General form of Snell’s law

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Problem 3.1a

3.1a A P-wave of amplitude $ A_{0} $ is incident at the angle $ \theta _{1} $ on a plane interface separating two solid media. This generates reflected and refracted P-waves and converted reflected and refracted S-waves. Amplitudes, angles of incidence and refraction, and directions of displacements of these waves are shown in Figure 3.1a.

Use Huygens’s principle to show that $ \theta _{1}=\theta _{1}^{\prime } $ and that


$ {\begin{aligned}{\frac {\sin \theta _{1}}{\alpha _{1}}}={\frac {\sin \delta _{1}}{\beta _{1}}}={\frac {\sin \theta _{2}}{\alpha _{2}}}={\frac {\sin \delta _{2}}{\beta _{2}}}=p,\end{aligned}} $ (3.1a)

where $ p $ is the raypath parameter. Equation (3.1a) is Snell’s law.

Background

When a wave is incident at the interface between two solid media, four boundary conditions must be satisfied (continuity of normal and tangental displacements and stresses). The velocities are determined by the densities and elastic constants while the angles of incidence, reflection, and refraction are fixed by the velocities [see equation (3.1a)]. So the only remaining parameters that one can adjust in order to satisfy the boundary conditions are the amplitudes of the four waves generated by the incident wave, the reflected and refracted P- and S-waves, $ A_{1} $, $ A_{2} $, $ B_{1} $, $ B_{2} $ as shown in Figure 3.1a.

Figure 3.1a.  Raypaths at solid-solid interface.

Huygens’s principle states that each point on a wavefront acts as a new point source radiating energy in all directions. Subsequent wavefronts can be located by swinging arcs with centers at points on the wavefront and radii equal to the distance traveled in a fixed time interval, the new wavefront being the envelope of the arcs. If the first wavefront and the reflector are planar, only two arcs are necessary, the new wavefront being tangent to the arcs.

Solution

In Figure 3.1b, $ AB $ is a wavefront of a planar P-wave approaching a planar interface. When the wavefront reaches the interface, point $ A' $ becomes a new source radiating energy upward and downward according to Huygens’s principle. When $ B' $ reaches the interface at $ R $, the distance $ B'R $ being $ \alpha _{1}\Delta t $, the wave reflected at $ A' $ has traveled upward the same distance $ \alpha _{1}\Delta t $. By drawing an arc with center $ A' $ and radius $ \alpha _{1}\Delta t $ and then drawing a line from $ R $ tangent to the arc, we get the reflected wavefront $ SR $. The angle of incidence is $ \theta _{1} $ and the angle of reflection is $ \theta _{1}^{\prime } $. In $ \Delta A'B'R $ and $ \Delta A'SR, $ the angles at $ B' $ and $ S $ are $ 90^{\circ } $ (because rays are perpendicular to wavefronts). Since the triangles have a common side $ A'R $, they are equal and $ \angle \theta _{1}=\angle \theta _{1}^{\prime } $, that is, the angle of incidence equals the angle of reflection (Law of reflection).

Figure 3.1b.  Snell’s law derivation.

In the case of the refracted wave, $ \theta _{2} $ is the angle of refraction and $ \left(\alpha _{2}\Delta t\right)/A'R=\sin \theta _{2} $ But $ \left(\alpha _{1}\Delta t\right)/A'R=\sin \theta _{1} $, so we have

$ {\begin{aligned}\Delta t/A'R=\sin \theta _{1}/\alpha _{1}=\sin \theta _{2}/\alpha _{2}.\end{aligned}} $

In Figure 3.1b, if we replace $ \theta _{1}^{\prime } $ in $ \Delta A'SR $ with $ \delta _{1} $ (compare with Figure 3.1a) and $ \theta _{2} $ in $ \Delta A'TR $ with $ \delta _{2} $, we arrive at

$ {\begin{aligned}\Delta t/A'R=\sin \delta _{1}/\beta _{1}=\sin \delta _{2}/\beta _{2}.\end{aligned}} $

Equating the four ratios of sines to velocities, we get equation (3.1a).

Problem 3.1b

3.1b Using the waveform $ e^{\mathrm {j} \omega [(lx+nz)/V-t]} $ (see problem 2.5b), where $ (l,n) $ are direction cosines of the ray, show that (omitting the factor $ e^{-\mathrm {j} \omega t} $) the incident, reflected, and refracted waves can be written


$ {\begin{aligned}\psi _{0}=A_{0}e^{\mathrm {j} \omega \zeta _{0}},\quad \psi _{1}&=A_{1}e^{\mathrm {j} \omega \zeta _{1}},\quad \psi _{2}=A_{2}e^{\mathrm {j} \omega \zeta _{2}},\end{aligned}} $ (3.1b)


$ {\begin{aligned}\psi _{1}^{'}&=B_{1}e^{\mathrm {j} \omega \zeta _{1}^{'}},\quad \psi _{2}^{'}=B_{2}e^{\mathrm {j} \omega \zeta _{2}^{'}},\end{aligned}} $ (3.1c)

where


$ {\begin{aligned}\zeta _{0}&=p\left(x-z\cot \theta _{1}\right),\quad \zeta _{1}=p\left(x+z\cot \theta _{1}\right),\quad \zeta _{2}=p\left(x-z\cot \theta _{2}\right),\end{aligned}} $ (3.1d)


$ {\begin{aligned}\zeta _{1}^{'}&=p\left(x+z\cot \delta _{1}\right),\quad \zeta _{2}^{'}=p\left(x-z\cot \delta _{2}\right).\end{aligned}} $ (3.1e)

Solution

We write $ \zeta =\left(lx\pm nz\right)/V $, where the plus sign is used for waves traveling upward (that is, in the positive $ z $-direction) and the minus for downward traveling waves. The velocity $ {\textit {V}} $ is $ \alpha $ for P-waves, $ \beta $ for S-waves. We note that for P-waves, $ l=\sin \theta _{i} $, $ n=\cos \theta _{i} $, $ i=1 $, 2. For S-waves we replace $ \theta $ with $ \delta $ so that $ l=\sin \delta _{i} $, $ n=\cos \delta _{i} $, $ i=1 $, 2.

Thus, for P-waves, $ \zeta _{i}=\left(x\sin \theta _{i}\pm z\cos \theta _{i}\right)/\alpha _{i}=p\left(x\pm z\cot \theta _{i}\right) $ while for S-waves we have $ \zeta _{i}^{'}=p\left(x\pm z\cot \delta _{i}\right) $. Inserting the amplitudes $ A_{i} $, $ B_{i} $, we get equations (3.1b,c,d,e).

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