Aberrations — chromatic, spherical (concepts): Exercises

2.5.9 · D4 · Physics › Optics › Aberrations — chromatic, spherical (concepts)

Level 1 — Recognition

L1.1 — Kaun sa colour sabse paas focus karta hai?

Ek simple converging glass lens ko white light se illuminate kiya gaya hai. Blue, yellow, aur red light har ek axis par ek focus banate hain. Teeno focal distances ko lens se, sabse chhote se shuru karke rank karo.

Recall Solution

WHAT we use: $f \propto \frac{1}{n - 1}$ lensmaker's equation se. WHY: glass dispersive hota hai, isliye $n_{F} > n_{D} > n_{C}$ (blue index sabse bada, red sabse chhota). Bada $n$ matlab bada $(n - 1)$ , yani chhota $f$ . Answer: blue sabse paas, phir yellow, phir red sabse door: $f_{F} < f_{D} < f_{C} .$

L1.2 — Kaun sa defect hai yeh?

Har statement ke liye bolo ki yeh chromatic (C) ya spherical (S) aberration describe karta hai.

Single wavelength ke liye bhi appear hota hai.
$n$ ke $λ$ ke saath change hone se hota hai.
Aperture band karne se theek hota hai.
Achromatic doublet se theek hota hai.
Axis se door rays ke liye sabse zyada hota hai.

Recall Solution

S — spherical monochromatic hota hai (parent §2).
C — colour dependence hi chromatic ki definition hai.
S — stopping down marginal rays ko hata deta hai, jo SA ke offenders hain (colour split ko nahi).
C — doublet do dispersive powers ko balance karta hai.
S — SA ray height $h$ ke saath badhta hai; yeh ek marginal-ray effect hai.

Level 2 — Application

L2.1 — Longitudinal chromatic spread

Ek thin crown lens hai jiska $f = 200 mm$ hai, aur $n_{C} = 1.514$ , $n_{D} = 1.517$ , $n_{F} = 1.522$ hain. (a) Dispersive power $ω$ aur (b) red aur blue foci ke beech axial separation $f_{C} - f_{F}$ nikalo.

Recall Solution

(a) WHAT: $ω = \frac{n _{F} - n _{C}}{n _{D} - 1}$ apply karo. WHY: $ω$ $f$ ki per unit fractional focal spread hai. $ω = \frac{1.522 - 1.514}{1.517 - 1} = \frac{0.008}{0.517} = 0.01547.$ (b) WHAT: $\frac{f _{C} - f _{F}}{f} = ω$ use karo (parent ka boxed longitudinal result). WHY: yeh fractional spread ko ek real length mein convert karta hai. $f_{C} - f_{F} = ω f = 0.01547 \times 200 = 3.09 mm .$ Blue ka focus lens se ~3.1 mm paas hota hai red se.

L2.2 — Boxed formula se focal shift

$\frac{df}{f} = - \frac{d n}{n - 1}$ use karte hue, agar red→blue jaane par index $d n = + 0.006$ badhta hai, aur lens ka $n - 1 = 0.52$ aur $f = 150 mm$ hai, toh $df$ nikalo aur batao $f$ kis direction mein jaata hai.

Recall Solution

WHAT: boxed differential mein plug karo. WHY: yeh directly ek chhote index change ko ek chhote focal change se relate karta hai. $df = - f \frac{d n}{n - 1} = - 150 \times \frac{0.006}{0.52} = - 1.73 mm .$ Minus sign ka matlab hai $f$ chhota hota hai — bluer light paas focus karti hai, bilkul expected hai.

Level 3 — Analysis

L3.1 — Ek achromatic doublet design karo

Tum ek doublet banana chahte ho jiska total focal length $F = + 500 mm$ ho ( $P = + 2.0 D$ , $F$ metres mein lete hue). Crown glass ka $ω_{1} = 0.021$ hai, flint ka $ω_{2} = 0.038$ hai. Individual powers $P_{1}$ (crown) aur $P_{2}$ (flint) nikalo, aur unki focal lengths bhi.

Recall Solution

WHAT we set up: contact mein do thin lenses $P = P_{1} + P_{2}$ obey karte hain, aur achromatic condition hai $ω_{1} P_{1} + ω_{2} P_{2} = 0$ (parent ka boxed result). Dekho Achromatic Doublet. WHY do equations: hamare paas do unknowns $P_{1}, P_{2}$ hain; ek equation power fix karti hai, doosri colour correction fix karti hai. $P_{1} + P_{2} = 2.0, ω_{1} P_{1} + ω_{2} P_{2} = 0.$ Doosri se, $P_{2} = - \frac{ω _{1}}{ω _{2}} P_{1} = - \frac{0.021}{0.038} P_{1} = - 0.5526 P_{1}$ . Pehli mein substitute karo: $P_{1} (1 - 0.5526) = 2.0 \Rightarrow P_{1} = \frac{2.0}{0.4474} = + 4.470 D,$ $P_{2} = 2.0 - 4.470 = - 2.470 D .$ Focal lengths: $f_{1} = 1/ P_{1} = + 0.2237 m = + 223.7 mm$ (converging crown), $f_{2} = 1/ P_{2} = - 0.4049 m = - 404.9 mm$ (diverging flint). Sanity: $P_{1} > 0$ , $P_{2} < 0$ — opposite signs, jaisa $ω > 0$ force karta hai. ✓

L3.2 — Marginal rays jaldi kyun cross karte hain

$sin θ = θ - \frac{θ ^{3}}{6} + \dots$ expansion use karte hue, ek line mein explain karo: (a) kyun leading correction third-order hai, (b) kyun yeh marginal rays ko zyada bend karta hai, (c) blur aperture radius $h$ ke saath kaise scale karta hai.

Figure — Aberrations — chromatic, spherical (concepts)

Recall Solution

(a) $θ$ ke baad agla term $- θ^{3} /6$ hai — paraxial (small-angle) formula mein pehla neglected term, isliye "third-order" ya Seidel kehte hain. Koi $θ^{2}$ term nahi hai kyunki $sin$ odd function hai. (b) True $sin θ$ , $θ$ se chhota hota hai, isliye Snell's law linear formula se predict se thoda zyada bending axis ki taraf deliver karta hai. Marginal rays (bada $θ$ ) yeh sabse zyada feel karte hain aur axis jaldi cross karte hain → focus paas hota hai. (c) Kyunki $θ \propto h$ (ray height) Paraxial Approximation regime mein, $θ^{3}$ term $h^{3}$ ki tarah scale karta hai. Figure dekho: baahri (coral) rays near point par cross karti hain, andar ki (mint) rays paraxial focus par cross karti hain.

Level 4 — Synthesis

L4.1 — Aperture aur circle of least confusion

Lateral spherical-aberration blur radius $b = c h^{3}$ ki tarah badhta hai ek constant $c$ ke liye. Ek lens full aperture $h_{0}$ par $0.80 mm$ ka blur deta hai. (a) $h_{0} /2$ par kitna blur bachega? (b) $0.10 mm$ blur reach karne ke liye kitne factor se stop down karna hoga?

Recall Solution

(a) WHAT: cube law $b \propto h^{3}$ use karo. WHY: correction term $θ^{3} \sim h^{3}$ hai. $b (\frac{h _{0}}{2}) = 0.80 \times (\frac{1}{2})^{3} = 0.80 \times \frac{1}{8} = 0.10 mm .$ (b) Set karo $b / b_{0} = (h / h_{0})^{3} = 0.10/0.80 = 1/8$ , isliye $h / h_{0} = (1/8)^{1/3} = 1/2$ . Half radius tak stop down karo (ek aperture-radius halving = area terms mein teen f-stops).

L4.2 — Dono aberrations combine karo

White light L2.1 wale single crown lens se guzarti hai ( $f_{D} = 200 mm$ , $ω = 0.01547$ ) jisme bhi spherical aberration hai jo full aperture par longitudinal SA $= 4.0 mm$ deta hai. (a) On-axis kaun bada hai: chromatic spread $f_{C} - f_{F}$ ya longitudinal SA? (b) Agar aperture ko half radius tak stop karo, toh kya comparison badal jaata hai? Explain karo.

Recall Solution

(a) Chromatic $f_{C} - f_{F} = 3.09 mm$ (L2.1 se); longitudinal SA $= 4.0 mm$ . Full aperture par SA bada hai ( $4.0 > 3.09$ ). (b) WHAT badalta hai: longitudinal SA bhi $h^{2}$ ki tarah scale karta hai (yeh $\propto h^{2}$ hai jabki lateral blur $\propto h^{3}$ hai). $h$ ko half karne par $SA \to 4.0 \times (1/2)^{2} = 1.0 mm$ ho jaata hai. Chromatic spread aperture se nahi badalta (yeh ek focal-length shift hai, $h$ se independent), isliye yeh $3.09 mm$ rehta hai. Conclusion: stop down karne ke baad, chromatic ab dominate karta hai ( $3.09 > 1.0$ ). Isi liye fast lenses SA ko shape se aur colour ko doublet se fight karte hain — alag defects ke liye alag levers.

Level 5 — Mastery

L5.1 — Parabola sphere ko beat karta hai (numeric)

Ek concave mirror ka radius $R = 400 mm$ hai aur paraxial focal length $f = R /2 = 200 mm$ hai. Height $h = 60 mm$ par hit karne wali marginal ray ke liye, spherical mirror ka marginal focus paraxial focus se approximately itna short fall karta hai: $Δ \approx \frac{h ^{2}}{2 R} \cdot \frac{1}{2} = \frac{h ^{2}}{4 R} .$ (a) $Δ$ compute karo. (b) Batao ki ek paraboloid is on-axis parallel beam ke liye $Δ = 0$ kyun dega, aur woh ek aberration bhi cite karo jise yeh fix nahi karta.

Recall Solution

(a) WHAT: given longitudinal-SA estimate mein numbers plug karo. WHY: yeh spherical mirror ke focus ke leading $h^{2}$ falloff ko capture karta hai. $Δ = \frac{h ^{2}}{4 R} = \frac{6 0 ^{2}}{4 \times 400} = \frac{3600}{1600} = 2.25 mm .$ Marginal rays paraxial focus se 2.25 mm paas focus karti hain. (b) Ek paraboloid defined hai on-axis point at infinity se focus tak equal optical path ke locus ke roop mein, isliye sab axial parallel rays ek point par in phase pahunchte hain — zero SA, koi $θ^{3}$ error nahi. Lekin yeh coma fix nahi karta: off-axis parallel bundles abhi bhi comet-shaped blurs mein smear hote hain.

L5.2 — Full doublet design (open synthesis)

Ek cemented doublet design karo jiska focal length $F = + 300 mm$ ho aur jo $C$ aur $F$ lines ke liye achromatic ho. Diya hai: crown $ω_{1} = 0.020$ , flint $ω_{2} = 0.036$ . $P_{1}, P_{2}, f_{1}, f_{2}$ nikalo. Phir residual chromatic spread compute karo agar flint ka $ω_{2}$ ko $+ 5%$ se mis-specify kiya gaya ho (actually $0.0378$ ) — yaani kya doublet phir bhi nearly achromatic hai?

Recall Solution

Step 1 — target power. $P = 1/ F = 1/0.300 = 3.333 D$ . Step 2 — do conditions solve karo $P_{1} + P_{2} = 3.333$ , $ω_{1} P_{1} + ω_{2} P_{2} = 0$ : $P_{2} = - \frac{ω _{1}}{ω _{2}} P_{1} = - \frac{0.020}{0.036} P_{1} = - 0.5556 P_{1},$ $P_{1} (1 - 0.5556) = 3.333 \Rightarrow P_{1} = \frac{3.333}{0.4444} = + 7.500 D,$ $P_{2} = 3.333 - 7.500 = - 4.167 D .$ Focal lengths: $f_{1} = 1/7.500 = + 133.3 mm$ (crown), $f_{2} = 1/ (- 4.167) = - 240.0 mm$ (flint). Step 3 — sensitivity check. Residual colour power hai: $d P = ω_{1} P_{1} + ω_{2}^{'} P_{2} = 0.020 (7.500) + 0.0378 (- 4.167) = 0.1500 - 0.1575 = - 0.0075 D .$ Same power ke ek single crown lens se compare karo, jiska colour power $ω_{1} P = 0.020 \times 3.333 = 0.0667 D$ hoga. Doublet ka residual $∣ - 0.0075 ∣$ us ka lagbhag 11% hai — ek order of magnitude chhota. Conclusion: $5%$ glass error ke saath bhi doublet strongly (though not perfectly) achromatic hai. Real designs isse re-optimise karte hain ya is "secondary spectrum" ko khatam karne ke liye teesra element add karte hain.

Recall One-line self-test (reveal)

Stopping down spherical aberration ko cut karta hai lekin chromatic ko untouched chhodta hai — sach ya jhooth? ::: Sach; SA aperture height $h$ ke saath scale karta hai, chromatic ek $λ$ -driven focal shift hai jo $h$ se independent hai.