Snell refraction and total internal reflection
Track frequency, angle, and critical conditions at a dielectric interface without confusing ray directions with energy fractions.
On this page
The question for today
What you’ll understand
- Apply Snell’s law with angles measured from the normal.
- Determine whether a real transmitted ray exists.
- Explain the evanescent field associated with total internal reflection.
A transparent interface can split incident light into reflected and transmitted fields. Geometry determines their directions before a full electromagnetic calculation determines their amplitudes. Refraction is especially revealing when light travels toward a lower-index medium: the transmitted direction moves farther from the normal until no real propagation angle remains. We will identify that boundary carefully, including which index belongs in the numerator, and distinguish the absence of a transmitted ray from the absence of any field beyond the interface.
Phase matching fixes the angles
Assume a flat, stationary interface between homogeneous, isotropic, lossless dielectrics with positive indices. The temporal oscillation must match on both sides, so frequency is unchanged. The tangential spatial phase must also agree at every interface point. Because the wavevector magnitude is , its tangential component supplies the index-weighted sine relation. Reflection preserves the index and reverses the normal propagation component, giving equal incident and reflected angles to the normal.
Increasing index reduces wavelength at fixed frequency, and the transmitted ray bends toward the normal. Decreasing index increases wavelength and bends an existing transmitted ray away from it. These statements concern phase-index ray geometry in isotropic media. They do not calculate how much power crosses the boundary. Even normal incidence generally reflects some power because the electromagnetic impedances differ; a direction law cannot decide that fraction or its polarization dependence.
The critical angle is a one-way condition
A real transmitted direction requires the computed sine to be at most one. If the incident medium has higher index, the limiting transmitted direction is tangent to the interface. Solving for its incident angle gives the critical angle. The ratio is the lower transmitted index divided by the higher incident index. Reversing that ratio produces a number greater than one and signals an incorrect assignment, rather than a mysterious angle that a calculator cannot evaluate.
For incidence strictly beyond the critical angle there is no propagating transmitted plane wave in an infinite lower-index medium. An ideal isolated lossless interface reflects all normal incident energy flux. At the critical angle itself the limiting transmitted direction is grazing, so it is useful to reserve “beyond critical” for the regime with an exponentially decaying normal field. From low index to high index no critical angle occurs; even very oblique incidence has an allowed refracted direction.
A missing ray can leave a field
Beyond critical incidence the conserved tangential wavevector exceeds the total wavevector magnitude permitted in medium two. Its normal component becomes imaginary, producing an evanescent field that decays away from the boundary. Such a field is required by the electromagnetic boundary conditions. For the ideal single interface it carries no net time-averaged power normally away into the second half-space, although its existence matters for phase shifts, near-field coupling, and interaction with nearby matter.
Bringing a third medium within the decay distance can permit energy transfer through the thin low-index gap, often called frustrated total internal reflection. Absorption also changes the ideal unit-reflectance conclusion. A ray diagram alone cannot calculate these effects. The diagnostic habit is to ask separately whether a real transmitted direction exists, whether a field exists, and whether net normal energy flows. Those are related but distinct questions, resolved at different levels of approximation.
Work it through
Two rays leaving glass
Glass of index 1.5 borders air of index 1. A first internal ray has ; a second has . Find the transmitted direction when one exists, and classify the second ray.
For glass to air, Snell’s law multiplies the incident sine by 1.5. The first transmitted sine is , so its transmitted angle is 30 degrees from the normal.
The critical internal sine is , corresponding to approximately 41.8 degrees. This test uses the angle inside glass, not the angle outside it.
For the second ray, the proposed transmitted sine is . Since no real angle has this sine, the ray undergoes total internal reflection in the stated ideal geometry.
The first ray emerges at 30 degrees. The second has no propagating transmitted ray; a decaying field can still exist in air.
The living notebook
Explore the model
Interactive lab · vectors
Resolve a direction into components
Treat the vector as a geometric wavevector analogy. Reset, set Angle = 30° and Magnitude = 2, and predict y projection 1. For a second fresh run Reset, set Angle = 90° and Magnitude = 2, and predict y projection 2, the largest possible tangential component when the x direction is normal.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA tangential component cannot exceed a real vector’s magnitude. The lab does not apply Snell’s law, change refractive index, or display evanescent waves.
Pause & reason
Check your understanding
Your turn
Try it yourself
A ray travels from index 1.25 into index 1.5 with . Find , state whether total internal reflection is possible in this direction, and find the wavelength in medium two for .
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the incident-to-transmitted index ratio for refraction and the transmitted index alone for wavelength.
Compare with the solution
Snell’s law gives , so the angle is 30 degrees. Total internal reflection cannot occur from lower to higher positive index. The wavelength is in the second medium. Frequency remains unchanged, and the smaller transmitted angle agrees with bending toward the normal.
Keep the idea
Take it with you
- Frequency and tangential phase match across a stationary interface.
- Total internal reflection requires incidence from higher to lower index beyond the internal critical angle.
- No transmitted ray does not imply zero field in the second medium.
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