---
title: "Snell’s law and the critical angle for seismic waves"
description: "Waves leaving a flat boundary keep the incident horizontal slowness. That rule fixes every reflected, transmitted and converted angle and each critical angle."
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language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/seismic-imaging/snell-law-and-critical-angles/"
last_modified: "2026-10-30"
---

[Seismic imaging · A course with experiments](https://sandboxphysics.com/stories/seismic-imaging/)

L1 · Chapter 2

# Snell’s law and the critical angle for seismic waves

A ray crossing into faster rock bends away from the vertical. Bend it far enough and the transmitted wave runs along the boundary instead of entering the rock: that incidence is the critical angle. One number, the horizontal slowness, stays the same for every reflected, transmitted and converted wave, so it fixes all their angles at once.

Sandbox Physics · 2026-10-30

In this chapter

[All eight chapters](https://sandboxphysics.com/stories/seismic-imaging/)

Before you start

Chapter one. Trigonometry, finding a minimum by setting a derivative to zero, and the idea of a complex exponential for an oscillation.

## The quickest path bends at the boundary

Measure every angle from the vertical, the normal to a flat boundary. A ray starts at point P in rock with speed c1 and ends at point Q in rock with speed c2. Inside each rock it runs straight, so the only freedom is where it crosses the boundary.

Fermat’s principle says the ray takes a path of stationary travel time. Write the time as a function of the crossing point, set its derivative to zero, and the two square-root terms turn into the sines of the two angles. The result is Snell’s law. The MIT notes for 12.201 do this in Section 4.14, Eqs. 4.91 to 4.94.

$$
t(x)=\frac{\sqrt{a^2+x^2}}{c_1}+\frac{\sqrt{b^2+(c-x)^2}}{c_2},\qquad \frac{dt}{dx}=0\ \Rightarrow\ \frac{\sin i_1}{c_1}=\frac{\sin i_2}{c_2}\equiv p
$$

## One ray parameter for every wave

The quantity p is the ray parameter. It is the horizontal component of the slowness, the inverse of the speed at which a wavefront sweeps along the boundary. Every wave created at the boundary must keep step with the incident wave along it, so all of them share the same p, including the S waves converted from an incident P wave. Section 4.15 of the MIT notes writes this as Eq. 4.95.

The ray parameter also stays fixed through a stack of flat layers, and it equals the slope of the travel-time curve along the surface (Eq. 4.98). Two stations a known distance apart that time the same arrival therefore measure p directly.

$$
p=\frac{\sin i_1}{\alpha_1}=\frac{\sin j_1}{\beta_1}=\frac{\sin i_2}{\alpha_2}=\frac{\sin j_2}{\beta_2}=\frac{dT}{dx}
$$

## Past the critical angle the wave cannot enter

When the lower rock is faster, the transmitted angle is larger than the incident one. At the critical incidence the transmitted sine reaches one and the transmitted ray runs along the boundary. Beyond it Snell’s law asks for a sine larger than one.

The wave still exists, but its vertical slowness becomes imaginary. Its amplitude then decays with depth below the boundary instead of oscillating, and it carries no energy down into the lower rock. Intermezzo 4.4 of the MIT notes works this out for a wave along a free surface (Eqs. 4.104 to 4.106). Saavedra and Moro show in their Section 3.2 that the reflected wavelet changes shape past the critical angle because it picks up a phase shift.

A wave travelling along the boundary at the lower rock’s speed keeps sending energy back up at the critical angle. Receivers above record it as the head wave. Far enough away it arrives before the direct wave, because it spent most of its path in the fast rock.

$$
\sin i_{\rm c}=\frac{\alpha_1}{\alpha_2},\qquad q_2=\sqrt{\frac{1}{\alpha_2^2}-p^2}=\mathrm i\sqrt{p^2-\frac{1}{\alpha_2^2}}\quad\text{for}\quad p>\frac{1}{\alpha_2}
$$

Worked example

## Angles and head-wave times for one boundary

The first preset of the interface Lab puts rock with a P speed of 2000 m/s and an S speed of 1000 m/s over rock with 4000 m/s and 2200 m/s. The transmitted P wave has a critical angle where the sine equals 2000 over 4000, which is 30.0 degrees. The transmitted S wave is faster than the incident P wave too, so it has its own critical angle, where the sine is 2000 over 2200: 65.4 degrees.

At the Lab’s default incidence of 40 degrees the ray parameter is the sine of 40 degrees over 2000 m/s, 0.3214 s/km. The reflected S wave leaves at 18.7 degrees and the transmitted S wave at 45.0 degrees. The transmitted P wave would need a sine of 1.29, so it is evanescent: the Lab shows zero transmitted P energy.

Now put a source and a line of receivers on a flat surface 750 m above this boundary. The head wave travels down at the critical angle, along the boundary at 4000 m/s and up again. Its arrival time is the distance divided by 4000 m/s plus a fixed delay of 0.650 s. It first appears 866 m from the source and overtakes the direct wave beyond 2.60 km.

$$
t_{\rm head}(x)=\frac{x}{\alpha_2}+\frac{2h\cos i_{\rm c}}{\alpha_1}=\frac{x}{4000\ \mathrm{m/s}}+0.650\ \mathrm s,\qquad x_{\rm cross}=\frac{0.650\ \mathrm s}{1/2000-1/4000\ \mathrm{s/m}}=2.60\ \mathrm{km}
$$

An experiment to think through

## Predict, change, explain

1. In the interface Lab, choose Past the critical angle and sweep the incidence angle from 10 to 40 degrees. Predict the angle at which the transmitted P energy drops to zero.
2. Lower the P speed of the lower rock to 3000 m/s and sweep the angle again.
3. Explain where the critical angle moved and why, using the sine rule above.

[Split one P wave into four at a rock boundary](https://sandboxphysics.com/labs/seismic-interface-conversion/)

The Labs use flat, isotropic, elastic or acoustic rock with made-up but realistic speeds; none of them reproduces a real survey. Each Lab states its model and limits in its notes and can export the data behind its readouts.

## Try explaining it yourself

### 1. Saavedra and Moro model ultrasound going from skin, 1615 m/s, into skull bone, 2500 m/s. Where is the P critical angle?

Read an explained answer

The sine of the critical angle is 1615 over 2500, which is 0.646, so the angle is 40.2 degrees. Their Figure 3 notes the change past about 40 degrees.

### 2. A ray leaves the top of a stack of layers with speeds 2, 3 and 5 km/s at 20 degrees. At what angle does it travel in each layer?

Read an explained answer

The ray parameter is the sine of 20 degrees over 2 km/s, 0.171 s/km. In the 3 km/s layer the sine is 0.513 and the angle 30.9 degrees; in the 5 km/s layer the sine is 0.855 and the angle 58.8 degrees.

$$
p=\frac{\sin20^\circ}{2}=0.171\ \mathrm{s/km},\quad i=\arcsin(3p)=30.9^\circ,\quad \arcsin(5p)=58.8^\circ
$$

### 3. Two stations 10 km apart along the surface time the same P arrival 1.0 s apart. The rock under them has a P speed of 6 km/s. At what angle did the wave arrive?

Read an explained answer

The ray parameter is the time difference over the distance, 0.10 s/km. The sine of the incidence angle is the speed times p, 0.60, so the wave came up at 36.9 degrees from the vertical.

## Sources to read alongside

1. [MIT OpenCourseWare, 12.201 Essentials of Geophysics (Fall 2004), lecture notes, Chapter 4: Seismology](https://ocw.mit.edu/courses/12-201-essentials-of-geophysics-fall-2004/b78f19037066b01c0977aa9dda43b225_ch4.pdf)

   Section 4.6 separates the elastic wave equation into P and S waves and gives their speeds (Eqs. 4.34 and 4.37). Sections 4.14 to 4.16 derive Snell’s law from Fermat’s principle (Eqs. 4.91 to 4.96) and show that the ray parameter is the slope of the travel-time curve (Eq. 4.98). Section 4.19 notes that in travel-time tomography the matrix entries are the ray lengths in each block.

2. [H. G. Saavedra and R. Moro, Zoeppritz equations: from seismology to medical exploration, Revista ECIPerú 22(2) (2025), arXiv:2510.26668v1](https://arxiv.org/abs/2510.26668)

   Equation 1 gives the normal-incidence reflection coefficient from the two acoustic impedances. Appendix 2 writes the four continuity conditions for an incident P wave (A2.1 to A2.8) and for an incident SV wave (A2.9 to A2.12). Section 3.2 and Figure 3 show the phase change of reflected wavelets past the critical angle, about 40 degrees for skin over bone.

[Previous Why a solid rock carries two kinds of seismic wave](https://sandboxphysics.com/stories/seismic-imaging/p-and-s-waves/) [Next How a P wave shares its energy at a rock boundary](https://sandboxphysics.com/stories/seismic-imaging/mode-conversion-and-energy-flux/)