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

2.5.4 · D3 · Physics › Optics › Snell's law — derivation from Fermat's principle

Yeh page parent derivation ki "koi surprise nahi bacha" companion hai. Rule hamesha wahi ek line hai:

Kuch bhi compute karne se pehle, teen plain-word reminders taaki koi symbol undefined na reh jaaye:

Ab hum map karte hain har tarah ki situation jo yeh single equation de sakti hai. Agar koi case neeche ki table mein nahi hai, toh woh Snell's law ke liye exist hi nahi karta.

The scenario matrix

Har refraction problem in cells mein se ek (ya blend) hoti hai. Last column us example ka naam deta hai jo ise best explain karta hai.

#	Case class	Kya special hai	Example
A	Rare → dense ( $n_{1} < n_{2}$ )	ray normal ki taraf bend karti hai, $θ_{2} < θ_{1}$	Ex 1
B	Dense → rare ( $n_{1} > n_{2}$ )	ray normal se door bend karti hai, $θ_{2} > θ_{1}$	Ex 2
C	Normal incidence ( $θ_{1} = 0$ )	degenerate: ray seedhi jaati hai, koi bend nahi	Ex 3
D	Grazing incidence ( $θ_{1} \to 9 0^{\circ}$ )	limiting: denser medium mein sabse bada possible bend	Ex 4
E	Critical angle ( $θ_{2} = 9 0^{\circ}$ )	existence ki boundary; iske baad → koi refraction nahi	Ex 5
F	Beyond critical ( $sin θ_{2} > 1$ )	total internal reflection, "koi answer nahi" hi answer hai	Ex 6
G	Same index ( $n_{1} = n_{2}$ )	degenerate: optically koi boundary nahi, $θ_{2} = θ_{1}$	Ex 7
H	Indices nahi, speeds di hain	Snell ko $v$ se rewrite karo; ek translation twist	Ex 8
I	Real-world word problem	pool/coin apparent-depth style geometry	Ex 9
J	Exam twist: angle surface se diya	trap — pehle $9 0^{\circ} - ϕ$ convert karo	Ex 10

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

The examples

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

Worked example Air se water mein

Air mein light ( $n_{1} = 1.00$ ) water ( $n_{2} = 1.33$ ) par $θ_{1} = 4 5^{\circ}$ par girrti hai. $θ_{2}$ nikalo.

Forecast: water optically denser hai (bada $n$ ), toh guess karo $θ_{2}$ $4 5^{\circ}$ se chhota hoga — ray normal ki taraf tuck karegi.

Snell likho: $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ . Yeh step kyun? Yahi ek relation hai jo dono angles ko link karta hai.
Unknown sine isolate karo: $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}} = \frac{1.00 \times sin 4 5 ^{\circ}}{1.33} = \frac{0.7071}{1.33} = 0.5317$ . Yeh step kyun? Hum $sin θ_{2}$ isolate karte hain kyunki $θ_{2}$ chahiye.
Sine undo karo: $θ_{2} = arcsin (0.5317) = 32. 1^{\circ}$ . Yeh step kyun? $arcsin$ jawab deta hai "kis angle ka sine yeh hai?" — yeh $sin$ ko undo karta hai.

Verify: $32. 1^{\circ} < 4 5^{\circ}$ ✔ (normal ki taraf bend, forecast se match). Cross-check: $1.33 \times sin 32. 1^{\circ} = 1.33 \times 0.5317 = 0.707 = 1.00 \times sin 4 5^{\circ}$ ✔.

Ex 2 — Cell B: dense → rare (normal se door bend)

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

Worked example Seedha andar jaata beam

Ek laser air ( $n_{1} = 1.0$ ) se glass ( $n_{2} = 1.5$ ) mein exactly normal ke along enter karti hai, $θ_{1} = 0^{\circ}$ .

Forecast: seedha andar shine kar rahe hain — koi "sideways" nahi hai bend karne ke liye, toh guess karo $θ_{2} = 0^{\circ}$ , koi bend nahi.

Snell: $1.0 sin 0^{\circ} = 1.5 sin θ_{2}$ . Yeh step kyun? Degenerate input plug karo rather than answer assume karo.
$sin 0^{\circ} = 0$ , toh $0 = 1.5 sin θ_{2} \Rightarrow sin θ_{2} = 0$ . Yeh step kyun? Left side par zero right side ko bhi zero force karta hai.
$θ_{2} = arcsin (0) = 0^{\circ}$ . Yeh step kyun? $[0, 9 0^{\circ}]$ mein sirf wahi angle jiska sine $0$ hai woh $0^{\circ}$ hai.

Verify: Ray seedhi bina deviate hue nikal jaati hai — issi liye seedha paani mein neeche dekhne par jo dikhta hai woh shift nahi hota. Glass ke andar speed abhi bhi girti hai (yaad karo $v = c / n$ upar ki definitions se: bada $n$ ⇒ chhota $v$ ), lekin direction nahi girti. ✔

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

Worked example Glass mein surface ko skim karta hua

Air mein light ( $n_{1} = 1.0$ ) glass ( $n_{2} = 1.5$ ) ko $θ_{1} = 89. 9^{\circ}$ par graze karti hai (almost surface ke along). $θ_{2}$ nikalo, aur is glass mein possible sabse bada $θ_{2}$ .

Forecast: sabse extreme sideways entry. Ek maximum $θ_{2}$ hona chahiye jis tak woh pahunch sake — guess karo yeh $θ_{1} = 9 0^{\circ}$ se set hota hai.

$sin θ_{2} = \frac{1.0 sin 89. 9 ^{\circ}}{1.5} = \frac{0.99999}{1.5} = 0.66666$ . Yeh step kyun? Hum $θ_{1}$ ko uske physical ceiling tak push karte hain $θ_{2}$ ka ceiling find karne ke liye.
$θ_{2} = arcsin (0.66666) = 41. 8^{\circ}$ . Yeh step kyun? Actual refracted angle pane ke liye invert karo.
Absolute max as $θ_{1} \to 9 0^{\circ}$ : $sin θ_{2}^{m a x} = \frac{n _{1}}{n _{2}} = \frac{1}{1.5} = 0.6667 \Rightarrow θ_{2}^{m a x} = 41. 8^{\circ}$ . Yeh step kyun? $sin 9 0^{\circ} = 1$ numerator ka sabse bada value hai, toh yahi is glass mein air se sabse bada refracted angle hai.

Verify: Yeh exactly s01 figure mein teal dot hai: teal curve ka ceiling $sin θ_{2} = 0.667$ hai, yaani $41. 8^{\circ}$ . Air se is glass mein har ray ek cone ke andar enter hoti hai jiska half-angle $41. 8^{\circ}$ hai. Notice karo yeh number Ex 5 ke critical angle ke barabar hai — refraction aur total internal reflection same equation ke mirror images hain. ✔

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

Worked example Barely escape karta hua

Light glass ( $n_{1} = 1.5$ ) se air ( $n_{2} = 1.0$ ) ki taraf travel karti hai. Kis $θ_{1} = θ_{c}$ par refracted ray surface ko skim karti hai, $θ_{2} = 9 0^{\circ}$ ?

Forecast: dense→rare bends away, toh kisi incidence par ray surface ke flat along ho jaayegi. Guess karo $θ_{c}$ thoda $4 0^{\circ}$ se upar hai.

$θ_{2} = 9 0^{\circ}$ set karo: $1.5 sin θ_{c} = 1.0 sin 9 0^{\circ} = 1.0$ . Yeh step kyun? $θ_{2} = 9 0^{\circ}$ woh last moment hai jab refracted ray exist karti hai — critical ki definition yahi hai.
$sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.6667$ . Yeh step kyun? Incidence sine solve karo; note karo yeh indices ka ratio hai, rare over dense.
$θ_{c} = arcsin (0.6667) = 41. 8^{\circ}$ . Yeh step kyun? Sine invert karo.

Verify: Wahi $41. 8^{\circ}$ jaise Ex 4 mein — entry cone aur escape cone match karte hain, jaisa hona chahiye. Neeche ka figure exactly yahi geometry draw karta hai: plum incident ray teal glass se upar aati hai, aur orange refracted ray boundary ke flat along hai $θ_{2} = 9 0^{\circ}$ par, chhota plum arc $θ_{c} = 41. 8^{\circ}$ dotted normal se mark karta hai. Dekho Critical Angle aur Total Internal Reflection. ✔

Ex 6 — Cell F: beyond critical ("koi answer nahi" hi answer hai)

Worked example Limit se aage refract karne ki koshish

Wahi glass→air ( $n_{1} = 1.5$ , $n_{2} = 1.0$ ), lekin ab $θ_{1} = 5 0^{\circ}$ , jo $θ_{c} = 41. 8^{\circ}$ se bada hai. $θ_{2}$ nikalo.

Forecast: hum critical angle se past hain — guess karo koi refracted ray nahi hai.

$sin θ_{2} = \frac{1.5 sin 5 0 ^{\circ}}{1.0} = 1.5 \times 0.766 = 1.149$ . Yeh step kyun? Blindly Snell apply karo aur dekho number ka kya hota hai.
$sin θ_{2} = 1.149 > 1$ . Yeh step kyun? Sine kabhi $1$ se zyada nahi ho sakta — yeh signal karta hai koi real angle equation solve nahi karta.
Isliye koi refracted ray exist nahi karti: saari light glass mein wapas reflect hoti hai — total internal reflection. Yeh step kyun? Impossible sine mathematics ka woh tarika hai jisse hume batata hai ki light nahi nikal sakti.

Verify: $arcsin (1.149)$ real angles ke liye undefined hai ✔. s01 figure par yeh woh jagah hai jahan orange curve apne intersection se plum line ke saath aage nikal chuki hai — usse height $1$ se upar jaana hoga. Exactly issi liye optical fibres light trap karte hain: andar wali rays har baar wall par $θ_{c}$ se past strike karti hain. ✔

Ex 7 — Cell G: same index (optically koi boundary nahi)

Worked example Same glass par identical glass

Ek hi same glass ke do blocks ( $n_{1} = n_{2} = 1.5$ ) bina gap ke press kiye hain. Light join par $θ_{1} = 6 0^{\circ}$ par strike karti hai. $θ_{2}$ nikalo.

Forecast: agar media identical hain, toh light ke liye koi real interface nahi hai — guess karo $θ_{2} = 6 0^{\circ}$ , seedha through.

Snell: $1.5 sin 6 0^{\circ} = 1.5 sin θ_{2}$ . Yeh step kyun? Degenerate equal-index case directly test karo.
Common $1.5$ cancel karo: $sin θ_{2} = sin 6 0^{\circ}$ . Yeh step kyun? Dono sides ko shared index se divide karne se sines isolate ho jaate hain.
$θ_{2} = 6 0^{\circ}$ . Yeh step kyun? Equal sines (dono $[0, 9 0^{\circ}]$ mein) matlab equal angles.

Verify: Koi bending nahi — "no optical boundary" se consistent. Issi liye matching index ke liquid mein daali object gayab lagti hai: koi refraction nahi, koi visible edge nahi. ✔

Ex 8 — Cell H: speeds diye hain, indices nahi

Worked example Snell disguise mein

Medium 1 mein light $v_{1} = 2.0 \times 1 0^{8} m/s$ ki speed se chalti hai; medium 2 mein $v_{2} = 1.5 \times 1 0^{8} m/s$ se. Ek ray boundary par $θ_{1} = 4 0^{\circ}$ par strike karti hai. $θ_{2}$ nikalo. (Yahan $v$ har medium ke andar light ki speed hai, upar ki definitions se.)

Forecast: $v_{2} < v_{1}$ matlab medium 2 mein light slower hai, yaani yeh optically denser hai ( $n = c / v$ bada), toh guess karo $θ_{2} < 4 0^{\circ}$ .

Snell ko speeds se rewrite karo. Kyunki $n = c / v$ (jahan $c$ vacuum speed hai, ek shared constant): $\frac{c}{v _{1}} sin θ_{1} = \frac{c}{v _{2}} sin θ_{2}$ ; $c$ cancel karo aur paate hain $\frac{sin θ _{1}}{v _{1}} = \frac{sin θ _{2}}{v _{2}}$ . Yeh step kyun? Problem speeds deti hai, indices nahi — pehle translate karo, aur $c$ cancel hota hai kyunki dono sides par same hai.
Solve karo: $sin θ_{2} = sin θ_{1} \cdot \frac{v _{2}}{v _{1}} = sin 4 0^{\circ} \times \frac{1.5}{2.0} = 0.6428 \times 0.75 = 0.4821$ . Yeh step kyun? Speed form use karke unknown sine isolate karo.
$θ_{2} = arcsin (0.4821) = 28. 8^{\circ}$ . Yeh step kyun? Sine invert karo.

Verify: $28. 8^{\circ} < 4 0^{\circ}$ ✔ — slower medium, normal ki taraf bend, forecast se match. Dekho Refractive Index. ✔

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

Worked example Pool ki tali mein sikka

Ek sikka $h = 1.20 m$ ki true depth par water ( $n_{1} = 1.33$ , neeche paani; air $n_{2} = 1.00$ upar) ke neeche rakha hai. Almost seedha neeche dekhne par woh kitna deep lagta hai?

Forecast: paani nikalne wali rays ko normal se door bend karta hai, aankh ko fool karke sikka shallower lagta hai — guess karo $h_{app} < 1.20 m$ .

Pehle figure dekho. Ek ray coin $C$ se nikalta hai, ek horizontal step $s$ paani se surface point $P$ tak travel karta hai, phir air mein refract hota hai. Aankh air ray ko backward extend karti hai (dashed) jahan se woh lagta hai aana — seedha coin ke upar ek point par lekin shallower apparent depth $h_{app}$ par. True ray (depth $h$ se) aur apparent ray (depth $h_{app}$ se) dono $P$ par same horizontal step $s$ share karte hain, toh unke tangents hi compare karte hain.

$P$ par, normal se water-side angle $θ_{1}$ hai jahan $tan θ_{1} = s / h$ ; air-side angle $θ_{2}$ hai jahan $tan θ_{2} = s / h_{app}$ . Yeh step kyun? Dono right triangles horizontal $s$ share karte hain; tangent (opposite $s$ over adjacent depth) har angle ko uski depth se link karta hai.
Near-vertical viewing ke liye dono angles tiny hain, aur small angles ke liye $sin θ \approx tan θ$ . Snell $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ ban jaata hai $n_{1} tan θ_{1} \approx n_{2} tan θ_{2}$ . Yeh step kyun? $sin$ ko $tan$ se replace karna (sirf small angles ke liye valid) depth expressions feed karne deta hai.
Tangents substitute karo: $n_{1} \frac{s}{h} = n_{2} \frac{s}{h _{app}}$ . $s$ cancel ho jaata hai, bachta hai $\frac{n _{1}}{h} = \frac{n _{2}}{h _{app}}$ , toh $h_{app} = h \cdot \frac{n _{2}}{n _{1}} = 1.20 \times \frac{1.00}{1.33}$ . Yeh step kyun? Shared $s$ cancel hota hai, aur rearrange karne se apparent depth isolate hoti hai.
$h_{app} = 1.20 \times 0.7519 = 0.902 m$ . Yeh step kyun? Numbers plug karo.

Verify: $0.902 m < 1.20 m$ ✔ — sikka lagbhag $0.30 m$ shallower lagta hai, exactly woh roz ka illusion. Units: metres × (dimensionless ratio) = metres ✔.

Ex 10 — Cell J: exam twist (angle surface se diya)

Worked example Classic trap

Air ( $n_{1} = 1.0$ ) mein ray water surface se $ϕ = 2 5^{\circ}$ ka angle banati hai (normal se nahi). Water ka $n_{2} = 1.33$ hai. Refraction angle $θ_{2}$ nikalo (normal se, as usual).

Forecast: $2 5^{\circ}$ surface se matlab ek shallow, grazing ray hai — uska normal se angle bada hoga. Guess karo $θ_{1}$ $6 5^{\circ}$ ke paas hoga.

Normal mein convert karo: $θ_{1} = 9 0^{\circ} - ϕ = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Yeh step kyun? Snell's law normal se define hota hai; surface angle aur normal angle complements hain.
Snell: $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}} = \frac{1.0 sin 6 5 ^{\circ}}{1.33} = \frac{0.9063}{1.33} = 0.6814$ . Yeh step kyun? Ab standard equation sahi incidence angle ke saath apply hoti hai.
$θ_{2} = arcsin (0.6814) = 42. 9^{\circ}$ . Yeh step kyun? Sine invert karo.

Verify: Agar convert karna bhool jaate aur $2 5^{\circ}$ plug karte, toh milta $sin θ_{2} = sin 2 5^{\circ} /1.33 = 0.318$ , yaani $18. 5^{\circ}$ — ek galat, bahut chhota answer. Complement step hi bachata hai. Dekho parent note "angle from surface" mistake. ✔

Recall Matrix ka ek-line summary

Incidence dial ko $0^{\circ}$ se $9 0^{\circ}$ tak ghao: normal incidence ⇒ koi bend nahi; rare→dense hamesha refract karta hai (normal ki taraf bend, cone $θ_{2}^{m a x}$ par capped); dense→rare door bend karta hai critical angle tak, uske baad total internal reflection; equal indices ⇒ koi bend nahi; aur hamesha surface-angles ko pehle normal-angles mein convert karo.

Kis cell mein "koi real answer nahi" hai, aur kyun?

Cell F (beyond critical):

sin θ_{2} > 1

impossible hai, toh light totally internally reflect hoti hai.

Dense→rare ke liye, last refracting angle kaun sa hota hai?

Critical angle

θ_{c} = arcsin (n_{2} / n_{1})

, jahan

θ_{2} = 9 0^{\circ}

Normal incidence mein koi bending kyun nahi hoti?

sin 0^{\circ} = 0

sin θ_{2} = 0

force karta hai, toh

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