
Optional unit · institution-dependent · Interactive lesson
Total internal reflection
A dense-to-rare boundary reflects a few percent of the light at first and then all of it. Learn to find the angle where that changeover happens from two refractive indices, to tell which side of the boundary you have to be on for it to happen, and to use it the way prism and fibre designers do.
UPII-13.03 · Lesson 3 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Past one angle, a boundary stops letting light out.
The key idea
Total internal reflection is what happens when light already inside a denser medium meets a boundary with a rarer one beyond the critical angle: no refracted ray exists, and the boundary sends the entire beam back.
Example. In water (n = 1.33) against air the critical angle is arcsin(1.00/1.33) = 48.8°, so a ray reaching the surface at 50.0° is turned back completely, while one at 47.0° still escapes, leaving at 76.6° to the normal.
The complete physical model and assumptions
Snell's law is an equation about sines, and a sine cannot exceed one. Send light from index n₁ into index n₂ and the transmitted ray obeys n₁ sin θ₁ = n₂ sin θ₂. When n₁ > n₂ the refracted ray leans away from the normal, so as θ₁ grows the transmitted ray swings towards the surface, and at the angle where sin θ₂ reaches 1 it lies flat along the boundary. Push one degree further and the equation demands a sine larger than one: there is no real transmitted ray at all, and every bit of the incident power turns back into the dense medium. That threshold, θc = arcsin(n₂/n₁), is the critical angle, and the reflection above it is the only perfectly efficient mirror in optics — no metal coating comes close. The ray model then stops one step short of the truth. Solving the electromagnetic boundary conditions at the same surface leaves a field on the far side even above θc: an evanescent wave whose amplitude falls as e(−z/d) over a few hundred nanometres and which carries no net power away from an isolated boundary. Bring a second surface within that distance and power crosses after all, which is how beam-splitter cubes and optical fingerprint readers work — and a reminder that "total" is a claim about a flat, clean, isolated interface, not about empty space beyond it.
1Where Snell's law runs out of angles
Refraction at a flat boundary obeys n₁ sin θ₁ = n₂ sin θ₂, both angles measured from the normal. Rearranged for the transmitted ray, sin θ₂ = (n₁/n₂) sin θ₁. When n₁ > n₂ that prefactor exceeds one, so θ₂ is always larger than θ₁ and the transmitted ray leans towards the surface. Increase θ₁ and θ₂ closes on 90°. The equation reaches its own limit when sin θ₂ = 1, which happens at n₁ sin θc = n₂, so sin θc = n₂/n₁ and θc = arcsin(n₂/n₁). For crown glass against air that is arcsin(1.00/1.52) = 41.1°; for water against air, 48.8°; for diamond, whose index is 2.42, only 24.4°. Beyond θc no real angle has the sine the equation is asking for, so the transmitted ray does not merely grow faint — it does not exist, and the surface returns the whole beam.
2Total reflection runs in one direction only
The ratio n₂/n₁ can only be a sine if it is at most 1, so a critical angle exists only when n₁ > n₂: the light must already be inside the denser medium. Run the same boundary the other way, air into crown glass, and sin θ₂ = sin θ₁/1.52 ≤ 0.658, so the refracted ray inside the glass never exceeds 41.1° from the normal however you aim the incoming beam. The same number appears, now as a ceiling on refraction rather than a threshold for reflection — the two statements are one fact read in both directions, which is ray reversibility. What matters is the pair of indices, not density in the everyday sense: glass at 1.52 against water at 1.33 still reflects totally, but only past 61.0°, and a fibre core barely denser than its cladding has a critical angle of 81.5°, within 8.5° of grazing.
3Total is not a manner of speaking
A boundary already reflects some light below the critical angle; total reflection is the end of a trend, not the start of one. Head-on, the reflected fraction at a glass-air surface is R₀ = ((1.52 − 1)/(1.52 + 1))² = 0.043, about 4%. That fraction climbs slowly as θ₁ grows, rises steeply over the last few degrees before θc, and equals exactly 1 at θc and everywhere beyond. No coating does that. A good aluminium mirror returns about 90% of visible light and a protected silver one perhaps 97%, and every bounce multiplies the loss: a two-mirror periscope keeps 0.90² = 0.81 of what enters it, while the two total reflections inside a prism periscope cost nothing whatever. The prism is not free — it still gives up about 4% at each of its own entry and exit faces unless they are coated — but the bounces themselves are exact. A right-angle prism in air reflects totally at 45° provided sin 45° ≥ 1/n, that is n ≥ √2 = 1.414, which crown glass at 1.52 comfortably satisfies.
4The field that reaches across the boundary
The electromagnetic boundary conditions do not let a field stop dead at a surface, so even above θc something exists in the rarer medium. It does not propagate away from the surface: its amplitude decays as e(−z/d) with distance z from the boundary, where d = λ₀ / (2π √(n₁² sin²θ₁ − n₂²)). This evanescent wave runs along the interface and, averaged over a cycle, carries no net power across it, so energy conservation and the word total both survive. The depth is small and shrinks as you move past θc: for crown glass against air at λ₀ = 589 nm and θ₁ = 45.0°, d = 238 nm. Bring a second glass surface within a few d and the evanescent field reaches it, a genuine transmitted ray reappears on the far side, and the reflection is frustrated. That is the mechanism of beam-splitter cubes, optical fingerprint readers, and microscopes that illuminate only the first hundred nanometres of a cell.
5Trapping light in a fibre: the acceptance cone
A step-index fibre is a high-index core inside a slightly lower-index cladding, and any ray meeting the wall past θc bounces along it indefinitely. What limits the fibre is the cone it will accept at its end face. A ray entering at θₐ in air refracts to θᵣ inside the core and then strikes the wall at 90° − θᵣ, so guiding demands 90° − θᵣ ≥ θc; combining that with n₀ sin θₐ = ncore sin θᵣ gives n₀ sin θₐ ≤ √(ncore² − nclad²), the numerical aperture. With a core of 1.4682 and a cladding of 1.4519, NA = √(2.15561 − 2.10801) = 0.2182, an acceptance half-angle of 12.6° in air. Light arriving steeper refracts into the cladding and is gone within centimetres. A wide cone has a price: the steepest guided ray travels 1/cos θᵣ as far as the axial one, 11.2 m extra per kilometre, arriving about 55 ns late and smearing the pulse.
6Where the ray model gives way
Total internal reflection is a statement about a flat, clean, unbounded boundary, and real surfaces fall short in specific ways. A bend tilts the wall relative to the rays, and below a critical bend radius the rays at the outer wall arrive inside θc and escape, which is why a kinked fibre leaks. Anything touching the surface changes n₂: a fingerprint replaces air at 1.00 with oil near 1.47, pushing θc for crown glass from 41.1° to 75.3° and killing the reflection at 45°. The reflected beam is also displaced sideways by a fraction of a wavelength, the Goos-Hanchen shift, because the field spends time in the evanescent skin — invisible in a ray diagram, measurable on a bench. And because n depends on wavelength, so does θc: in crown glass violet sees 40.9° and red 41.3°, so a boundary set at 41.0° reflects violet totally while red still escapes.
Key equations
Read the units, assumptions and limits before substituting.
Critical angle at a boundary
sin θc = n₂/n₁, with n₁ > n₂
Sapphire at 1.77 against air gives θc = arcsin(1.00/1.77) = 34.4°.
n₁ is the index of the medium the light is already in, n₂ the index beyond; both dimensionlessReflected fraction at normal incidence
R₀ = ((n₁ − n₂)/(n₁ + n₂))²
Glass to air returns 0.043 of the power head-on, against exactly 1 above θc.
R₀ is the fraction of incident power reflected at θ₁ = 0, dimensionlessEvanescent penetration depth
d = λ₀ / (2π √(n₁² sin²θ₁ − n₂²))
Tells you how close a second surface must come to frustrate the reflection.
λ₀ is the vacuum wavelength in m; d in m is the distance over which the amplitude falls by a factor eAcceptance cone of a step-index fibre
n₀ sin θₐ = √(ncore² − nclad²)
Light inside that cone is guided; anything steeper leaks into the cladding.
θₐ is the half-angle of the entry cone in a medium of index n₀; the right-hand side is the numerical apertureRight-angle prism used in air
n ≥ 1/sin 45° = √2 ≈ 1.414
Crown glass at 1.52 clears it, which is why binoculars fold light with prisms.
n is the prism index with air outside, so that a hypotenuse face struck at 45° reflects totallyInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Hold n₁ at 1.50 and raise θ₁ through 41.8°: the refracted ray swings to grazing, then vanishes the moment sin θ₂ passes 1. Then lower n₁ and watch the critical angle climb towards grazing, or raise it and watch total reflection start far earlier.
Live results
- CRITICAL ANGLE θc
- 41.8 °
- n₁ sin θ₁ (= sin θ₂)
- 0.750
- MARGIN θ₁ − θc
- -11.8 °
- EVANESCENT SKIN (0 = NONE)
- 0.00 λ₀
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Incidence angle θ₁ in the dense medium: 30 °; Index n₁ of the dense medium: 1.50
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
A boundary is not a switch that is off below θc and on above it, and total does not mean that nothing is there. Below the critical angle some light already reflects — about 4% head-on at a glass-air surface, and far more in the last degrees before θc — so partial reflection does not begin at θc, it finishes there. Above θc there is still a field in the rarer medium: the evanescent wave, decaying over a couple of hundred nanometres. It carries no net power away from an isolated boundary, which is exactly why the reflection is genuinely total, but it is physically present, and a second surface placed within a wavelength lets power cross the gap even though no refracted ray exists inside it. The reflection is total because of what the evanescent field does, not because the far side is empty.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A ray inside a flint-glass paperweight (n = 1.62) reaches the flat top face, with air above it, at 38.0° to the normal. Find the critical angle for this boundary, decide whether the ray escapes, and if it does, find the angle of the emerging ray.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The light is in the denser medium, so a critical angle exists: sin θc = n₂/n₁ = 1.00/1.62 = 0.6173.
- θc = arcsin(0.6173) = 38.1°.
- The ray arrives at 38.0°, just below θc, so Snell's law still has a solution and the ray escapes — but only just.
- sin θ₂ = (n₁/n₂) sin θ₁ = 1.62 × sin 38.0° = 1.62 × 0.6157 = 0.9974, so θ₂ = arcsin(0.9974) = 85.8°, almost flat along the surface.
- Tilt the ray one degree further: 1.62 × sin 39.0° = 1.019, greater than 1, so no refracted ray exists and the face reflects the beam totally.
Answer. θc = 38.1°; the ray escapes at 85.8° from the normal, grazing the surface. At 39.0° it would be totally reflected.
MediumExample 2
A step-index silica fibre has a core of index 1.4682 and a cladding of index 1.4519, and its polished end face sits in air (n₀ = 1.000). Find (a) the critical angle at the core-cladding wall, (b) the numerical aperture, (c) the acceptance half-angle in air, and (d) how much farther than 1.00 km of axis the steepest guided ray travels.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) At the wall the light is in the core, so sin θc = nclad/ncore = 1.4519/1.4682 = 0.98890 and θc = 81.45°. Guided rays must hit the wall within 8.55° of grazing.
- (b) A ray entering at θₐ refracts to θᵣ and strikes the wall at 90° − θᵣ. Setting 90° − θᵣ = θc in 1.000 sin θₐ = 1.4682 sin θᵣ gives n₀ sin θₐ = √(ncore² − nclad²) = √(2.15561 − 2.10801) = √0.04760 = 0.2182.
- (c) θₐ = arcsin(0.2182) = 12.6°, the half-angle of the acceptance cone; light arriving steeper refracts into the cladding and is lost.
- (d) That steepest guided ray runs at θᵣ = arcsin(0.2182/1.4682) = 8.55° to the axis, so over 1.00 km of fibre it covers 1000 m ÷ cos 8.55° = 1011.2 m.
- Its extra path is 11.2 m; at speed c/1.4682 = 2.042 × 10⁸ m s⁻¹ that is about 55 ns of delay per kilometre, which is what limits the pulse rate.
Answer. θc = 81.45°, NA = 0.2182, acceptance half-angle 12.6°, and the steepest guided ray travels 11.2 m farther per kilometre.
HardExample 3
A right-angle crown-glass prism (n = 1.520) in air totally reflects a 633 nm laser beam striking its hypotenuse face at 45.0°. A second identical prism is then brought up parallel to that face. (a) Confirm the reflection is total. (b) Find the penetration depth of the evanescent field. (c) Find the gap at which the evanescent amplitude has fallen to 1.0% of its value at the surface.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) sin θc = 1.000/1.520 = 0.6579, so θc = 41.1°. The beam arrives at 45.0°, which is 3.9° past the critical angle, so no refracted ray exists in the air gap.
- (b) Use d = λ₀ / (2π √(n₁² sin²θ₁ − n₂²)). The bracket first: n₁² sin²45.0° = 2.3104 × 0.5000 = 1.1552, minus n₂² = 1.000 leaves 0.1552, whose square root is 0.3940.
- So d = 633 nm ÷ (2π × 0.3940) = 633 nm ÷ 2.475 = 256 nm, about 0.40 vacuum wavelengths.
- (c) The amplitude follows e(−z/d), so 0.010 = e(−z/d) gives z = d ln(100) = 256 nm × 4.605 = 1.18 × 10³ nm = 1.18 μm.
- A gap of a few micrometres therefore leaves the reflection total for any practical purpose, while closing it below about 0.2 μm lets a large share of the beam cross — a beam splitter whose ratio you tune by turning a screw.
Answer. θc = 41.1°, so the reflection at 45.0° is total; d = 256 nm, and the amplitude is down to 1.0% at a gap of 1.18 μm.
Quick check
Select an answer and explain it before checking.
Review & continue
Total internal reflection is what happens when light already inside a denser medium meets a boundary with a rarer one beyond the critical angle: no refracted ray exists, and the boundary sends the entire beam back.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.