Snell's law — derivation from Fermat's principle

WHAT is Fermat's Principle?

Optical path length (OPL): $\text{OPL} = n \cdot d$ , where $n$ is the refractive index and $d$ is the geometric distance.
WHY $n$ appears: speed of light in a medium is $v = c/n$ . Time to cross distance $d$ is $t = d/v = nd/c$ . So minimizing time $\Leftrightarrow$ minimizing $nd$ = OPL (since $c$ is constant).

HOW to derive Snell's law

We send light from point $A$ in medium 1 (index $n_1$ ) to point $B$ in medium 2 (index $n_2$ ). The boundary is a flat horizontal line. Let the ray cross the boundary at horizontal position $x$ .

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

Setup (Why these variables?): We let $x$ be the crossing point — it is the one free choice light makes. Everything else (heights $a$ , $b$ , total width $w$ ) is fixed by where $A$ and $B$ are.

Path length in medium 1: $\ell_1 = \sqrt{a^2 + x^2}$
Path length in medium 2: $\ell_2 = \sqrt{b^2 + (w-x)^2}$

Total time (Why $n/c$ ? because $t = nd/c$ ): $T(x) = \frac{n_1}{c}\sqrt{a^2 + x^2} + \frac{n_2}{c}\sqrt{b^2 + (w-x)^2}$

Minimize: light picks the $x$ that makes $T$ stationary, so set $\dfrac{dT}{dx}=0$ .

Why this step? Stationary = the derivative vanishes; this is the mathematical translation of "least time."

$\frac{dT}{dx} = \frac{n_1}{c}\cdot\frac{x}{\sqrt{a^2+x^2}} + \frac{n_2}{c}\cdot\frac{-(w-x)}{\sqrt{b^2+(w-x)^2}} = 0$

Why this step? Each square root differentiates by the chain rule: $\frac{d}{dx}\sqrt{a^2+x^2}=\frac{x}{\sqrt{a^2+x^2}}$ .

Now read the geometry: $\sin\theta_1 = \frac{x}{\sqrt{a^2+x^2}}, \qquad \sin\theta_2 = \frac{w-x}{\sqrt{b^2+(w-x)^2}}$

Why this step? $\theta_1$ is the angle from the normal (vertical). The opposite side is the horizontal run $x$ ; the hypotenuse is $\ell_1$ . So $\sin\theta_1 = x/\ell_1$ — pure right-triangle trig.

Substitute, cancel $c$ : $\frac{n_1}{c}\sin\theta_1 - \frac{n_2}{c}\sin\theta_2 = 0$

Worked Examples

Worked example Example 3 — Speeds, not indices

A medium where light travels at $v_1 = 2\times10^8\,$ m/s; another at $v_2 = 1.5\times10^8\,$ m/s. Light hits at $40^\circ$ . Find $\theta_2$ .

Step 1: Since $n = c/v$ , Snell becomes $\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$ . Why? $n_1\sin\theta_1=n_2\sin\theta_2 \Rightarrow \frac{c}{v_1}\sin\theta_1=\frac{c}{v_2}\sin\theta_2$ , cancel $c$ . Step 2: $\sin\theta_2 = \sin40^\circ \cdot \frac{v_2}{v_1} = 0.643 \times \frac{1.5}{2} = 0.482$ . Step 3: $\theta_2 = 28.8^\circ$ . Slower medium → bends toward normal. ✔

Common Mistakes

Recall Feynman: explain it to a 12-year-old

Light is in a hurry but it can't go fast through thick stuff like water or glass. So when it has to cross from fast air into slow glass, it's clever: it spends less of its trip inside the slow glass and more in the fast air. That makes its path bend at the surface, like a lifeguard who runs farther on the beach before splashing into the slow water. The exact amount of bending follows one neat rule: $n_1\sin\theta_1 = n_2\sin\theta_2$ .

Flashcards

What does Fermat's principle state?

Light travels the path for which the travel time (optical path length) is stationary, usually minimum.

Why does minimizing time equal minimizing

n\cdot d

Because

t = d/v = nd/c

, and

c

is constant, so minimizing time ⇔ minimizing the optical path length

nd

State Snell's law.

n_1\sin\theta_1 = n_2\sin\theta_2

, angles measured from the normal.

In the derivation, what is the single free variable?

The horizontal crossing point

x

on the boundary; setting

dT/dx=0

gives Snell's law.

How does

\sin\theta_1

appear from the geometry?

\sin\theta_1 = x/\sqrt{a^2+x^2}

= (horizontal run)/(path length) in a right triangle with the normal.

Entering a denser medium, does light bend toward or away from the normal?

Toward the normal (larger

n

⇒ smaller

\sin\theta

⇒ smaller angle).

What is the critical angle condition?

When

\theta_2 = 90^\circ

\sin\theta_c = n_2/n_1

(going from dense to rare).

Snell's law in terms of speeds?

\dfrac{\sin\theta_1}{v_1} = \dfrac{\sin\theta_2}{v_2}

, since

n = c/v

From which line is the angle of incidence measured?

From the normal (perpendicular to the surface), not the surface itself.

Connections

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bilkul simple hai: light hamesha woh raasta choose karti hai jisme time sabse kam lage — distance kam nahi, time kam. Jab light kisi dense medium (glass, paani) me jaati hai to uski speed kam ho jaati hai ( $v = c/n$ ). Isliye light thoda chalaak ban jaati hai: slow medium me kam chalti hai aur fast medium me zyada, taaki total time bach jaaye. Yahi chालakी ki wajah se boundary par ray mud jaati hai (bend hoti hai).

Derivation me hum sirf ek cheez free rakhte hain — woh point jahan ray boundary cross karti hai, usse $x$ bolte hain. Total time $T(x)$ likhte hain dono parts ka ( $n_1 \ell_1/c + n_2 \ell_2/c$ ), aur minimum time ke liye $dT/dx = 0$ set karte hain. Calculus karne ke baad jo terms aate hain woh exactly $\sin\theta_1$ aur $\sin\theta_2$ ban jaate hain (right-triangle geometry se). Result: $n_1\sin\theta_1 = n_2\sin\theta_2$ — yahi Snell's law hai.

Yaad rakhne wali baat: angle hamesha normal se naapo, surface se nahi. Aur ek rule — "Slow medium me jao to normal ki taraf jhuko". Jaise denser glass me ghuste hi ray normal ke paas aa jaati hai. Yeh poora concept fibre optics, lenses, prism, mirage — sab me kaam aata hai, isliye base strong rakho.

Snell's law — derivation from Fermat's principle

WHAT is Fermat's Principle?

HOW to derive Snell's law

Worked Examples

Common Mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Optics

Connections