Reflection and Refraction — Question 5

Question 5

Stationary travel time generates Snells law

Point A is at (0,h1) in index n1 and point B at (L,-h2) in index n2, with positive heights and L. A ray crosses the plane y=0 at (x,0), where zero is less than x and x is less than L.

1. Write the travel time and differentiate it with respect to x.

2. Use the stationary-time condition to obtain Snells law, and show the stationary point is a minimum.

Question 5 - Solution

Stationary travel time generates Snells law

(a) Method and calculation

$$T(x)=\frac{n_1}{c}\sqrt{x^2+h_1^2}+\frac{n_2}{c}\sqrt{(L-x)^2+h_2^2}.$$$$cT^\prime=\frac{n_1x}{\sqrt{x^2+h_1^2}}-\frac{n_2(L-x)}{\sqrt{(L-x)^2+h_2^2}}.$$

(b) Interpretation and further calculation

$$T^\prime=0\Rightarrow\boxed{n_1\sin\theta_1=n_2\sin\theta_2}.$$$$\boxed{cT^{\prime\prime}=\frac{n_1h_1^2}{(x^2+h_1^2)^{3/2}}+\frac{n_2h_2^2}{[(L-x)^2+h_2^2]^{3/2}}>0}.$$

Physical check

The positive second derivative makes the path the unique minimum of this two-segment travel-time function.