Aberrations — chromatic, spherical (concepts): Visual walkthrough

2.5.9 · D2 · Physics › Optics › Aberrations — chromatic, spherical (concepts)

Step 1 — Ek lens parallel rays ke bundle ke saath kya karta hai

KYA. Socho koi cheez bahut door se aa rahi ho — Suraj, koi tara. Uski rays ek bundle of parallel lines ki tarah aati hain. Ek converging lens unhe itna modhta hai ki sab ek hi point par cross kar jaayein.

YE YAHAN SE KYUN SHURU KARTE HAIN. Chromatic aberration ki poori kahaani is baare mein hai ki woh crossing point kahan baithta hai. Toh pehle use naam dena zaroori hai. Lens se us crossing point tak ki doori focal length hai, jise $f$ likhte hain. Crossing point ko focus kehte hain.

PICTURE. Figure mein, seedhi blue lines left se aati hain. Lens ke baad woh jhuk jaati hain aur axis par ek dot par milti hain. Lens se dot tak ki horizontal doori $f$ label hai.

Step 2 — Woh rule jo $f$ set karta hai: lensmaker's equation

KYA. Ek lens light ko kitni strongly modhta hai yeh do cheezein tay karti hain: uske glass surfaces ki shape, aur glass light ko kitna slow karta hai air ke muqable mein. Lensmaker's equation ise package karti hai:

$bending strength \frac{1}{f} = glass factor (n - 1) shape factor (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Jab bhi koi piece aaye, use main turant naam deta hoon:

$n$ = refractive index — ek simple number jo batata hai ki glass mein light air ke muqable mein kitni slow travel karti hai. Air ka $n = 1$ hota hai; glass ka lagbhag $n = 1.5$ . Dekho Dispersion and Refractive Index.
$(n - 1)$ = air ke upar ka excess. Agar glass bilkul air jaisa hota ( $n = 1$ ) toh yeh zero hota aur lens kuch bhi nahi karta — jo bilkul sahi hai.
$R_{1}, R_{2}$ = dono curved surfaces ke radii — har face kitna domed hai. Yeh pure geometry hai: ek baar glass grind ho jaaye, yeh kabhi nahi badlte.

EK LETTER MEIN COMBINE KYUN KARTE HAIN. Kyunki $R_{1}$ aur $R_{2}$ lens ban jaane ke baad frozen hote hain, hum unhe ek single shape factor mein lump kar lete hain:

$K = \frac{1}{R _{1}} - \frac{1}{R _{2}} (ek constant, grinder ne decide kiya)$

Ab equation beautifully simple hai:

$\frac{1}{f} = (n - 1) K$

PICTURE. Figure mein wahi lens do baar dikhta hai: left par ek mota lens (bada $K$ , chhota $f$ ), right par ek patla wala (chhota $K$ , bada $f$ ). Grinding ke baad sirf ek cheez $f$ badal sakti hai — woh hai $n$ — aur $n$ humein dhoka dene waala hai.

Step 3 — Chhupa hua jhooth: $n$ ek number nahi hai

KYA. White light colours ka ek mixture hai. Glass har colour ko equally slow nahi karta. Yeh blue ko red se zyada slow karta hai. Toh ek $n$ nahi hai, balki puri family hai: $n_{red} < n_{yellow} < n_{blue}$ .

YEH KYUN MATTER KARTA HAI. Step 2 par wapas dekho: $f$ depend karta hai $n$ par. Agar blue ke liye $n$ bada hai, toh blue ke liye $(n - 1)$ bada hai, toh blue ke liye $1/ f$ bada hai, toh $f$ blue ke liye chhota hai. Blue ka crossing point lens ke zyada paas hai. Wahi colour-se-alag-hona chromatic aberration hai.

PICTURE. Figure mein white light andar jaati hai; lens ke baad rays ek chhote spectrum mein fan ho jaati hain. Blue ray pehle axis cross karti hai (chhota $f$ ), red ray baad mein (bada $f$ ). Yellow beech mein baithti hai.

Step 4 — "Blue pehle focus karta hai" ko formula mein badalna: hum differentiate kyun karte hain

KYA. Hum jaanna chahte hain: agar $n$ ek tiny amount se nudge kare (red → blue), toh $f$ kitna shift karta hai? Yeh ek rate-of-change ka sawaal hai — "input change ke per output change" — aur is sawaal ke liye bilkul bana tool hai derivative $\frac{df}{d n}$ .

DERIVATIVE KYUN, DO NUMBERS PLUG IN KARKE SUBTRACT KYUN NAHI? Hum $f_{red}$ aur $f_{blue}$ alag-alag compute karke subtract kar sakte hain. Lekin derivative ek clean, general formula deta hai jo kisi bhi small colour change ke liye kaam karta hai, aur sign reveal karta hai (kis direction mein $f$ move karta hai) ek hi line mein. Change $n_{F} - n_{C}$ chhota hai (lagbhag $0.008$ ), toh ise tiny nudge $d n$ maanna honest hai.

Move, term by term. Step 2 se shuru karo aur $n$ ko vary hone do jabki $K$ frozen rehta hai:

$\frac{1}{f} = (n - 1) K$

Dono sides ka derivative $n$ ke respect mein lo. Left par, $\frac{d}{d n} (\frac{1}{f}) = - \frac{1}{f ^{2}} \frac{df}{d n}$ (woh minus sign puri kahaani hai — yahi se "closer" aata hai). Right par, $\frac{d}{d n} [(n - 1) K] = K$ kyunki $K$ constant hai:

$- \frac{1}{f ^{2}} df = K d n$

Ab $K$ ko $K = \frac{1}{( n - 1 ) f}$ se replace karo (sirf Step 2 se rearrange kiya), taaki geometry gayab ho jaaye aur sirf measurable cheezein bachein:

$- \frac{1}{f ^{2}} df = \frac{1}{( n - 1 ) f} d n$

Dono sides ko $- f$ se multiply karo ( $f$ ka ek power khatam hota hai, sign flip hota hai):

$\frac{df}{f} = - \frac{d n}{n - 1}$

Har symbol padhte hain:

$\frac{df}{f}$ = focal length mein fractional change (ek percentage-jaisi number).
$d n$ = do colours ke beech index mein change.
minus = $n$ mein upar matlab $f$ mein neeche. Blue ( $+ d n$ ) → chhota $f$ . Bilkul Step 3 ki picture, ab algebra mein.

PICTURE. Figure $f$ ko $n$ ke against plot karta hai: ek downward-sloping curve. Ek tiny step right ( $+ d n$ ) ek step down produce karta hai ( $- df$ ). Slope arrow red mein draw hai taaki "negative" cheekhta rahe.

Step 5 — Poori spread measure karna: dispersive power $ω$

KYA. Step 4 ne infinitesimal nudge handle ki. Ab hum total red-to-blue spread ke liye poochhte hain. Hum set karte hain $d n = n_{F} - n_{C}$ (blue index minus red index — poora colour gap) aur middle colour denominator ke liye use karte hain, $n - 1 \approx n_{D} - 1$ :

$\frac{f _{C} - f _{F}}{f} = \frac{n _{F} - n _{C}}{n _{D} - 1} \equiv ω$

Right-hand side ka apna naam hai, dispersive power $ω$ (dekho Dispersion and Refractive Index). Term by term:

numerator $n_{F} - n_{C}$ = blue aur red kitne door hain (glass ka spread).
denominator $n_{D} - 1$ = glass ki average bending strength.
$ω$ = spread ÷ strength = "yeh glass kitna rainbow-y hai, relative to kitna bend karta hai."

RATIO KYUN, SIRF $n_{F} - n_{C}$ KYUN NAHI. Ek strongly bending lens jisme small colour gap ho woh bura ya achha ho sakta hai ek weak lens ke muqable mein jisme bada gap ho — image ke liye jo matter karta hai woh gap hai relative to bending. Divide karne se ek pure number milta hai jo glasses ko fairly compare karta hai.

PICTURE. Do glass samples side by side: crown glass (chhota $ω$ , tight colours) aur flint glass (bada $ω$ , fanned colours). Same aata white ray, bahut alag fans.

Step 6 — Picture mein ilaaj: jhooth ko uski mirror image se cancel karo

KYA. Do lenses ko back-to-back rakho: ek converging crown ( $P_{1} > 0$ ) aur ek diverging flint ( $P_{2} < 0$ ). "Power" $P$ simply $1/ f$ hai — ek lens kitna strongly bend karta hai. Powers add hoti hain: total $P = P_{1} + P_{2}$ .

YEH KYUN KAAM KAR SAKTA HAI. Har lens colours ko apni amount se smear karta hai. Step 5 se, ek single lens ki colour-smear hai $d P = P ω$ (fractional spread $ω$ uski power se multiply). Agar hum dono smears ko equal aur opposite banayein, woh cancel ho jaayein jabki net bending bacha rahe.

Total colour-smear ko zero set karo:

$d P = d P_{1} + d P_{2} = P_{1} ω_{1} + P_{2} ω_{2} = 0$

Term by term:

$P_{1} ω_{1}$ = crown lens ka colour error.
$P_{2} ω_{2}$ = flint lens ka colour error.
sum ko zero set karna = "dono rainbows ek doosre ko khatam kar dete hain."

Rearrange karo:

$ω_{1} P_{1} + ω_{2} P_{2} = 0 \Rightarrow \frac{P _{1}}{P _{2}} = - \frac{ω _{2}}{ω _{1}}$

EK LENS DIVERGE KYUN KARNA CHAHIYE. Dono dispersive powers $ω_{1}, ω_{2}$ positive hain (glass kabhi colours ko un-spread nahi karta). $ω_{1} P_{1} = - ω_{2} P_{2}$ hold karne ke liye, $P_{1}$ aur $P_{2}$ ke opposite signs hone chahiye — ek converge kare, ek diverge. Yahi hai Achromatic Doublet.

PICTURE. Crown blue-below-red bend karta hai; flint (flipped) red-below-blue matching amount se bend karta hai. Pair ke baad blue aur red rays dobara parallel hokar ek focus par milti hain.

Recall Quick self-check

Hum flint ko thoda sa diverge kyun nahi kar sakte? ::: Magnitudes pinned hain: $ω_{1} P_{1} = - ω_{2} P_{2}$ exact ratio fix karta hai. Thoda kam aur colour bachta hai; zyada aur woh over-correct ho jaata hai (colours doosre taraf cross karte hain).

Ek-picture summary

KYA compress karta hai. Ek frame mein sab kuch: white light ek single lens se takrati hai → colours split hote hain ( $f_{blue} < f_{red}$ , Step 4 ke minus sign se) → spread $ω$ se measure hoti hai (Step 5) → ek second, diverging lens ise undo karta hai (Step 6). Causes ki chain left se right chalti hai.

Recall Feynman retelling — poora walkthrough simple words mein

Lens ka kaam hai parallel sunlight pakad ke ek dot mein daalna; us dot tak ki reach focal length $f$ hai (Step 1). Lens kitna hard pull karta hai yeh uski shape aur glass ki light-slowing capacity se set hota hai — ek number $n$ (Step 2). Yahan sneaky part hai: glass blue light ko red se zyada slow karta hai (Step 3). Slow-more matlab bend-more matlab chhoti reach, isliye blue paas land karta hai aur red door — dot ek tiny smeared rainbow ban jaata hai. Exactly kitna, yeh kehne ke liye hum poochhte hain "reach kitni fast change hoti hai jab colour-number nudge karta hai upar?" — woh ek derivative hai, aur woh ek minus sign ke saath nikalta hai, jo math-speak hai "upar mein $n$ , neeche mein $f$ " ke liye (Step 4). Full red-to-blue nudge ko ek number mein bundle karne se milta hai dispersive power $ω$ : glass kitna rainbow-y hai compared to kitna bend karta hai (Step 5). Finally, trick: pehle lens ke peechhe ek diverging lens lagao jo extra-rainbow-y glass se bana ho. Woh colours ko opposite taraf utni hi amount se spread karta hai, dono rainbows ek doosre ko maarte hain, lekin net converging bachta hai — aur dot phir clean hai (Step 6).

Recall

Recall Khud test karo

Lensmaker's equation mein kaunsa term colour dependence carry karta hai? ::: $(n - 1)$ , kyunki $n = n (λ)$ . $\frac{df}{f} = - \frac{d n}{n - 1}$ mein, minus sign physically kya matlab hai? ::: Bada index (redder→bluer) chhota focal length deta hai — blue pehle focus karta hai. $ω$ ek ratio kyun hai, sirf $n_{F} - n_{C}$ kyun nahi? ::: Colour spread ko overall bending strength $(n_{D} - 1)$ ke against compare karne ke liye — ek pure, fair number. Achromat mein diverging lens kyun hona chahiye? ::: Dono $ω$ positive hain, isliye $ω_{1} P_{1} + ω_{2} P_{2} = 0$ force karta hai $P_{1}, P_{2}$ ke opposite signs hone ke liye.