The formula

Snell's law
n₁·sin θ₁ = n₂·sin θ₂
Index of refraction
n = c / v
Critical angle, valid only when n₁ > n₂
sin θ_c = n₂ / n₁

What the symbols mean

SymbolMeaningUnit
n₁, n₂Indices of refraction: air 1.00, water 1.33, diamond 2.42dimensionless
nIndex of refraction of a medium, written without a subscript in the defining formdimensionless
θ₁Angle of incidence, measured from the normal to the surfacedegrees
θ₂Angle of refraction, measured from the same normaldegrees
cSpeed of light in vacuum3.00 × 10⁸ m/s
vSpeed of light in the mediumm/s
θ_cCritical angle for total internal reflectiondegrees

When it applies

  • At the boundary between two transparent materials, with both angles measured from the normal rather than from the surface.
  • Light passing into a higher index bends toward the normal; going the other way it bends away.
  • When n₁ > n₂ and the angle of incidence passes the critical angle, there is no refracted ray at all and the light totally internally reflects.

Worked example

Problem. A ray of light in air strikes the surface of a pond at 40° from the normal. At what angle does it travel inside the water?

  1. Identify the media and the known angle: n₁ = 1.00 for air, n₂ = 1.33 for water, θ₁ = 40°.
  2. Write Snell's law: (1.00)(sin 40°) = (1.33)(sin θ₂).
  3. sin 40° = 0.643, so sin θ₂ = 0.643 / 1.33 = 0.483.
  4. Take the inverse sine: θ₂ = 28.9°, which rounds to 29°.

Answer. The ray travels at about 29° from the normal inside the water, bent toward the normal because water slows light down.

Common mistakes

  • Measuring the angle from the surface rather than from the normal, which turns a 40° incidence into 50°.
  • Leaving the calculator in radian mode.
  • Pairing an index with the angle in the wrong medium. Each n goes with the angle on its own side of the boundary.
  • Expecting an answer when light meets a lower-index medium beyond the critical angle. There is no refracted ray to find.

Related formulas

Sources