Total internal reflection — critical angle derivation: Visual walkthrough

Step 0 — The four things we must draw first

Before any physics, let us agree on the cast of characters. Nothing here is a formula yet — just labelled pictures.

Look at the figure: the yellow dotted line is the normal. The blue ray comes in below the boundary and the angle it opens against the normal is $θ_{1}$ . The refracted ray (green) leaves above and opens angle $θ_{2}$ . Angles hug the normal, not the surface — that single habit prevents half of all sign mistakes.

Step 1 — The only physics we import: Snell's Law

WHAT. We import one equation from Snell's Law: $index of medium 1 n_{1} opens the incident ray sin θ_{1} = index of medium 2 n_{2} opens the refracted ray sin θ_{2}$

Read it term by term: the slow-down number times the sine of the angle is the same on both sides of the boundary. That product is a conserved bookkeeping quantity — whatever medium 1 has, medium 2 must match.

WHY this tool and not another? We need a rule that connects the two angles across a boundary. Reflection alone can't tell us when refraction stops; only a law linking $θ_{1}$ and $θ_{2}$ can. Snell's law is the unique such link. It drops out of light slowing to $v = c / n$ — where $v$ is the light's speed inside the medium and $c$ is the speed of light in vacuum (a fixed universal number, about $3 \times 1 0^{8} m/s$ , the same everywhere) — together with wavefronts staying continuous across the boundary; see Refraction of Light.

WHY sine and not tan or cosine? Sine measures the sideways lean of the ray away from the normal — exactly the quantity that must stay matched when wavefronts glide along the boundary. Cosine would measure the forward part (unmatched); tan mixes both. Sine is the honest bookkeeper here.

The figure shows the two shaded rectangles $n_{1} sin θ_{1}$ and $n_{2} sin θ_{2}$ having equal area — that equality is Snell's law made visible.

Step 2 — Choose the direction that makes trapping possible

WHAT. We send light from the denser medium into the rarer one: $n_{1} > n_{2}$ . (Denser = bigger index = slower light.)

WHY. Rearrange Snell to isolate the outgoing angle: $sin θ_{2} = bigger > 1 \frac{n _{1}}{n _{2}} sin θ_{1}$

Because $n_{1} > n_{2}$ , the fraction $\frac{n _{1}}{n _{2}}$ is greater than 1. It amplifies $sin θ_{1}$ . That is the whole seed of trapping: the outgoing angle is being pushed upward, and $sin$ can only reach $1$ before it runs out of room.

Red arrow in the figure: $θ_{2}$ (green ray) is drawn wider than $θ_{1}$ (blue ray). The ray bends away from the normal.

Step 3 — Turn up the dial and watch $θ_{2}$ race ahead

WHAT. Slowly increase $θ_{1}$ from $0°$ upward and track $θ_{2}$ .

WHY. From Step 2, $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1}$ is always a stretched copy of $sin θ_{1}$ . So as $θ_{1}$ climbs, $θ_{2}$ climbs faster and reaches the top ( $90°$ ) first. That " $θ_{2}$ arrives first" is the pivotal fact.

The figure shows three incident rays (blue, growing) and their refracted partners (green, growing faster and flattening toward the surface). Notice the green ray lying almost flat while the blue ray is still leaning — $θ_{2}$ is winning the race.

Step 4 — The tipping point: define the critical angle

WHAT. The critical angle $θ_{c}$ is the special value of $θ_{1}$ at which the refracted ray just grazes the surface, i.e. $θ_{2} = 90°$ .

WHY $90°$ ? Because $90°$ is the largest possible refraction angle — the refracted ray lying flat along the boundary. There is no " $91°$ " for a real ray to escape into. So $θ_{2} = 90°$ is the last frame before refraction dies.

At this exact incidence (yellow ray), the green refracted ray lies flat on the surface — grazing, about to vanish. This is the frame we build the formula from.

Step 5 — Substitute and solve (term by term)

WHAT. Put $θ_{1} = θ_{c}$ and $θ_{2} = 90°$ into Snell's law.

$n_{1} sin θ_{c} = n_{2} = 1 sin 90°$

Here $sin 90° = 1$ because — look back at Step 0 — a ray flat along the surface makes a $90°$ angle with the normal, and the sine of $90°$ is its maximum, $1$ . This is why the grazing frame is so clean: it collapses the right side to just $n_{2}$ .

Solve by dividing both sides by $n_{1}$ :

The figure is a right triangle whose hypotenuse is $n_{1}$ and whose vertical side is $n_{2}$ : $sin θ_{c} = \frac{opposite}{hyp} = \frac{n _{2}}{n _{1}}$ — the formula drawn as a shape.

Step 6 — Push past $θ_{c}$ : why the light gets trapped

WHAT. Set $θ_{1} > θ_{c}$ and ask Snell what $θ_{2}$ should be.

$sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} > \frac{n _{1}}{n _{2}} sin θ_{c} = \frac{n _{1}}{n _{2}} \cdot \frac{n _{2}}{n _{1}} = 1$

So Snell demands $sin θ_{2} > 1$ .

WHY that's impossible. The sine of any real angle lives between $- 1$ and $+ 1$ . There is no real angle whose sine exceeds $1$ . So no refracted ray can exist. With nowhere to escape, 100% of the light energy reflects back into the denser medium — this is Total Internal Reflection.

Left panel: $θ_{1} = θ_{c}$ , grazing green ray. Right panel: $θ_{1} > θ_{c}$ — the green ray is gone, replaced by a red reflected ray bouncing straight back. This is what powers Optical Fibres and the periscope prisms in Prisms and Total Internal Reflection.

Step 7 — The degenerate and edge cases (never leave a gap)

Recall Forecast-then-verify: two glass media

Predict $θ_{c}$ for $n_{1} = 1.6$ into $n_{2} = 1.4$ . Verify: $sin θ_{c} = 1.4/1.6 = 0.875 \Rightarrow θ_{c} = sin^{- 1} (0.875) \approx 61.0°$ . Both indices matter — never assume the second medium is air.

The one-picture summary

This single figure stacks the whole story: a fan of incident rays (blue) below the boundary; each escapes (green) bending away — until one grazes flat at $θ_{c}$ (yellow) — and every ray steeper than that reflects (red). The formula $sin θ_{c} = n_{2} / n_{1}$ is printed at the tipping ray.

Recall Feynman retelling — the whole walkthrough in plain words

Picture yourself as a beam of light living in glass, trying to escape into the air above. The air lets light move faster, so as you cross the boundary you swing away from the straight-up line — and the more slanted you already are, the more violently you swing. There's a catch: you can only swing until you're lying flat along the surface. Snell's law is the rulebook, and it says your outgoing tilt is a stretched copy of your incoming tilt (stretched by $n_{1} / n_{2}$ , a number bigger than 1). So your outgoing tilt hits "flat" before your incoming tilt does. The incoming angle at that exact moment is the critical angle, and plugging "flat = $90°$ " into the rulebook gives $sin θ_{c} = n_{2} / n_{1}$ . Lean in any harder and the rulebook asks for a tilt beyond flat — which doesn't exist — so you're stuck, and you bounce right back inside. That bounce is total internal reflection.

Connections

Snell's Law — the single imported rule (Step 1).
Refraction of Light — bending away from the normal (Steps 2–3).
Refractive Index — the slow-down number $n$ (Step 0).
Optical Fibres — trapping put to work (Step 6).
Prisms and Total Internal Reflection — $45° > θ_{c}$ in glass.
Mirage and Atmospheric Refraction — gradual version of the same trapping.
Brewster's Angle — a different named angle ( $tan θ_{B} = n_{2} / n_{1}$ ), about polarization not trapping.