# Reflection and refraction laws and Fermat’s principle

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.3a

Use Fermat’s principle of stationary time to derive the law of reflection.

### Background

In the solution of problem 3.1a we showed that the angle of incidence equals the angle of reflection and that, for the angle of refraction ${\displaystyle \theta _{2}}$, ${\displaystyle {\rm {\;sin\;}}\theta _{2}=(V_{2}/V_{1}){\rm {\;sin\;}}\theta _{1}}$ [see also equation (3.1a)].

These are the laws of reflection and refraction. Fermat’s principle of least time (more accurately, of stationary time) states that wave travel between any two points is along the path for which the traveltime is either a maximum or a minimum value (i.e., the derivative of the traveltime equals zero) compared with the traveltimes along adjacent paths.

Figure 6.3a.  Deriving Snell’s law.

### Solution

In Figure 6.3a, the source ${\displaystyle S}$ and the receiver ${\displaystyle R}$ have coordinates ${\displaystyle (0,\;h_{1})}$ and ${\displaystyle (a,\;h_{2})}$. The traveltime for a wave from ${\displaystyle S}$ to ${\displaystyle R}$ with reflecting point ${\displaystyle M(x,\;0)}$ is

{\displaystyle {\begin{aligned}t=(1/V)\{(x^{2}+h_{1}^{2})^{1/2}+[(a-x)^{2}+h_{2}^{2}]^{1/2}\}.\end{aligned}}}

To find the point ${\displaystyle M}$ for which the value of ${\displaystyle t}$ is stationary, we differentiate ${\displaystyle t}$ with respect to ${\displaystyle x}$ and set the result equal to zero. Thus,

{\displaystyle {\begin{aligned}{\frac {{\rm {d}}t}{{\rm {d}}x}}=(1/V)\left\{{\frac {x}{(x^{2}+h_{1}^{2})^{1/2}}}-{\frac {(a-x)}{[(a-x)^{2}+h_{2}^{2}]^{1/2}}}\right\}=0.\end{aligned}}}

The two terms in the brackets are the sines of the angles ${\displaystyle \theta _{1}}$ and ${\displaystyle \theta _{1}^{'}}$; hence, ${\displaystyle {\rm {\;sin\;}}\theta _{1}={\rm {\;sin\;}}\theta _{2}^{'}.}$

## Problem 6.3b

Repeat part (a) for the refracted path SMQ, in Figure 5.3a.

### Solution

The traveltime for the path SMQ is

{\displaystyle {\begin{aligned}t=(x^{2}+h_{1}^{2})^{1/2}/V_{1}+[(b-x)^{2}+h_{3}^{2}]^{1/2}/V_{2}.\end{aligned}}}

Differentiation gives

{\displaystyle {\begin{aligned}{\frac {{\rm {d}}t}{{\rm {d}}x}}={\frac {x}{V_{1}(x^{2}+h_{1}^{2})^{1/2}}}-{\frac {(b-x)}{V_{2}[(b-x)^{2}+h_{3}^{2}]^{1/2}}}=0,\end{aligned}}}

that is, ${\displaystyle {\rm {\;sin\;}}\theta _{1}/V_{1}={\rm {\;sin\;}}\theta _{2}/V_{2}}$.

## Problem 6.3c

Repeat parts (a), (b) for reflected and refracted converted S-waves.

### Solution

If we replace the angles ${\displaystyle \theta }$ with the angles ${\displaystyle \delta }$ and use the S-wave velocities ${\displaystyle \beta }$, the foregoing proofs are otherwise unchanged.