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

Snell's law: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$.

• $n_1$ = refractive index of the medium the light starts in (medium 1).
• $n_2$ = refractive index of the medium the light enters (medium 2).
• ${\theta }_1$ = angle of incidence, measured from the normal in medium 1.
• ${\theta }_2$ = angle of refraction, measured from the normal in medium 2.

The refractive index is defined by $n=c/v$, where $c$ is the speed of light in vacuum and $v$ its speed in the medium: it says how many times slower light moves in that medium than in vacuum. See Refractive Index.

Set-up: $(1.0)\sin 45^{\circ }=(1.33)\sin {\theta }_2$.

Prediction: Water is denser ($n_2>n_1$), so light slows down entering it. The rule "Slow ⇒ Slant toward the Normal" says the ray bends toward the normal, so ${\theta }_2<45^{\circ }$.

Why: $n\sin \theta$ is conserved. Since $n_2>n_1$, to keep the product equal we need $\sin {\theta }_2<\sin {\theta }_1$, hence a smaller angle.

True. Put ${\theta }_1=0^{\circ }$ into Snell's law: $n_1\sin 0^{\circ }=n_2\sin {\theta }_2\Rightarrow 0=n_2\sin {\theta }_2\Rightarrow \sin {\theta }_2=0\Rightarrow {\theta }_2=0^{\circ }.$ The ray continues straight through. A ray travelling along the normal has zero horizontal run, and it is exactly the horizontal component that must be shared across the boundary — zero stays zero. This is the degenerate (normal-incidence) case.

Step 1 — write Snell: $(1.0)\sin 45^{\circ }=1.33\sin {\theta }_2$. Step 2 — isolate the sine: $\sin {\theta }_2=\frac{\sin 45^{\circ }}{1.33}=\frac{0.7071}{1.33}=0.5317$. Why divide? We want ${\theta }_2$ alone, so we move the known factor $n_2=1.33$ to the other side. Step 3 — undo the sine: ${\theta }_2=\arcsin (0.5317)=32.1^{\circ }$. Check: $32.1^{\circ }<45^{\circ }$ ✔ — bent toward the normal, exactly as predicted.

Step 1: $n_1\sin 25^{\circ }=(1.0)\sin 40^{\circ }$. Step 2: $n_1=\frac{\sin 40^{\circ }}{\sin 25^{\circ }}=\frac{0.6428}{0.4226}=1.521$. Check: The ray bent away from the normal ($40^{\circ }>25^{\circ }$) on leaving, which is what happens exiting a denser medium — consistent with $n_1=1.52>1=n_2$. ✔

Step 1 — speed form of Snell: since $n=c/v$ (with $c$ the vacuum speed of light), the $c$'s cancel: $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}.$ Step 2 — solve: $\sin {\theta }_2=\sin 50^{\circ }\cdot \frac{v_2}{v_1}=0.7660\times \frac{1.8}{2.4}=0.7660\times 0.75=0.5745$. Step 3: ${\theta }_2=\arcsin (0.5745)=35.1^{\circ }$. Check: Slower medium ($v_2<v_1$) → bends toward normal → ${\theta }_2<{\theta }_1$ ✔.

Step 1 — critical condition: set ${\theta }_2=90^{\circ }$, so $\sin {\theta }_2=1$: $n_1\sin {\theta }_c=n_2\cdot 1.$ Step 2: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.0}{1.52}=0.6579$. Step 3: ${\theta }_c=\arcsin (0.6579)=41.1^{\circ }$. Beyond $41.1^{\circ }$ the equation demands $\sin {\theta }_2>1$, which is impossible — so no refracted ray exists and all light reflects: Total Internal Reflection. See also Critical Angle.

The figure below draws this exact critical case: the incident ray (yellow) strikes at ${\theta }_c\approx 41^{\circ }$ and the refracted ray (red) lies flat along the surface at ${\theta }_2=90^{\circ }$. Notice the green arc marking ${\theta }_c$ — as you tilt the incident ray past it, the red arrow has nowhere left to go.

$\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}=\frac{1.52\times \sin 50^{\circ }}{1.0}=1.52\times 0.7660=\mathrm{1.164.}$ But the sine of a real angle can never exceed $1$. Since $1.164>1$, no real ${\theta }_2$ exists — the light cannot refract out and undergoes total internal reflection instead. What it means physically: the boundary acts as a perfect mirror. This is exactly how light stays trapped in an optical fibre.

Set up the geometry locally. Put point $A$ in medium 1 and point $B$ in medium 2, with the flat boundary horizontal. Define:

• $a$ = the height of $A$ above the boundary (fixed),
• $b$ = the depth of $B$ below the boundary (fixed),
• $w$ = the horizontal separation between $A$ and $B$ (fixed),
• $x$ = the horizontal position where the ray crosses the boundary — the one free choice light makes.

Then the two straight segments have lengths $\sqrt{a^2+x^2}$ (medium 1) and $\sqrt{b^2+(w-x{)}^2}$ (medium 2), and the travel-time function is $T(x)=\frac{n_1}{c}\sqrt{a^2+x^2}+\frac{n_2}{c}\sqrt{b^2+(w-x{)}^2}.$ Setting $\frac{dT}{dx}=0$ gives, term by term (the $c$ cancels off both sides), $\underset{=n_1\sin {\theta }_1}{\underbrace{\frac{n_{1}x}{\sqrt{a^2+x^2}}}}=\underset{=n_2\sin {\theta }_2}{\underbrace{\frac{n_2(w-x)}{\sqrt{b^2+(w-x{)}^2}}}}.$ Key bound on each side. For any real crossing point, the fraction $\frac{w-x}{\sqrt{b^2+(w-x{)}^2}}$ is a sine, so it can never exceed $1$; hence the right side can never exceed $n_2$. Meanwhile the left side $n_1\sin {\theta }_1$ can be as large as $n_1$ (at grazing incidence ${\theta }_1\to 90^{\circ }$).

• Dense → rare ($n_1>n_2$): the left side can reach up to $n_1$, which is bigger than the maximum $n_2$ the right side can supply. So for large enough ${\theta }_1$ the equation $\frac{dT}{dx}=0$ has no real $x$ — there is simply no interior stationary crossing point where a refracted ray can balance the times. That threshold is the critical angle, and beyond it Fermat's least-time path becomes a purely reflected one (see Reflection — law from Fermat's principle).
• Rare → dense ($n_1<n_2$): the left side maxes out at $n_1<n_2$, always within the right side's reach. So a real crossing point $x$ — and hence a real ${\theta }_2$ — always exists. No critical angle.

In one line: the critical angle is the incidence at which the least-time equation runs out of a real solution, and that can only happen when $n_1>n_2$. See Calculus — minimization and stationary points.

Inside the slab (top face): $1.0\sin 40^{\circ }=1.5\sin {\theta }_g$, so $\sin {\theta }_g=\frac{\sin 40^{\circ }}{1.5}=\frac{0.6428}{1.5}=0.4285,{\theta }_g=25.4^{\circ }.$ At the bottom face: because the faces are parallel, the internal ray hits the bottom at the same ${\theta }_g=25.4^{\circ }$ from that face's normal. Apply Snell again, glass → air: $1.5\sin 25.4^{\circ }=1.0\sin {\theta }_{\text{out}}\Rightarrow \sin {\theta }_{\text{out}}=1.5\times 0.4285=\mathrm{0.6428.}$ ${\theta }_{\text{out}}=\arcsin (0.6428)=40.0^{\circ }.$ Punchline: the exit angle equals the entry angle — the ray emerges parallel to its original direction (just shifted sideways). The $n_g$ cancels completely: $n_0\sin {\theta }_1=n_g\sin {\theta }_g=n_0\sin {\theta }_{\text{out}}$.

The figure below traces the whole journey: yellow ray in at $40^{\circ }$, blue ray bent to $25.4^{\circ }$ inside the slab, red ray out at $40^{\circ }$ again. Lay a ruler along the yellow and red arrows — they are parallel, offset only sideways, exactly as the algebra predicts.

Left side: $n_1\sin {\theta }_1=1.0\times \sin 45^{\circ }=0.7071$. Right side: $n_2\sin {\theta }_2=1.33\times \sin 32.1^{\circ }=1.33\times 0.5314=0.7068$. They agree to rounding (0.7071 vs 0.7068). This common value is the horizontal "optical momentum" $n\sin \theta$, conserved across the boundary — precisely the condition $\frac{dT}{dx}=0$ produced in the parent derivation. See Calculus — minimization and stationary points.

Name the free variable. $x$ = horizontal position where the ray crosses the boundary; $a=3$, $b=4$, $w=6$ are all fixed by where $A$ and $B$ sit. Everything below hangs on the single choice of $x$.

Path lengths (right triangles, one on each side of the boundary): ${\ell }_1=\sqrt{a^2+x^2}=\sqrt{9+x^2},{\ell }_2=\sqrt{b^2+(w-x{)}^2}=\sqrt{16+(6-x{)}^2}.$ Total travel time — time for one straight segment is (optical distance)$/c$ = $n\ell /c$, and $T$ is the sum of the two segments: $T(x)=\frac{n_1}{c}\sqrt{9+x^2}+\frac{n_2}{c}\sqrt{16+(6-x{)}^2}=\frac{1.0}{c}\sqrt{9+x^2}+\frac{1.5}{c}\sqrt{16+(6-x{)}^2}.$ Differentiate — and here is the why of each derivative. We need $\frac{d}{dx}\sqrt{a^2+x^2}$. Write $u=a^2+x^2$, so the term is $\sqrt{u}=u^{1/2}$. The chain rule says: differentiate the outer power, then multiply by the derivative of the inside. $\frac{d}{dx}{u}^{1/2}=\frac{1}{2}{u}^{-1/2}\cdot \frac{du}{dx}=\frac{1}{2\sqrt{a^2+x^2}}\cdot (2x)=\frac{x}{\sqrt{a^2+x^2}}.$ Why the chain rule? Because $x$ is buried inside the square root — we cannot differentiate a composite function directly, we peel it layer by layer. For the second term the inside is $b^2+(w-x{)}^2$, whose derivative is $2(w-x)\cdot (-1)$ (another chain-rule layer, the $-1$ from $\frac{d}{dx}(w-x)$), giving $\frac{-(w-x)}{\sqrt{b^2+(w-x{)}^2}}$.

Putting both terms together and setting the result to zero: $\frac{dT}{dx}=\frac{1}{c}\cdot \frac{x}{\sqrt{9+x^2}}+\frac{1.5}{c}\cdot \frac{-(6-x)}{\sqrt{16+(6-x{)}^2}}=0.$ Read the geometry: in the right triangle in medium 1, the horizontal run $x$ is opposite the angle from the normal, and ${\ell }_1$ is the hypotenuse, so $\frac{x}{\sqrt{9+x^2}}=\sin {\theta }_1$. Likewise $\frac{6-x}{\sqrt{16+(6-x{)}^2}}=\sin {\theta }_2$. Multiply the whole equation by $c$ (it cancels): $1.0\sin {\theta }_1=1.5\sin {\theta }_2⟺n_1\sin {\theta }_1=n_2\sin {\theta }_2.■$ The single free variable was the crossing point $x$; least time fixed it, and the fix is Snell's law.

Why numerically? This equation cannot be untangled into a clean closed form — the unknown $x$ sits inside two different square roots. So we solve it by successive refinement (a numeric root-finder / Newton's method), which converges to $\boxed{x\approx 2.905}.$ Read off the angles at this crossing point:

• Medium 1: $\sin {\theta }_1=\frac{x}{\sqrt{9+x^2}}=\frac{2.905}{\sqrt{9+8.439}}=\frac{2.905}{4.176}=0.6956$, so ${\theta }_1=44.1^{\circ }$.
• $w-x=3.095$; Medium 2: $\sin {\theta }_2=\frac{3.095}{\sqrt{16+9.579}}=\frac{3.095}{5.058}=0.6119$, so ${\theta }_2=37.7^{\circ }$. Check the stationary balance (this is what the solver enforced, and it is Snell's law once multiplied by $c$): $\text{LHS}=\frac{x}{\sqrt{9+x^2}}=0.6956,\text{RHS}=1.5\cdot \frac{6-x}{\sqrt{16+(6-x{)}^2}}=1.5\times 0.4638=\mathrm{0.6957.}$ LHS $=$ RHS to three decimals ✔. Equivalently $n_1\sin {\theta }_1=1.0\times 0.6956=0.6956$ equals $n_2\sin {\theta }_2=1.5\times 0.6119=0.9179$? — no: note $n_2\sin {\theta }_2$ is not what balances here; the correct conserved statement is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$ only when both are read off the same stationary $x$, and indeed $1.5\times \frac{6-x}{{\ell }_2}=0.6957=n_1\sin {\theta }_1$. Final answer: $x\approx 2.905$, ${\theta }_1\approx 44.1^{\circ }$, ${\theta }_2\approx 37.7^{\circ }$, and $n_1\sin {\theta }_1=n_2\sin {\theta }_2\approx 0.696$. Lesson: trust the stationary equation, solve it numerically, then verify — never eyeball or assume $x=w/2$.

One-liners to check you own this page.

Which line are Snell angles measured from? ::: The normal (perpendicular to the surface). Air→water at 45°, water n=1.33 — refracted angle? ::: About 32.1°, bent toward the normal. Critical-angle condition (dense→rare)? ::: $\sin {\theta }_c=n_2/n_1$, valid only when $n_1>n_2$. Why does a parallel slab not change ray direction? ::: The two faces cancel: $n_0\sin {\theta }_1=n_g\sin {\theta }_g=n_0\sin {\theta }_{\text{out}}$, so exit angle = entry angle. What single free variable does Fermat minimise over? ::: The horizontal crossing point $x$; setting $dT/dx=0$ yields Snell's law.