Find the threshold for total internal reflection

Find a change of regime

Set second index to 1.00. At 40 degrees incidence the required transmitted sine is 0.964181, giving a real angle of 74.618568 degrees. At 45 degrees incidence the required sine is 1.060660. No real angle has that sine: a propagating transmitted ray cannot satisfy the boundary condition. The browser removes that ray and reports total internal reflection.

Do not round a sine above one down to one to force an answer. That would turn a regime change into a misleading grazing ray. Do not replace the missing angle with zero either: zero means normal transmission, a different prediction.

Derive the threshold

At critical angle θc, the transmitted direction reaches 90 degrees from the normal, along the interface. Since sin 90° = 1, Snell’s law gives sin θc = n₂/n₁, so θc = inverse sine(n₂/n₁) when n₁ is greater than n₂. For 1.50 into 1.00, this is inverse sine(2/3) = 41.810315 degrees.

Total internal reflection requires two conditions: approach a lower-index medium, and incidence greater than the critical angle. Exactly at the threshold, the limiting transmitted direction lies along the boundary with no propagation component into medium 2. Above it, no real transmitted ray direction exists. These are different boundary cases.

The five-degree slider cannot select 41.810315 degrees exactly. Captures at 40 and 45 degrees bracket the threshold; the ledger computes the ideal value. Its extra digits come from mathematics, not a more precise physical instrument. A bracket supports an interval, not every displayed decimal.

Change the second medium

At 60 degrees incidence compare second indices 1.00, 1.40 and 2.00. At 1.00 the critical angle is 41.810315 degrees and total internal reflection occurs. At 1.40 the critical angle is 68.960530 degrees: 60 degrees is below it, so transmission is possible. At 2.00 there is no critical angle in this direction because the incident index is lower. Increasing incidence within the control range cannot produce total internal reflection in that last case.

For this ideal lossless boundary, total internal reflection returns all incident power to medium 1. Ordinary reflection does not generally return all power. The equal drawn ray lengths never encode power fractions.

Distinguish a ray from the entire field

“No propagating transmitted ray” does not mean the electromagnetic field is identically zero outside. Total internal reflection involves an evanescent field, which decays away from the interface. This geometric model does not calculate that field or interactions with a nearby third medium. The medium-two speed and wavelength in the table remain bulk reference properties at the given frequency; they do not establish that a ray is propagating there.

Optical fibers can guide light using internal reflection, but one flat boundary does not model a full fiber’s geometry, losses or supported modes. As a paper extension, draw two parallel boundaries and trace repeated reflections. Label the additional geometry assumptions.

Transfer to a new boundary

A separate medium has index 1.60 and borders index 1.20. Find the critical angle. Decide whether 45 and 55 degrees incidence give total internal reflection, then explain what changes if travel reverses.

Check the critical-angle transfer

The critical angle is inverse sine(1.20/1.60) = inverse sine(0.75) = 48.590378 degrees. Incidence at 45 degrees is below critical; 55 degrees is above critical and gives total internal reflection. Reversing travel makes the incident index lower, so there is no total-internal-reflection threshold in that direction.

OpenStax: total internal reflection explains the threshold condition. Return to refraction to check the sine calculation before interpreting a missing solution.