Snell's law — derivation from Fermat's principle: Worked examples

This page is the "no surprise left behind" companion to the parent derivation. The rule is always the same one line:

Before we compute anything, three plain-word reminders so no symbol sneaks in undefined:

Now let us map out every kind of situation this single equation can hand you. If a case is not in the table below, it does not exist for Snell's law.

The scenario matrix

Every refraction problem is one (or a blend) of these cells. The last column names the example that nails it.

#	Case class	What is special	Example
A	Rare → dense ( $n_{1} < n_{2}$ )	ray bends toward normal, $θ_{2} < θ_{1}$	Ex 1
B	Dense → rare ( $n_{1} > n_{2}$ )	ray bends away from normal, $θ_{2} > θ_{1}$	Ex 2
C	Normal incidence ( $θ_{1} = 0$ )	degenerate: ray goes straight, no bend	Ex 3
D	Grazing incidence ( $θ_{1} \to 9 0^{\circ}$ )	limiting: biggest possible bend into a denser medium	Ex 4
E	Critical angle ( $θ_{2} = 9 0^{\circ}$ )	boundary of existence; beyond it → no refraction	Ex 5
F	Beyond critical ( $sin θ_{2} > 1$ )	total internal reflection, "no answer" is the answer	Ex 6
G	Same index ( $n_{1} = n_{2}$ )	degenerate: no boundary optically, $θ_{2} = θ_{1}$	Ex 7
H	Given speeds not indices	rewrite Snell with $v$ ; a translation twist	Ex 8
I	Real-world word problem	pool/coin apparent-depth style geometry	Ex 9
J	Exam twist: angle from surface	trap — convert $9 0^{\circ} - ϕ$ first	Ex 10

Figure — Snell's law — derivation from Fermat's principle

The examples

Ex 1 — Cell A: rare → dense (bend toward normal)

Ex 2 — Cell B: dense → rare (bend away from normal)

Ex 3 — Cell C: normal incidence (degenerate, $θ_{1} = 0$ )

Worked example Straight-in beam

A laser enters glass ( $n_{2} = 1.5$ ) from air ( $n_{1} = 1.0$ ) exactly along the normal, $θ_{1} = 0^{\circ}$ .

Forecast: shining straight in — there is no "sideways" to bend toward, so guess $θ_{2} = 0^{\circ}$ , no bend.

Snell: $1.0 sin 0^{\circ} = 1.5 sin θ_{2}$ . Why this step? Plug the degenerate input in rather than assume the answer.
$sin 0^{\circ} = 0$ , so $0 = 1.5 sin θ_{2} \Rightarrow sin θ_{2} = 0$ . Why this step? A zero on the left forces the right side to zero.
$θ_{2} = arcsin (0) = 0^{\circ}$ . Why this step? The only angle in $[0, 9 0^{\circ}]$ with sine $0$ is $0^{\circ}$ .

Verify: The ray passes straight through undeviated — this is why looking straight down into water does not shift what you see. The speed inside the glass still drops (recall $v = c / n$ from the definitions above: bigger $n$ ⇒ smaller $v$ ), but the direction does not. ✔

Ex 4 — Cell D: grazing incidence (limiting $θ_{1} \to 9 0^{\circ}$ )

Worked example Skimming the surface into glass

Light in air ( $n_{1} = 1.0$ ) grazes glass ( $n_{2} = 1.5$ ) at $θ_{1} = 89. 9^{\circ}$ (almost along the surface). Find $θ_{2}$ , and the largest $θ_{2}$ ever possible entering this glass.

Forecast: the most extreme sideways entry. There must be a maximum $θ_{2}$ it can reach — guess it is set by $θ_{1} = 9 0^{\circ}$ .

$sin θ_{2} = \frac{1.0 sin 89. 9 ^{\circ}}{1.5} = \frac{0.99999}{1.5} = 0.66666$ . Why this step? We push $θ_{1}$ to its physical ceiling to find the ceiling on $θ_{2}$ .
$θ_{2} = arcsin (0.66666) = 41. 8^{\circ}$ . Why this step? Invert to get the actual refracted angle.
The absolute max as $θ_{1} \to 9 0^{\circ}$ : $sin θ_{2}^{m a x} = \frac{n _{1}}{n _{2}} = \frac{1}{1.5} = 0.6667 \Rightarrow θ_{2}^{m a x} = 41. 8^{\circ}$ . Why this step? $sin 9 0^{\circ} = 1$ is the biggest the numerator can be, so this is the largest refracted angle glass will ever show from air.

Verify: This is exactly the teal dot in the s01 figure: the teal curve's ceiling is $sin θ_{2} = 0.667$ , i.e. $41. 8^{\circ}$ . Every ray from air enters this glass within a cone of half-angle $41. 8^{\circ}$ . Notice this number equals the critical angle of Ex 5 — refraction and total internal reflection are mirror images of the same equation. ✔

Ex 5 — Cell E: the critical angle ( $θ_{2} = 9 0^{\circ}$ )

Worked example Just barely escaping

Light travels from glass ( $n_{1} = 1.5$ ) toward air ( $n_{2} = 1.0$ ). For which $θ_{1} = θ_{c}$ does the refracted ray skim the surface, $θ_{2} = 9 0^{\circ}$ ?

Forecast: dense→rare bends away, so at some incidence the ray will lie flat along the surface. Guess $θ_{c}$ is a bit above $4 0^{\circ}$ .

Set $θ_{2} = 9 0^{\circ}$ : $1.5 sin θ_{c} = 1.0 sin 9 0^{\circ} = 1.0$ . Why this step? $θ_{2} = 9 0^{\circ}$ is the last moment a refracted ray exists — the definition of critical.
$sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.6667$ . Why this step? Solve for the incidence sine; note it equals the ratio of indices, rare over dense.
$θ_{c} = arcsin (0.6667) = 41. 8^{\circ}$ . Why this step? Invert the sine.

Verify: Same $41. 8^{\circ}$ as Ex 4 — the entry cone and the escape cone match, as they must. The figure below draws this exact geometry: the plum incident ray rises through the teal glass, and the orange refracted ray lies flat along the boundary at $θ_{2} = 9 0^{\circ}$ , with the little plum arc marking $θ_{c} = 41. 8^{\circ}$ from the dotted normal. See Critical Angle and Total Internal Reflection. ✔

Ex 6 — Cell F: beyond critical ("no answer" is the answer)

Worked example Trying to refract past the limit

Same glass→air ( $n_{1} = 1.5$ , $n_{2} = 1.0$ ), but now $θ_{1} = 5 0^{\circ}$ , which is bigger than $θ_{c} = 41. 8^{\circ}$ . Find $θ_{2}$ .

Forecast: we are past the critical angle — guess there is no refracted ray at all.

$sin θ_{2} = \frac{1.5 sin 5 0 ^{\circ}}{1.0} = 1.5 \times 0.766 = 1.149$ . Why this step? Blindly apply Snell and watch what happens to the number.
$sin θ_{2} = 1.149 > 1$ . Why this step? A sine can never exceed $1$ — this signals no real angle solves the equation.
Therefore no refracted ray exists: all the light reflects back into the glass — total internal reflection. Why this step? The impossible sine is the mathematics telling us the light cannot leave.

Verify: $arcsin (1.149)$ is undefined for real angles ✔. On the s01 figure this is where the orange curve has already passed its intersection with the plum line — it would need to go above height $1$ . This is precisely why optical fibres trap light: rays inside strike the wall past $θ_{c}$ every time. ✔

Ex 7 — Cell G: same index (optically no boundary)

Ex 8 — Cell H: given speeds, not indices

Worked example Snell in disguise

Medium 1 lets light travel at $v_{1} = 2.0 \times 1 0^{8} m/s$ ; medium 2 at $v_{2} = 1.5 \times 1 0^{8} m/s$ . A ray hits the boundary at $θ_{1} = 4 0^{\circ}$ . Find $θ_{2}$ . (Here $v$ is the speed of light inside each medium, from the definitions above.)

Forecast: $v_{2} < v_{1}$ means light is slower in medium 2, i.e. it is optically denser ( $n = c / v$ larger), so guess $θ_{2} < 4 0^{\circ}$ .

Rewrite Snell with speeds. Since $n = c / v$ (with $c$ the vacuum speed, a shared constant): $\frac{c}{v _{1}} sin θ_{1} = \frac{c}{v _{2}} sin θ_{2}$ ; cancel $c$ to get $\frac{sin θ _{1}}{v _{1}} = \frac{sin θ _{2}}{v _{2}}$ . Why this step? The problem hands speeds, not indices — translate first, and $c$ cancels because it is the same on both sides.
Solve: $sin θ_{2} = sin θ_{1} \cdot \frac{v _{2}}{v _{1}} = sin 4 0^{\circ} \times \frac{1.5}{2.0} = 0.6428 \times 0.75 = 0.4821$ . Why this step? Isolate the unknown sine using the speed form.
$θ_{2} = arcsin (0.4821) = 28. 8^{\circ}$ . Why this step? Invert the sine.

Verify: $28. 8^{\circ} < 4 0^{\circ}$ ✔ — slower medium, bends toward normal, matching the forecast. See Refractive Index. ✔

Ex 9 — Cell I: real-world word problem (apparent depth)

Worked example The coin at the bottom of the pool

A coin sits at true depth $h = 1.20 m$ under water ( $n_{1} = 1.33$ , water below; air $n_{2} = 1.00$ above). Looking almost straight down, how deep does it appear?

Forecast: water bends the escaping rays away from the normal, fooling your eye into thinking the coin is shallower — guess $h_{app} < 1.20 m$ .

Look at the figure first. A ray leaves the coin $C$ , travels up through the water a horizontal step $s$ to the surface point $P$ , then refracts into air. Your eye extends the air ray backward (dashed) to where it seems to come from — a point directly above the coin but at the shallower apparent depth $h_{app}$ . Both the true ray (from depth $h$ ) and the apparent ray (from depth $h_{app}$ ) share the same horizontal step $s$ at $P$ , so their tangents are what we compare.

At $P$ , the water-side angle from the normal is $θ_{1}$ with $tan θ_{1} = s / h$ ; the air-side angle is $θ_{2}$ with $tan θ_{2} = s / h_{app}$ . Why this step? Both right triangles share the horizontal $s$ ; the tangent (opposite $s$ over adjacent depth) links each angle to its depth.
For near-vertical viewing both angles are tiny, and for small angles $sin θ \approx tan θ$ . Snell $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ becomes $n_{1} tan θ_{1} \approx n_{2} tan θ_{2}$ . Why this step? Replacing $sin$ by $tan$ (valid only for small angles) lets us feed in the depth expressions.
Substitute the tangents: $n_{1} \frac{s}{h} = n_{2} \frac{s}{h _{app}}$ . The $s$ cancels, leaving $\frac{n _{1}}{h} = \frac{n _{2}}{h _{app}}$ , so $h_{app} = h \cdot \frac{n _{2}}{n _{1}} = 1.20 \times \frac{1.00}{1.33}$ . Why this step? The shared $s$ cancels, and rearranging isolates the apparent depth.
$h_{app} = 1.20 \times 0.7519 = 0.902 m$ . Why this step? Plug the numbers.

Verify: $0.902 m < 1.20 m$ ✔ — the coin looks about $0.30 m$ shallower, exactly the everyday illusion. Units: metres × (dimensionless ratio) = metres ✔.

Ex 10 — Cell J: exam twist (angle given from the surface)

Worked example The classic trap

A ray in air ( $n_{1} = 1.0$ ) makes an angle of $ϕ = 2 5^{\circ}$ with the water surface (not the normal). Water has $n_{2} = 1.33$ . Find the refraction angle $θ_{2}$ (from the normal, as usual).

Forecast: $2 5^{\circ}$ from the surface is a shallow, grazing ray — its angle from the normal is large. Guess $θ_{1}$ near $6 5^{\circ}$ .

Convert to the normal: $θ_{1} = 9 0^{\circ} - ϕ = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Why this step? Snell's law is defined from the normal; the surface angle and normal angle are complements.
Snell: $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}} = \frac{1.0 sin 6 5 ^{\circ}}{1.33} = \frac{0.9063}{1.33} = 0.6814$ . Why this step? Now the standard equation applies with the correct incidence angle.
$θ_{2} = arcsin (0.6814) = 42. 9^{\circ}$ . Why this step? Invert the sine.

Verify: If you had forgotten to convert and plugged $2 5^{\circ}$ , you would get $sin θ_{2} = sin 2 5^{\circ} /1.33 = 0.318$ , i.e. $18. 5^{\circ}$ — a wrong, much smaller answer. The complement step is what saves you. See the parent note "angle from surface" mistake. ✔

Recall One-line summary of the matrix

Turn the incidence dial from $0^{\circ}$ to $9 0^{\circ}$ : normal incidence ⇒ no bend; rare→dense always refracts (bends toward normal, cone capped at $θ_{2}^{m a x}$ ); dense→rare bends away until the critical angle, past which it is total internal reflection; equal indices ⇒ no bend; and always convert surface-angles to normal-angles first.

Which cell has "no real answer," and why?

Cell F (beyond critical):

sin θ_{2} > 1

is impossible, so the light totally internally reflects.

For dense→rare, what marks the last refracting angle?

The critical angle

θ_{c} = arcsin (n_{2} / n_{1})

, where

θ_{2} = 9 0^{\circ}

Why does normal incidence give no bending?

sin 0^{\circ} = 0

forces

sin θ_{2} = 0

, so

θ_{2} = 0^{\circ}

Connections

Snell's law — derivation from Fermat's principle (parent)
Fermat's Principle
Refractive Index
Total Internal Reflection
Critical Angle
Optical Path Length
Calculus — minimization and stationary points