Aberrations — chromatic, spherical (concepts): Worked examples

2.5.9 · D3 · Physics › Optics › Aberrations — chromatic, spherical (concepts)

Ye page aberrations topic ka drill ground hai. Parent note ne ideas build kiye; yahan hum har tarah ke case grind karte hain jo ek problem mein aa sakte hain. Har answer se pehle aap forecast karoge — pehle guess karo, phir khud check karo.

Jo kuch bhi chahiye woh parent note mein pehle se earn kiya hua hai. Kuch symbols baar baar aate hain, toh chaliye unhe ek ek line mein re-anchor karte hain:

Definition Woh symbols jo hum baar baar use karte hain

$f$ = focal length: thin lens se uss jagah ki distance jahan parallel rays milti hain. Zyada $f$ = kamzor bending. Converging lens ka $f > 0$ hota hai; diverging lens ka $f < 0$ .
$P = \frac{1}{f}$ = power: lens kitni strongly bends karti hai. Dioptres (D) mein measure hoti hai jab $f$ metres mein ho. Converging $\Rightarrow P > 0$ ; diverging $\Rightarrow P < 0$ .
$n$ = glass ka refractive index; $n_{C}, n_{D}, n_{F}$ red, yellow, blue light ke liye iske values hain.
$K = \frac{1}{R _{1}} - \frac{1}{R _{2}}$ = shape (curvature) constant: lens ke do surface radii $R_{1}, R_{2}$ se fixed hota hai. Yeh lensmaker's equation $\frac{1}{f} = (n - 1) K$ mein appear karta hai aur colour ke saath nahi badlta.
$θ$ = woh angle jo ek ray axis ke saath banati hai jab woh kisi surface se takrati hai; central rays ke liye chota, edge rays ke liye bada.
$ω = \frac{n _{F} - n _{C}}{n _{D} - 1}$ = dispersive power: glass rang kitna smear karta hai per unit bending. Hamesha positive. ( $C$ =red, $D$ =yellow, $F$ =blue Fraunhofer lines.)

Scenario matrix

Har aberration problem in cells mein se ek hoti hai. Neeche ke worked examples us cell ke saath tagged hain jo woh cover karti hai.

Cell	Case class	Kya tricky banata hai	Example
A	Basic chromatic: $f_{C} - f_{F}$ dhundo	$\frac{df}{f} = ω$ mein plug karo	Ex 1
B	$df$ ka sign (red vs blue), converging aur diverging	kaun sa colour nearer hai? kya sign flip hota hai?	Ex 2
C	Degenerate: $ω = 0$ (koi dispersion nahi)	zero-input — kya bachta hai?	Ex 3
D	Achromatic doublet design	opposite-sign powers, 2 equations solve karo	Ex 4
E	Limiting: $ω_{1} = ω_{2}$ (same glass)	kya achromatise kar sakte ho?	Ex 5
F	Spherical: aperture cube-law	blur $\propto h^{3}$ scaling	Ex 6
G	Spherical: circle of least confusion	screen kahan rakhein	Ex 7
H	Parabola vs sphere (limiting shape)	axis par zero SA, real-world	Ex 8
I	Exam twist: mixed — C hai ya S?	aberration diagnose karo	Ex 9

Cells A–E chromatic hain (colour matter karta hai). Cells F–H spherical hain (single colour, ray height matter karti hai). Cell I aapko unhe alag karne par force karti hai.

Cell A — Basic chromatic spread

Steps.

Dispersive power compute karo $ω = \frac{n _{F} - n _{C}}{n _{D} - 1} = \frac{1.526 - 1.517}{1.520 - 1} = \frac{0.009}{0.520}$ . Yeh step kyun? $ω$ exactly fractional focal spread hai — parent note ka boxed result $\frac{f _{C} - f _{F}}{f} = ω$ . Sab kuch isi par depend karta hai.
$ω = 0.009/0.520 = 0.017308$ (dimensionless). Kyun? Yeh $n$ -values ka ratio hai, isliye isme koi units nahi hain — yeh ek pure fraction hai.
$f$ se multiply karo: $f_{C} - f_{F} = ω f = 0.017308 \times 20 cm = 0.3462 cm \approx 3.46 mm$ . Kyun? $\frac{f _{C} - f _{F}}{f} = ω$ ko rearrange karne se actual distance milti hai.

Verify: Units check — $ω$ unitless hai, toh $ω f$ ko cm $f$ se milta hai. Magnitude check — ~1.7% fraction of 20 cm kuch mm hoga, jo forecast se match karta hai. Red ka focus ( $f_{C}$ ) farther hota hai; blue ka ( $f_{F}$ ) nearer hota hai, toh $f_{C} - f_{F} > 0$ . ✓

Cell B — Kaun sa colour nearer focus karta hai (dono lens signs)

Steps — Part (a) (converging).

Yaad karo $f \propto \frac{1}{n - 1}$ . Zyada $n$ ⇒ chota $f$ ⇒ focus lens ke nearer. Yeh step kyun? Yahi poori sign story hai: zyada bending meeting point ko andar kheenchti hai.
Kyunki $n_{F} > n_{D} > n_{C}$ (blue > yellow > red), focal lengths order hoti hain $f_{F} < f_{D} < f_{C}$ . Kyun? $(n - 1)$ ka reciprocal inequality ko flip karta hai: sabse bada $n$ sabse chota $f$ deta hai.
Toh blue sabse nearest focus karta hai, red sabse farther. Numerically, $\frac{df}{f} = - \frac{d n}{n - 1}$ use karke $f_{D} = + 20$ cm ke saath (yellow focal length humara reference $f$ hai):
- Blue ka yellow se offset: $Δ f_{D F} = f_{D} \cdot \frac{n _{F} - n _{D}}{n _{D} - 1} = 20 \times \frac{0.006}{0.520} = 0.2308 cm$ (blue focus $\approx 2.31$ mm lens ke nearer hai).
- Red ka yellow se offset: $Δ f_{D C} = f_{D} \cdot \frac{n _{D} - n _{C}}{n _{D} - 1} = 20 \times \frac{0.003}{0.520} = 0.1154 cm$ (red focus $\approx 1.15$ mm farther hai). Kyun? Formula har colour step ke liye shift deta hai; har colour-gap ek alag $d n$ hai. Yahan $f_{D}$ positive hai, toh chota $f_{F}$ literally lens ke real image side par closer baith ta hai.

Steps — Part (b) (diverging, $f_{D} = - 20$ cm).

Relation $∣ f ∣ \propto \frac{1}{n - 1}$ abhi bhi hold karti hai, toh blue ka abhi bhi sabse chota magnitude focal length hai: $∣ f_{F} ∣ < ∣ f_{D} ∣ < ∣ f_{C} ∣$ . Kyun? Dispersion glass par act karta hai, lens sign par nahi — blue hamesha sabse zyada bend hota hai.
Lekin ab $f < 0$ hai: foci virtual hain aur object ki taraf wali side par hain. Blue, jiska sabse chota $∣ f ∣$ hai, uska virtual focus lens ke nearest hai; red ka virtual focus farthest hai. Offset magnitudes Part (a) se identical hain: $∣Δ f_{D F} ∣ = 0.2308$ cm, $∣Δ f_{D C} ∣ = 0.1154$ cm — sirf unki lens ki side flip hoti hai. Kyun? $\frac{df}{f} = - \frac{d n}{n - 1}$ ek fractional rule hai; $f = - 20$ cm ke saath, $df$ $f$ ka sign le leta hai, toh virtual foci converging case ke mirror images ki tarah reorder hote hain.

Verify: Part (a): $Δ f_{D F} + Δ f_{D C} = 0.2308 + 0.1154 = 0.3462$ cm $= f_{C} - f_{F}$ Ex 1 se ✓. Part (b): magnitudes Part (a) se exactly match karte hain (0.2308 aur 0.1154 cm) ✓, aur rule "blue sabse zyada bent ⇒ blue focus lens ke nearest" dono signs ke liye hold karta hai — yeh bending ka statement hai, is baat ka nahi ki image kis side land hoti hai. ✓

Cell C — Degenerate input: koi dispersion nahi

Steps.

Dispersive power: $ω = \frac{n _{F} - n _{C}}{n _{D} - 1} = \frac{1.500 - 1.500}{1.500 - 1} = \frac{0}{0.5} = 0$ . Yeh step kyun? Numerator $n_{F} - n_{C}$ colour spread ka poora cause hai. Ise kill karo aur $ω \to 0$ .
Longitudinal aberration: $f_{C} - f_{F} = ω f = 0 \times 20 = 0$ . Kyun? Koi dispersion nahi ⇒ koi focal spread nahi ⇒ sab colours ek hi point par milte hain.
Lekin lens abhi bhi kaam karti hai! Upar define kiya hua shape constant $K = \frac{1}{R _{1}} - \frac{1}{R _{2}}$ use karke lensmaker's equation se, $\frac{1}{f} = (n - 1) K = (0.5) K$ nonzero hai, toh yeh abhi bhi focus karti hai. Kyun? Dispersion (colour) aur refraction (bending) alag hain: $ω = 0$ sirf $n$ ki colour ke saath variation ko remove karta hai, bending ko nahi — $K$ aur $(n - 1)$ dono abhi bhi nonzero hain.

Verify: Single-index medium non-dispersive hota hai — yeh paraxial ideal hai jise parent note ki "first lie" refer karti hai. $ω = 0$ ke saath literally koi chromatic aberration correct karne ke liye nahi hai; koi doublet ki zaroorat nahi. ✓

Cell D — Ek achromatic doublet design karo

Steps.

Do conditions likho.
- Net power: $P_{1} + P_{2} = P = 2$ .
- Achromatic condition: $ω_{1} P_{1} + ω_{2} P_{2} = 0$ (parent ke boxed result se). Yeh step kyun? Do unknowns ke liye do equations chahiye: ek strength fix karti hai, ek colour error kill karti hai.
Achromatic condition se, $P_{1} = - \frac{ω _{2}}{ω _{1}} P_{2} = - \frac{0.030}{0.018} P_{2} = - \frac{5}{3} P_{2}$ . Kyun? Rearrange karne se $P_{1}$ isolate hota hai; minus sign pehle se batata hai ki dono powers ka opposite sign hai.
$P_{1} + P_{2} = 2$ mein substitute karo: $- \frac{5}{3} P_{2} + P_{2} = 2 \Rightarrow - \frac{2}{3} P_{2} = 2 \Rightarrow P_{2} = - 3 D$ (flint diverges). Phir $P_{1} = 2 - (- 3) = + 5 D$ (crown converges). Kyun? Back-substitution linear system finish karta hai. Crown ko over-converge karna chahiye taaki flint subtract kar sake.

Verify: Net: $P_{1} + P_{2} = 5 + (- 3) = 2$ D ✓. Achromatic: $ω_{1} P_{1} + ω_{2} P_{2} = 0.018 \times 5 + 0.030 \times (- 3) = 0.090 - 0.090 = 0$ ✓. Crown converging, flint diverging — "ek negative hona chahiye" se match karta hai. ✓

Cell E — Limiting case: identical glasses

Steps.

Equal $ω$ ke saath achromatic condition: $ω P_{1} + ω P_{2} = ω (P_{1} + P_{2}) = 0$ . Yeh step kyun? Shared $ω$ factor out karne se woh expose hota hai jo force hota hai.
Kyunki $ω \neq = 0$ , hume $P_{1} + P_{2} = 0$ chahiye. Kyun? Nonzero factor product ko zero nahi bana sakta, toh bracket vanish hona chahiye.
Lekin $P_{1} + P_{2} = 0$ matlab net power $P = 0$ — ek useless flat combination, $+ 2$ D lens nahi. Kyun? Ek glass use karke colour error aur power dono nahi rakh sakte. Doublet ka poora point hai do alag $ω$ 's.

Verify: Design require karta hai $ω_{1} \neq = ω_{2}$ . Yahi wajah hai ki real achromats crown + flint pair karte hain (ek low- $ω$ aur ek high- $ω$ glass). Same-glass ⇒ sirf $P = 0$ hi dono conditions satisfy karta hai. ✓

Cell F — Spherical aberration: cube law

Steps.

Scaling law likho: $\frac{b}{b _{0}} = (\frac{h}{h _{0}})^{3}$ . Yeh step kyun? Spherical aberration $sin θ$ mein neglected $- θ^{3} /6$ term se aata hai (jahan $θ$ ray ka axis se angle hai, upar define kiya). Height $h$ par ek ray surface se $θ \propto h$ angle par hit karti hai, toh blur $\propto θ^{3} \propto h^{3}$ . Yahi cube action mein hai.
(a) $h = 5$ mm ⇒ $(\frac{5}{10})^{3} = \frac{1}{8}$ , toh $b = 0.80 \times \frac{1}{8} = 0.10 mm$ . Kyun? $h$ half karna cube karke one-eighth ho jaata hai — dramatic, gentle nahi.
(b) $h = 2$ mm ⇒ $(\frac{2}{10})^{3} = \frac{8}{1000} = 0.008$ , toh $b = 0.80 \times 0.008 = 0.0064 mm = 6.4 μ m$ . Kyun? One-fifth aperture tak stop karne par blur $125 \times$ drop karta hai — yahi wajah hai ki lenses $f /8$ par razor-sharp hote hain.

Figure — Aberrations — chromatic, spherical (concepts)

Figure (s01): black curve blur $b$ ko aperture radius $h$ ke against dikhata hai; yeh axis ke near flat hai aur large $h$ par steeply upar jaata hai — cube ka signature. Teen red dots hamare answers ko $h = 10, 5, 2$ mm par mark karte hain. Dhyaan do ki $h = 2$ par red dot floor ke paas kaise hai: almost saara blur sabse outer rays mein hota hai.

Cell G — Circle of least confusion

Steps.

Longitudinal SA: $L_{p} - L_{m} = 100.0 - 98.0 = 2.0 mm$ . Yeh step kyun? Yeh "edge focus" aur "centre focus" ke beech axial gap hai — smear ka size.
Marginal rays lens ke nearer cross karti hain (98 vs 100 mm), yeh confirm karta hai ki $- θ^{3} /6$ term unhe axis ki taraf bend karta hai, toh woh jaldi milti hain. Kyun? Direction-of-error check — extra bending crossing ko inward kheencha hai.
Best screen $\frac{3}{4}$ raaste par paraxial se marginal ki taraf: $L_{best} = L_{p} - \frac{3}{4} (L_{p} - L_{m}) = 100.0 - 0.75 \times 2.0 = 100.0 - 1.5 = 98.5 mm$ . Kyun? Na to koi focus sabse chota spot deta hai — tightest bundle (least confusion) beech mein hota hai, marginal side ke closer.

Figure (s02): solid black lines steep marginal rays hain jo axis ko 98.0 mm par early cross karti hain; dashed black lines shallow paraxial rays hain jo 100.0 mm par baad mein cross karti hain. Red vertical line 98.5 mm mark karti hai — circle of least confusion. Observe karo ki ray bundle red line par sabse narrow hai, na ki kisi bhi focus dot par: wahi screen rakho.

Cell H — Parabola sphere ko beat karta hai (limiting shape)

Steps.

Parabola define hota hai un points ke locus ke roop mein jo ek focus aur ek flat wavefront se equidistant (optical path mein) hain. Yeh step kyun? "Focus tak equal path" exactly woh condition hai ki sab rays in-phase aaye aur ek point par mile — koi $θ^{3}$ error nahi, kisi bhi ray height $h$ ke liye.
Ek sphere sirf apne vertex ke near parabola approximate karta hai; sphere ki sag $R - R^{2} - h^{2}$ expand karne se pata chalta hai ki sphere aur parabola order $h^{2}$ tak agree karte hain lekin order $h^{4}$ par differ karte hain. Kyun? $h^{4}$ mismatch exactly spherical-aberration term hai — yeh ray height $h$ ke saath badhta hai.
Numerical taste: $R = 200$ mm (radius of curvature) aur $h = 40$ mm (ray height) ke liye, sag difference hai $Δ z = \frac{h ^{4}}{8 R ^{3}} = \frac{4 0 ^{4}}{8 \times 20 0 ^{3}} = \frac{2 , 560 , 000}{64 , 000 , 000} = 0.04 mm .$ Kyun? Yeh $0.04$ mm surface error woh hai jo ek parabola remove karta hai aur sphere rakh ta hai — ek star blur karne ke liye kaafi hai.

Figure (s03): dono curves vertex par saath shuru hote hain (neeche). Black curve sphere hai, red curve parabola hai. Woh centre ke near overlap karte hain lekin black sphere large $h$ par aage nikal jaata hai; red double arrow $h = 40$ mm par 0.04 mm sag gap mark karta hai. Woh tiny gap ek sharp star aur ek fuzzy star ke beech ka poora farq hai.

Cell I — Exam twist: aberration diagnose karo

Steps.

Defect (i): coloured fringes, aperture se unchanged ⇒ chromatic aberration. Yeh wavelength ke saath focal-length shift hai, marginal-ray effect nahi, toh iris ise fix nahi kar sakta. Yeh step kyun? Diagnostic key: colour + aperture-independent ⇒ chromatic. Fix = achromatic doublet, chota stop nahi.
Defect (ii): monochromatic haze jo aperture ke saath shrinkarti hai ⇒ spherical aberration (ek Seidel third-order effect). Fix = stop down / best-form bending / aspheric. Kyun? Single-colour + strongly aperture-dependent ⇒ spherical.
$f /2.8$ se $f /5.6$ jaane par aperture diameter half ho jaata hai, toh aperture radius $h \to h /2$ . Cube law se, SA blur ban jaata hai $\frac{b _{new}}{b _{old}} = (\frac{h /2}{h})^{3} = (\frac{1}{2})^{3} = \frac{1}{8} = 0.125.$ Kyun? Ek full f-stop diameter mein factor-of-two change hai; $h^{3}$ law phir ise cube karke $1/8$ kar deta hai. Toh haze wide-open value ka 12.5% ho jaata hai — ek 8× improvement — jabki defect (i) ke colour fringes isi move se untouched rehte hain.

Verify: Reduction factor $= (1/2)^{3} = 0.125$ , yaani blur 12.5% tak drop hota hai ✓. Dono defects bilkul alag cures par respond karte hain — aperture spherical ko kill karta hai lekin chromatic ko nahi — exactly parent note ka key contrast. ✓

Recall Self-test — answers cover karo

Kaun se cell mein $ω = 0$ hai aur kya bachta hai? ::: Cell C (degenerate); refraction/power bachta hai ( $K$ aur $n - 1$ nonzero), chromatic aberration gayab ho jaata hai. Ek doublet mein kaun sa glass diverge karta hai — crown ya flint? ::: High- $ω$ flint diverge karta hai; low- $ω$ crown converge karta hai (Ex 4: $P_{2} = - 3$ D flint, $P_{1} = + 5$ D crown). Kya "blue focuses nearest" diverging lens ke liye survive karta hai? ::: Haan — blue hamesha sabse zyada bent hota hai, toh uska (virtual) focus lens ke nearest hota hai; sirf side flip hoti hai (Ex 2b). Aperture half karne par SA blur kitne factor se change hota hai? ::: $(1/2)^{3} = 1/8$ (cube law, Ex 6). Kya stopping down coloured fringes fix karta hai? ::: Nahi — chromatic aperture-independent hai; doublet chahiye (Ex 9). Circle of least confusion kahan hota hai? ::: Foci ke beech, marginal focus ke nearer (Ex 7: 98.5 mm).