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

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

• $n_1$ = us medium ka refractive index jisme se light start karti hai (medium 1).
• $n_2$ = us medium ka refractive index jisme light enter karti hai (medium 2).
• ${\theta }_1$ = angle of incidence, medium 1 mein normal se measure kiya hua.
• ${\theta }_2$ = angle of refraction, medium 2 mein normal se measure kiya hua.

Refractive index define hoti hai $n=c/v$ se, jahan $c$ vacuum mein light ki speed hai aur $v$ us medium mein uski speed: yeh batata hai ki light us medium mein vacuum ke muqable mein kitni zyada slow chalti hai. Dekho Refractive Index.

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

Prediction: Water denser hai ($n_2>n_1$), isliye light usme enter karte waqt slow ho jaati hai. Rule "Slow ⇒ Slant toward the Normal" kehta hai ki ray normal ki taraf bend karti hai, isliye ${\theta }_2<45^{\circ }$.

Kyun: $n\sin \theta$ conserve hota hai. Kyunki $n_2>n_1$, product ko equal rakhne ke liye hume $\sin {\theta }_2<\sin {\theta }_1$ chahiye, isliye angle chhota hoga.

Sahi. ${\theta }_1=0^{\circ }$ ko Snell's law mein daalo: $n_1\sin 0^{\circ }=n_2\sin {\theta }_2\Rightarrow 0=n_2\sin {\theta }_2\Rightarrow \sin {\theta }_2=0\Rightarrow {\theta }_2=0^{\circ }.$ Ray seedhi nikal jaati hai. Jo ray normal ke saath travel karti hai uska horizontal run zero hota hai, aur bilkul yahi horizontal component hai jo boundary ke across share hona chahiye — zero zero hi rehta hai. Yeh degenerate (normal-incidence) case hai.

Step 1 — Snell likho: $(1.0)\sin 45^{\circ }=1.33\sin {\theta }_2$. Step 2 — sine isolate karo: $\sin {\theta }_2=\frac{\sin 45^{\circ }}{1.33}=\frac{0.7071}{1.33}=0.5317$. Divide kyun? Hume ${\theta }_2$ akela chahiye, isliye known factor $n_2=1.33$ dusri side le jaate hain. Step 3 — sine undo karo: ${\theta }_2=\arcsin (0.5317)=32.1^{\circ }$. Check: $32.1^{\circ }<45^{\circ }$ ✔ — normal ki taraf bend hua, bilkul jaisa predict kiya tha.

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: Ray nikalne par normal se door bend hui ($40^{\circ }>25^{\circ }$), jo denser medium se nikalte waqt hota hai — $n_1=1.52>1=n_2$ ke saath consistent hai. ✔

Step 1 — Snell ka speed form: kyunki $n=c/v$ (jahan $c$ vacuum mein light ki speed hai), $c$ cancel ho jaata hai: $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}.$ Step 2 — solve karo: $\sin {\theta }_2=\sin 50^{\circ }\cdot \frac{v_2}{v_1}=0.7660\times \frac{1.8}{2.4}=0.7660\times 0.75=0.5745$. Step 3: ${\theta }_2=\arcsin (0.5745)=35.1^{\circ }$. Check: Slower medium ($v_2<v_1$) → normal ki taraf bend → ${\theta }_2<{\theta }_1$ ✔.

Step 1 — critical condition: ${\theta }_2=90^{\circ }$ set karo, isliye $\sin {\theta }_2=1$: $n_1\sin {\theta }_c=n_2\cdot 1.$ Step 2: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.0}{1.52}=0.6579$. Step 3: ${\theta }_c=\arcsin (0.6579)=41.1^{\circ }$. $41.1^{\circ }$ se aage equation $\sin {\theta }_2>1$ maangti hai, jo impossible hai — isliye koi refracted ray exist nahi karti aur saari light reflect ho jaati hai: Total Internal Reflection. Dekho bhi Critical Angle.

Neeche ka figure bilkul yahi critical case draw karta hai: incident ray (yellow) ${\theta }_c\approx 41^{\circ }$ par strike karti hai aur refracted ray (red) ${\theta }_2=90^{\circ }$ par surface ke saath flat lie karti hai. Green arc ${\theta }_c$ mark karta hai — jab tum incident ray ko us se aage tilt karte ho, red arrow ke paas jaane ki jagah nahi bachti.

$\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}=\frac{1.52\times \sin 50^{\circ }}{1.0}=1.52\times 0.7660=\mathrm{1.164.}$ Lekin kisi bhi real angle ka sine $1$ se kabhi zyada nahi ho sakta. Kyunki $1.164>1$, koi real ${\theta }_2$ exist nahi karta — light refract hokar bahar nahi ja sakti aur total internal reflection hoti hai. Physically iska matlab: boundary ek perfect mirror ki tarah kaam karti hai. Exactly isi tarah light optical fibre mein trapped rehti hai.

Geometry locally set up karo. Point $A$ medium 1 mein aur point $B$ medium 2 mein rakho, flat boundary horizontal ho. Define karo:

• $a$ = boundary ke upar $A$ ki height (fixed),
• $b$ = boundary ke neeche $B$ ki depth (fixed),
• $w$ = $A$ aur $B$ ke beech horizontal separation (fixed),
• $x$ = woh horizontal position jahan ray boundary cross karti hai — ek free choice jo light karti hai.

Tab dono straight segments ki lengths hain $\sqrt{a^2+x^2}$ (medium 1) aur $\sqrt{b^2+(w-x{)}^2}$ (medium 2), aur travel-time function hai $T(x)=\frac{n_1}{c}\sqrt{a^2+x^2}+\frac{n_2}{c}\sqrt{b^2+(w-x{)}^2}.$ $\frac{dT}{dx}=0$ set karne par, term by term ($c$ dono sides se cancel ho jaata hai), $\underset{=n_1\sin {\theta }_1}{\underbrace{\frac{n_{1}x}{\sqrt{a^2+x^2}}}}=\underset{=n_2\sin {\theta }_2}{\underbrace{\frac{n_2(w-x)}{\sqrt{b^2+(w-x{)}^2}}}}.$ Har side par key bound. Kisi bhi real crossing point ke liye, fraction $\frac{w-x}{\sqrt{b^2+(w-x{)}^2}}$ ek sine hai, isliye yeh kabhi $1$ se zyada nahi ho sakta; isliye right side kabhi $n_2$ se zyada nahi ho sakti. Meanwhile left side $n_1\sin {\theta }_1$ utni badi ho sakti hai jitna $n_1$ (grazing incidence ${\theta }_1\to 90^{\circ }$ par).

• Dense → rare ($n_1>n_2$): left side $n_1$ tak pahunch sakti hai, jo right side ki maximum $n_2$ se badi hai. Isliye itne bade ${\theta }_1$ ke liye equation $\frac{dT}{dx}=0$ ka koi real $x$ nahi — simply koi interior stationary crossing point nahi jahan refracted ray times balance kar sake. Woh threshold critical angle hai, aur us se aage Fermat ka least-time path purely reflected ho jaata hai (dekho Reflection — law from Fermat's principle).
• Rare → dense ($n_1<n_2$): left side max $n_1<n_2$ hoti hai, hamesha right side ki reach ke andar. Isliye ek real crossing point $x$ — aur isliye ek real ${\theta }_2$ — hamesha exist karta hai. Koi critical angle nahi.

Ek line mein: critical angle woh incidence hai jis par least-time equation ka real solution khatam ho jaata hai, aur aisa sirf tab ho sakta hai jab $n_1>n_2$ ho. Dekho Calculus — minimization and stationary points.

Slab ke andar (top face): $1.0\sin 40^{\circ }=1.5\sin {\theta }_g$, isliye $\sin {\theta }_g=\frac{\sin 40^{\circ }}{1.5}=\frac{0.6428}{1.5}=0.4285,{\theta }_g=25.4^{\circ }.$ Bottom face par: kyunki faces parallel hain, internal ray bottom par us face ke normal se same ${\theta }_g=25.4^{\circ }$ par hit karti hai. Phir se Snell apply karo, 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: exit angle entry angle ke barabar hai — ray apni original direction ke parallel nikarti hai (bas sideways shift ke saath). $n_g$ completely cancel ho jaata hai: $n_0\sin {\theta }_1=n_g\sin {\theta }_g=n_0\sin {\theta }_{\text{out}}$.

Neeche ka figure poori journey trace karta hai: yellow ray $40^{\circ }$ par andar, blue ray slab ke andar $25.4^{\circ }$ par bend, red ray phir $40^{\circ }$ par bahar. Yellow aur red arrows ke saath ek ruler rakho — woh parallel hain, sirf sideways offset hain, bilkul jaisa algebra predict karta hai.

Left side: $n_1\sin {\theta }_1=1.0\times \sin 45^{\circ }=0.7071$. Right side: $n_2\sin {\theta }_2=1.33\times \sin 32.1^{\circ }=1.33\times 0.5314=0.7068$. Dono rounding tak agree karte hain (0.7071 vs 0.7068). Yeh common value horizontal "optical momentum" $n\sin \theta$ hai, boundary ke across conserved — bilkul wahi condition jo parent derivation mein $\frac{dT}{dx}=0$ ne produce ki thi. Dekho Calculus — minimization and stationary points.

Free variable naam do. $x$ = woh horizontal position jahan ray boundary cross karti hai; $a=3$, $b=4$, $w=6$ sab fixed hain is baat se ki $A$ aur $B$ kahan baithe hain. Neeche sab kuch $x$ ke ek choice par tikaa hai.

Path lengths (right triangles, boundary ke dono taraf ek ek): ${\ell }_1=\sqrt{a^2+x^2}=\sqrt{9+x^2},{\ell }_2=\sqrt{b^2+(w-x{)}^2}=\sqrt{16+(6-x{)}^2}.$ Total travel time — ek straight segment ka time (optical distance)$/c$ = $n\ell /c$ hota hai, aur $T$ dono segments ka sum hai: $T(x)=\frac{n_1}{c}\sqrt{9+x^2}+\frac{n_2}{c}\sqrt{16+(6-x{)}^2}=\frac{1.0}{c}\sqrt{9+x^2}+\frac{1.5}{c}\sqrt{16+(6-x{)}^2}.$ Differentiate karo — aur yahan har derivative ka kyun hai. Hume $\frac{d}{dx}\sqrt{a^2+x^2}$ chahiye. $u=a^2+x^2$ likho, to term $\sqrt{u}=u^{1/2}$ hai. Chain rule kehta hai: outer power differentiate karo, phir inside ke derivative se multiply karo. $\frac{d}{dx}{u}^{1/2}=\frac{1}{2}{u}^{-1/2}\cdot \frac{du}{dx}=\frac{1}{2\sqrt{a^2+x^2}}\cdot (2x)=\frac{x}{\sqrt{a^2+x^2}}.$ Chain rule kyun? Kyunki $x$ square root ke andar dabba hua hai — hum ek composite function directly differentiate nahi kar sakte, layer by layer peelte hain. Doosre term ke liye inside $b^2+(w-x{)}^2$ hai, jiska derivative $2(w-x)\cdot (-1)$ hai (ek aur chain-rule layer, $-1$ $\frac{d}{dx}(w-x)$ se), jo $\frac{-(w-x)}{\sqrt{b^2+(w-x{)}^2}}$ deta hai.

Dono terms saath rakhke aur result zero set karke: $\frac{dT}{dx}=\frac{1}{c}\cdot \frac{x}{\sqrt{9+x^2}}+\frac{1.5}{c}\cdot \frac{-(6-x)}{\sqrt{16+(6-x{)}^2}}=0.$ Geometry padho: medium 1 mein right triangle mein, horizontal run $x$ normal se angle ka opposite side hai, aur ${\ell }_1$ hypotenuse hai, isliye $\frac{x}{\sqrt{9+x^2}}=\sin {\theta }_1$. Isi tarah $\frac{6-x}{\sqrt{16+(6-x{)}^2}}=\sin {\theta }_2$. Poori equation ko $c$ se multiply karo (cancel ho jaata hai): $1.0\sin {\theta }_1=1.5\sin {\theta }_2⟺n_1\sin {\theta }_1=n_2\sin {\theta }_2.■$ Ek hi free variable tha crossing point $x$; least time ne use fix kiya, aur woh fix hai hi Snell's law.

Numerically kyun? Yeh equation ek clean closed form mein nahi suljhaya ja sakta — unknown $x$ do alag square roots ke andar baitha hai. Isliye hum ise successive refinement (numeric root-finder / Newton's method) se solve karte hain, jo converge karta hai $\boxed{x\approx 2.905}.$ Is crossing point par angles padho:

• Medium 1: $\sin {\theta }_1=\frac{x}{\sqrt{9+x^2}}=\frac{2.905}{\sqrt{9+8.439}}=\frac{2.905}{4.176}=0.6956$, isliye ${\theta }_1=44.1^{\circ }$.
• $w-x=3.095$; Medium 2: $\sin {\theta }_2=\frac{3.095}{\sqrt{16+9.579}}=\frac{3.095}{5.058}=0.6119$, isliye ${\theta }_2=37.7^{\circ }$. Stationary balance check karo (yahi solver ne enforce kiya, aur $c$ se multiply karne par yeh Snell's law hi hai): $\text{LHS}=\frac{x}{\sqrt{9+x^2}}=0.6956,\text{RHS}=1.5\cdot \frac{6-x}{\sqrt{16+(6-x{)}^2}}=1.5\times 0.4638=\mathrm{0.6957.}$ LHS $=$ RHS teen decimals tak ✔. Equivalently $n_1\sin {\theta }_1=1.0\times 0.6956=0.6956$ equals $n_2\sin {\theta }_2=1.5\times 0.6119=0.9179$? — nahi: note karo ki $n_2\sin {\theta }_2$ woh nahi jo yahan balance karta hai; correct conserved statement $n_1\sin {\theta }_1=n_2\sin {\theta }_2$ tab hi valid hai jab dono same stationary $x$ se read ho, aur indeed $1.5\times \frac{6-x}{{\ell }_2}=0.6957=n_1\sin {\theta }_1$. Final answer: $x\approx 2.905$, ${\theta }_1\approx 44.1^{\circ }$, ${\theta }_2\approx 37.7^{\circ }$, aur $n_1\sin {\theta }_1=n_2\sin {\theta }_2\approx 0.696$. Lesson: stationary equation par trust karo, numerically solve karo, phir verify karo — kabhi eyeball mat karo ya $x=w/2$ assume mat karo.

One-liners taaki check kar sako ki is page par tumhari pakad hai.

Snell angles kaunsi line se measure hoti hain? ::: Normal se (surface ke perpendicular). Air→water 45° par, water n=1.33 — refracted angle? ::: Lagbhag 32.1°, normal ki taraf bend. Critical-angle condition (dense→rare)? ::: $\sin {\theta }_c=n_2/n_1$, valid sirf jab $n_1>n_2$. Parallel slab ray direction kyun nahi badlta? ::: Dono faces cancel kar dete hain: $n_0\sin {\theta }_1=n_g\sin {\theta }_g=n_0\sin {\theta }_{\text{out}}$, isliye exit angle = entry angle. Fermat kis ek free variable ko minimise karta hai? ::: Horizontal crossing point $x$; $dT/dx=0$ set karne par Snell's law milti hai.