc(AO), the angles of BKO and BON have the same arc(BO) then;
The angle of incidence = s(AON) = s(AKO) = alpha
v1/sin(AKO) = hypotenuse
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