SME
Geophysics · Problem type
Critical angle and refraction of a seismic ray
Use this for “A ray in 1,500 m/s sediment meets 4,500 m/s rock at 10°. At what angle does it go on, and what is the critical angle?”
Diagram
sin
r
=
(
v₂
v₁
)
sin
i
=
(
4,500
1,500
)
×
sin
10°
=
0.5209,
r
=
31.4°:
bent
away
from
the
normal,
into
the
faster
layer.
Critical
angle:
its
sine
is
v₁
v₂
=
0.3333,
so
it
is
19.47°.
This is an example. Type your own values to start a new problem.
sin(
r
°)
=
v₂ (m/s)
v₁ (m/s)
sin(
i
°)
ic
Critical angle
calculated · °
Formulas
sin(
r
°)
=
v₂
v₁
×
sin(
i
°)
sin(31.3956°)
=
4,500
1,500
×
sin(10°)
sin(ic°)
=
v₁
v₂
sin(19.4712°)
=
1,500
4,500
Assumptions
•
The boundary is flat; the ray bends there by Snell’s law written with speeds.
•
Angles are measured from the normal, the line square to the boundary.
•
Into a faster layer, past the critical angle no ray gets through: it is all reflected.
Step-by-step
WE KNOW
v₁ = 1,500 m/s, v₂ = 4,500 m/s, i = 10°
FIND
refraction angle (r), critical angle (ic)
Step 1 · Find refraction angle (r)
Use sin(r°) = v₂ ÷ v₁ × sin(i°)
Snell’s law with speeds: the sine of the angle grows with the speed.
r
=
sin⁻¹(
v₂
v₁
×
sin(
i
°))
r
=
sin⁻¹(
4,500
1,500
×
sin(10°))
r
=
31.3956°
Step 2 · Find critical angle (ic)
Use sin(ic°) = v₁ ÷ v₂
At
the
critical
angle
the
ray
runs
along
the
boundary
(
r
=
90°),
so
sin
ic
=
v₁
v₂
.
ic
=
sin⁻¹(
v₁
v₂
)
ic
=
sin⁻¹(
1,500
4,500
)
ic = 19.4712°
Check
Put the numbers back in. Both sides should match.
sin(31.3956°)
=
4,500
1,500
×
sin(10°)
✓
sin(19.4712°)
=
1,500
4,500
✓
Topic check
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