4 Engineering Example 2

4.1 Refraction

The problem

A light ray is travelling in a medium ( A ) at speed c A . The ray encounters an interface with a medium ( B ) where the velocity of light is c B . Between two fixed points P in media A and Q in media B , find the path through the interface point O that minimizes the time of light travel (see Figure 18). Express the result in terms of the angles of incidence and refraction at the interface and the velocities of light in the two media.

Figure 18 :

{ Geometry of light rays at an interface}

The solution

The light ray path is shown as P O Q in the above figure where O is a point with variable horizontal position x . The points P and Q are fixed and their positions are determined by the constants a , b , d indicated in the figure. The total path length can be decomposed as P O + O Q so the total time of travel T ( x ) is given by

T ( x ) = P O c A + O Q c B . (1)

Expressing the distances P O and O Q in terms of the fixed coordinates a , b , d , and in terms of the unknown position x , Equation (1) becomes

T ( x ) = a 2 + x 2 c A + b 2 + ( d x ) 2 c B (2)

It is assumed that the minimum of the travel time is given by the stationary point of T ( x ) such that

d T d x = 0. (3)

Using the chain rule in ( HELM booklet  11.5) to compute (3) given (2) leads to

1 2 2 x c A a 2 + x 2 + 1 2 2 x 2 d c B b 2 + ( d x ) 2 = 0.

After simplification and rearrangement

x c A a 2 + x 2 = d x c B b 2 + ( d x ) 2 .

Using the definitions sin θ A = x a 2 + x 2 and sin θ B = d x b 2 + ( d x ) 2 this can be written as

sin θ A c A = sin θ B c B . (4)

Note that θ A and θ B are the incidence angles measured from the interface normal as shown in the figure. Equation (4) can be expressed as

sin θ A sin θ B = c A c B

which is the well-known law of refraction for geometrical optics and applies to many other kinds of waves. The ratio c A c B is a constant called the refractive index of medium ( B ) with respect to medium ( A ) .