There an analogy that vividly indicates why a beam of light would bend at the interface of materials in which light travels at different speeds. Imagine a column of soldiers marching twelve abreast on concrete whose path is going to take them into a field in which there is mud. Suppose the column of soldiers crosses the edge of the concrete at angle different from the perpendicular. The first soldiers to cross the edge of the concrete into the mud will be slowed down compared to the other soldiers still marching on the concrete. A rank of soldiers pivots as a result of the slower marching of those at one end of the rank who are marching in the mud compared to those at the other end who are still marching on the concrete. All of the turning of the rank occurs between the time the first soldier and last soldier in a rank hit the mud. This is illustrated below.

In the extreme case in which no progress at all can be made in the mud, one end of the rank is held fixed and the other end procedes until it is stopped by the mud. In this extreme case the rank assumes the position of interface and the deviation of its normal from the normal of the interface is zero.

Consider the diagram that shows what happens between the time that one end of a rank touches the new slower media and the time when the other end of the rand enters the new media.

In the diagram the angles ABC and CDA are supposed to be right angles. The angle of the rank with the horizontal in the diagram before encountering the slower medium is α. In the slower medium the angle is β

The time required to traverse the distance |BC| is |BC|/v₁. This is the same time required to traverse the distance |AD|; i.e., |AD|/v₂. These are equal; i.e.,

|BC|/v₁ = |AD|/v₂.

But

|BC| = |AC|sin(α) and |AD| = |AC|sin(β)
so
|AC|sin(α)/v₁ = |AC|sin(β)/v₂
and hence Snell's Law
sin(α)/sin(β) = v₁/v₂

Although α and β were defined with respect to the horizontal the angles of the normals to the lines with respect to the vertical are exactly equal to α and β.