Geometric optics — rectilinear propagation, reflection, refraction: Visual walkthrough

2.5.1 · D2 · Physics › Optics › Geometric optics — rectilinear propagation, reflection, refr

Step 1 — "Ray" kya hota hai, aur "path ke saath time" kya hota hai?

KYA. Ek ray woh patla seedha arrow hai jo dikhata hai ki light energy kis direction mein travel karti hai. Hum do points rakhte hain: ek start point $A$ jahan se light nikalti hai, aur ek end point $B$ jahan use pahunchna hai. Light ko $A$ se $B$ tak jana hai.

KYUN. "Least time" kehne se pehle hume ek trip ka time measure karna aana chahiye. Time bas distance divided by speed hai. Agar ek seedhe segment ki length $ℓ$ hai aur light $v$ speed se move karti hai, toh use cross karne ka time hai

$t = \frac{ℓ}{v}$

$ℓ$ ::: path ke seedhe tukde ki length (metres mein).
$v$ ::: us material mein light ki speed (metres per second mein).
$t$ ::: light ka us tukde par time (seconds mein).

PICTURE. Ek point $A$ , ek point $B$ , unke beech ek seedha arrow, jis par uski length aur speed likhi hai. Yahi aage aane wali har cheez ka atom hai.

Step 2 — Ek medium: least time = seedhi line

KYA. Maano poora space ek uniform material hai, toh speed $v$ har jagah same hai. $A$ se $B$ tak ke sabhi possible tedhe-medhe raston mein se kaun sa sabse kam time leta hai?

KYUN. Kyunki $v$ constant hai, $t = (total length) / v$ . $t$ ko chhota karna matlab length ko chhota karna. Do points ke beech sabse chhoti length seedhi line hoti hai — toh light seedhi jaati hai. Yahi rectilinear propagation hai, aur hum ne ise assume karne ki jagah derive kiya hai.

PICTURE. $A$ se $B$ tak teen candidate paths: do tedhe wale (lambe, slow) aur ek seedha wala (sabse chhota, fastest). Seedha wala jeetta hai.

Step 3 — Do mediums: crossing set up karo

KYA. Ab space ko ek horizontal boundary ( $x$ -axis) se do hisson mein baanto. Uske upar light $v_{1}$ speed se chalti hai; neeche alag speed $v_{2}$ se. Point $A = (0, a)$ upar baith hai; point $B = (d, - b)$ neeche. Light boundary cross karti hai kisi point $P = (x, 0)$ par jo hum choose karte hain.

KYUN. Yahi refraction ka asli sawaal hai: crossing point $P$ hi woh ek aazaadi hai jo light ke paas hai. $P$ ko left ya right khiskaao aur tum upar ki thodi journey ko neeche ki thodi journey se trade karte ho. Hum woh $P$ chahte hain jo total time sabse chhota kare.

$a$ ::: $A$ boundary ke kitna upar hai (ek fixed positive number).
$b$ ::: $B$ boundary ke kitna neeche hai (ek fixed positive number).
$d$ ::: $A$ aur $B$ ke beech horizontal gap.
$x$ ::: crossing point $P$ ki horizontal position — woh ek cheez jo hum vary karte hain.

PICTURE. Do-tone scene: top strip speed $v_{1}$ , bottom strip speed $v_{2}$ , slanted top leg $A \to P$ , slanted bottom leg $P \to B$ , aur movable dot $P$ sirf allowed stretch $0 \leq x \leq d$ par slide karta hua.

Step 4 — Total time ko ek formula mein likho

KYA. Har leg ek seedhi line hai, toh uski length ek right triangle ki hypotenuse hai (Pythagoras). Hum poochte hain kyun Pythagoras aur kuch nahi: kyunki legs hain hi seedhi (Step 2), aur ek seedhe segment ke horizontal aur vertical offsets ek right triangle ki do legs hain — uski length $(horizontal)^{2} + (vertical)^{2}$ hai.

Top leg $A \to P$ : horizontal offset $x$ , vertical offset $a$ , toh length $x^{2} + a^{2}$ . Bottom leg $P \to B$ : horizontal offset $d - x$ , vertical offset $b$ , toh length $(d - x)^{2} + b^{2}$ .

KYUN. Har length ko us medium ki speed se divide karo us leg ka time paane ke liye, phir add karo:

$T (x) = medium 1 mein time \frac{x ^{2} + a ^{2}}{v _{1}} + medium 2 mein time \frac{( d - x ) ^{2} + b ^{2}}{v _{2}} (0 \leq x \leq d)$

$x^{2} + a^{2}$ ::: top leg ki length ( $A$ se $P$ tak).
$(d - x)^{2} + b^{2}$ ::: bottom leg ki length ( $P$ se $B$ tak).
$v_{1}$ , $v_{2}$ se divide karna ::: har length ko time mein badalta hai, kyunki $t = ℓ / v$ .
$T (x)$ ::: total travel time ek function ke roop mein ki hum $P$ kahan rakhte hain, allowed stretch $0 \leq x \leq d$ par valid.

PICTURE. Do right triangles explicitly drawn, unke legs $x$ , $a$ aur $d - x$ , $b$ label kiye hue, aur dono hypotenuses highlight kiye hue.

Step 5 — "Least time" ka matlab hai slope zero hai

KYA. $T (x)$ ek smooth curve hai: chhote $x$ ke liye top leg chhoti hai par bottom leg lambi hai, bade $x$ ke liye ulta. Kahin beech mein $T$ ek minimum tak gir jaata hai. Valley ke bilkul neeche curve momentarily flat hoti hai.

KYUN yeh tool — derivative. Derivative $\frac{d T}{d x}$ ek machine hai jo har $x$ par $T$ ki slope report karta hai: agar tum $P$ ko thoda sa right nudge karo toh time kitni tezi se change hota hai. Hum yahi tool use karte hain aur koi nahi kyunki sawaal "time sabse chhota kahan hai?" exactly sawaal hai "slope zero kahan hai?" Minimum par, $P$ ko kisi bhi taraf nudge karna time nahi badalta (first order tak), toh

$\frac{d T}{d x} = 0.$

PICTURE. $T$ ka graph $x$ ke against allowed stretch $0 \leq x \leq d$ par (stretch ke dono ends marked hain): ek U-shaped valley, jiske lowest point par ek flat tangent line kiss kar rahi hai. Khaas baat yeh hai ki lowest point strictly $x = 0$ aur $x = d$ ke beech hai, toh yeh ek genuine physical crossing hai, endpoint nahi. Left taraf slope negative hai (downhill), right taraf positive (uphill); yeh exactly winning $P$ par zero se guzarta hai.

Step 6 — Differentiation karo

KYA. $T (x)$ ko term by term differentiate karo. $x^{2} + a^{2}$ ka derivative $\frac{x}{x ^{2} + a ^{2}}$ hai (chain rule), aur $(d - x)^{2} + b^{2}$ ka derivative $\frac{- ( d - x )}{( d - x ) ^{2} + b ^{2}}$ hai (extra minus andar ke $- x$ se aata hai).

KYUN. Poori slope ko zero set karo:

$\frac{d T}{d x} = \frac{x}{v _{1} x ^{2} + a ^{2}} - \frac{d - x}{v _{2} ( d - x ) ^{2} + b ^{2}} = 0$

Doosre term ko paar le jaao:

$\frac{x}{v _{1} x ^{2} + a ^{2}} = \frac{d - x}{v _{2} ( d - x ) ^{2} + b ^{2}}$

$\frac{x}{x ^{2} + a ^{2}}$ ::: top triangle ka horizontal offset ÷ hypotenuse.
$\frac{d - x}{( d - x ) ^{2} + b ^{2}}$ ::: bottom triangle ka horizontal offset ÷ hypotenuse.
$\frac{1}{v _{1}}$ , $\frac{1}{v _{2}}$ ::: har medium ki "time-cost per metre" — woh factors jo ise seedhi line nahi rehne dete.

PICTURE. Phir se do triangles, par ab horizontal leg over hypotenuse ratio har ek par highlight kiya hua — ek bada coloured wedge jo exactly equation ke fractions ko mark karta hai.

Step 7 — Sines pehchano (angles chhupi hui thi har waqt)

KYA. Normal draw karo — $P$ par boundary ke perpendicular vertical dashed line. Top leg aur normal ke beech ka angle $θ_{1}$ hai; bottom leg aur normal ke beech $θ_{2}$ .

KYUN yeh tool — sine. Ek right triangle mein angle ka sine hota hai opposite over hypotenuse. Top triangle dekho: $θ_{1}$ ke opposite wali side (us angle ke aage wali side) horizontal offset $x$ hai, aur hypotenuse $x^{2} + a^{2}$ hai. Toh

$sin θ_{1} = \frac{x}{x ^{2} + a ^{2}}, sin θ_{2} = \frac{d - x}{( d - x ) ^{2} + b ^{2}} .$

Yahi exactly Step 6 ke do ugly fractions hain! Sine yahan perfect tool hai kyunki yeh woh ek ratio hai jo "kitna $P$ sideways khiska" ko "ray normal se kitni slanted hai" mein convert karta hai. Substitute karo:

$\frac{sin θ _{1}}{v _{1}} = \frac{sin θ _{2}}{v _{2}}$

Ab refractive index ki definition use karo, $n = c / v$ . Yahan $c$ speed of light in vacuum hai — ek fixed universal constant, lagbhag $3 \times 1 0^{8} m/s$ , sabse fast speed jo koi cheez ja sakti hai. Kisi bhi material mein light slower hoti hai, speed $v < c$ par, toh $n = c / v$ hamesha $\geq 1$ hai; $n$ jitna bada, light utni slow. $n = c / v$ se hume milta hai $\frac{1}{v} = \frac{n}{c}$ . Boxed relation ke dono taraf $c$ multiply karo; $c$ cancel ho jaata hai:

$n_{1} sin θ_{1} = n_{2} sin θ_{2}$

$θ_{1}$ ::: aane wali ray ka normal se angle, medium 1 mein.
$θ_{2}$ ::: jaane wali ray ka normal se angle, medium 2 mein.
$c$ ::: vacuum mein light ki speed, ek universal constant ( $\approx 3 \times 1 0^{8}$ m/s).
$n_{1}, n_{2}$ ::: refractive indices; $n = c / v$ , bada $n$ = slow light.

PICTURE. Finished diagram: normal drawn, dono angles $P$ par marked, do sine-triangles shaded, aur Snell's law upar likha hua.

Step 8 — Fast → slow: ray TOWARD normal bend karti hai

KYA. Everyday case lo: light ek faster medium se ek slower medium mein ja rahi hai, yani $n_{1} < n_{2}$ (jaise air → glass, ya air → water). Snell's law $θ_{2}$ ke liye kya predict karta hai?

KYUN. $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ ko rearrange karo $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1}$ ke liye. Kyunki $n_{1} < n_{2}$ hai, factor $\frac{n _{1}}{n _{2}} < 1$ hai, toh $sin θ_{2} < sin θ_{1}$ , matlab $θ_{2} < θ_{1}$ : jaane wali ray kam slanted hai — yeh normal ki taraf bend karti hai. Yeh marching-band picture se match karta hai: row slow medium se takraati hai aur perpendicular ki taraf swing karti hai. Yeh least time se bhi match karta hai — light slow medium mein "kam metres" chahti hai, toh woh seedhe-neeche ke kareebi dive karti hai.

Worked number: air ( $n_{1} = 1$ ) → glass ( $n_{2} = 1.5$ ) par $θ_{1} = 30°$ . Toh $sin θ_{2} = \frac{1}{1.5} sin 30° = \frac{0.5}{1.5} = 0.3333$ , deता hai $θ_{2} \approx 19.47°$ . Waqai $19.47° < 30°$ : normal ki taraf bend. ✔

PICTURE. $θ_{1}$ par incoming ray fast (rarer) top medium mein aur chhote $θ_{2}$ par refracted ray normal ke paas slow (denser) bottom medium mein.

Step 9 — Baaki cases (koi scenario unshown mat chhodo)

KYA / KYUN / PICTURE, sab ek figure mein teen panels ke saath:

Panel A — same speed, $v_{1} = v_{2}$ (yani $n_{1} = n_{2}$ ). Snell's law deta hai $sin θ_{1} = sin θ_{2}$ , toh $θ_{1} = θ_{2}$ : ray seedhi guzarti hai, koi bend nahi. Yahi Step 2 general law ke andar chhupa hua hai — ek sanity check.
Panel B — head-on, $θ_{1} = 0$ . Ray normal ke saath aati hai. Toh $sin θ_{1} = 0 \Rightarrow sin θ_{2} = 0 \Rightarrow θ_{2} = 0$ : perpendicular enter karne wali light bilkul bend nahi karti, chahe do media kuch bhi hon. (Yeh slow zaroor hoti hai — speed badlti hai, direction nahi.)
Panel C — slow → fast ( $n_{1} > n_{2}$ ), Step 8 ka ulta. Ab $\frac{n _{1}}{n _{2}} > 1$ hai, toh $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} > sin θ_{1}$ , deता hai $θ_{2} > θ_{1}$ : ray normal se door bend karti hai. $θ_{1}$ badhao jab tak $θ_{2}$ $90°$ tak na pahunche — woh critical angle $θ_{c}$ hai jisme $sin θ_{c} = n_{2} / n_{1}$ . Glass→air ke liye, $sin θ_{c} = 1/1.5 = 0.6667$ , toh $θ_{c} \approx 41.81°$ . Usse aage, $sin θ_{2}$ 1 se zyada ho jaata, jo impossible hai, toh light nikal nahi sakti — woh totally internally reflect ho jaati hai (yahi Total internal reflection aur Optical fibres ko power deta hai).

Ek-picture summary

Ek figure poore safar ko compress karta hai: ek valley-shaped time curve (jiska minimum endpoints $x = 0$ aur $x = d$ ke beech baitha hai) jiska lowest point crossing point $P$ select karta hai; $P$ par bent ray apne normal aur do angles ke saath; aur boxed result $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ unhe ek saath baandhta hua — geometry aur calculus ek hi nazar mein.

Recall Feynman: poora walkthrough ek 12-saal ke bacche ko retell karo

Light ek runner hai jo sirf jaldi finish karne ki parwah karta hai, seedha jaane ki nahi. Ek flat field par (ek medium) sabse fast hai hi seedha, toh light seedhi jaati hai — yahi Step 1–2 hai. Ab field ke neeche ek swamp daalo: runner grass par daurta hai ( $v_{1}$ ) aur swamp mein dheere wade karta hai ( $v_{2}$ ). Agar woh bilkul seedha aim kare toh swamp mein bahut zyada wallowing ho jaaye; agar woh pehle swamp ke edge tak daude toh grass par distance waste ho. Start aur finish ke beech kahin — pehle ke pehle ya doosre ke baad kabhi nahi — woh sweet spot $P$ hai jahan total time sabse chhota hai (Steps 3–4). Use dhundhne ke liye hum poochte hain "kahan $P$ nudge karna help karna band kar deta hai?" — time-valley ka flat bottom, slope zero, derivative (Steps 5–6). Algebra clean karo, do messy fractions $sin θ_{1}$ aur $sin θ_{2}$ nikalta hai — up-down line se slants — aur speeds se divide karo ( $n = c / v$ ke saath, jahan $c$ empty space mein light ki top speed hai) toh Snell's law milta hai (Step 7). Special cases: grass→swamp woh seedhe-neeche ki taraf angle karta hai (Step 8, bend toward normal); equal ground → seedha; head-on → seedha andar; aur swamp se bahut steep bahar sprinit karna → woh nikal nahi sakta aur bounce back karta hai, total internal reflection (Step 9). Same runner, same rule — least time — har baar.

Connections

Fermat's principle — least-time idea jo yeh poora page unpack karta hai.
Total internal reflection, Optical fibres — Step 9, Panel C action mein.
Wavefronts and Huygens' principle — same bending ke peeche wave-level "kyun wavefront pivot karta hai".
Mirrors and Lenses, Dispersion — jahan Snell's law aage apply hota hai.