Visual walkthrough — Resolving power — Rayleigh criterion

2.5.16 · D2 · Physics › Optics › Resolving power — Rayleigh criterion

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Step 1 — "Point source" kya hota hai aur woh point kyun nahi rehta?

WHAT. Ek point source woh object hai jo itna door (ek star) ya itna chhota (slide par ek dot) hota hai ki jo bhi light hume chahiye woh ek hi jagah se aati hai. Woh light ek hole se bhejo — ek lens, ek pupil, ek telescope opening — aur tum expect kar sakte ho ki doosri taraf ek single bright point banega.

WHY it fails. Light ek wave hai. Ek wave jo hole ki edge se guzarti hai woh uske aas-paas mur jaati hai — yeh bending diffraction hai. Sirf ek infinitely wide hole hi wave ko bina chhue guzarne deta. Har real hole ki edges hoti hain, isliye har real image of a point thodi si smeared blob hoti hai, point nahi.

PICTURE. Neeche: ek flat wavefront (seedhi red line) ek hole of width $D$ par aati hai. Doosri taraf woh ab flat nahi hai — woh fan out karti hai. Hole jitna narrow, fan utna wide. Woh fanning resolution ki dushman hai.

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Step 2 — EK source ka diffraction pattern banao

WHAT. Woh fan-out light ek screen par girti hai. Woh uniform smear nahi banati — woh ek bright centre banati hai jiske aas-paas dark aur bright rings hoti hain. Yahi Airy pattern hai.

WHY rings. Wavefront ke alag-alag parts screen par kisi ek point tak pahunchne ke liye thoda alag-alag distance travel karte hain. Jahan woh distances ek canceling way mein puri wavelength se differ karti hain, waves destructively interfere karti hain aur darkness milti hai — ek dark ring. Dark rings ke beech, light bachti hai — bright rings.

PICTURE. Intensity (brightness) centre se angle $\theta$ ke against plot ki gayi hai. Uuncha central peak central maximum hai; pehli jagah jahan woh zero hit karta hai woh first minimum (pehli dark ring) hai, red mein mark ki gayi. Is page ki har cheez us red zero ki position par depend karti hai.

Us curve ki poori shape ke liye Airy disc and Bessel functions dekho.

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Step 3 — Pehle easy case mein ${\theta }_{\min }$ nikalte hain: slit

WHAT. Round hole tackle karne se pehle, aasaan slit — width $a$ ka lamba patla rectangular gap — solve karo. Hum uski pehli dark fringe ka angle chahte hain.

WHY the slit first. Slit one-dimensional hai, isliye hum haath se rays pair kar sakte hain. Ek baar yeh samajh lo ki zero kyun aata hai, circular case sirf usi idea ka ek mushkil version hai.

PICTURE. Slit ko top half aur bottom half mein baanto. Top half ki har ray ko exactly $a/2$ neeche wali ray se pair karo (red pair). Jab ek pair ke beech path difference half a wavelength, $\lambda /2$, ke barabar hoti hai, woh pair cancel ho jaati hai. Har pair saath mein cancel hoti hai ⇒ total darkness.

Ek pair ki ek ray jo extra distance travel karti hai woh hai $\frac{a}{2}\sin \theta$. Ise $\lambda /2$ set karo:

$\underset{\text{half-slit spacing}}{\underbrace{\frac{a}{2}}}\underset{\text{tilts path by this factor}}{\underbrace{\sin \theta }}=\underset{\text{half-wave}\to \text{cancel}}{\underbrace{\frac{\lambda }{2}}}$

Dono sides ko 2 se multiply karo:

$\underset{\text{slit width}}{\underbrace{a}}\underset{\text{direction to screen point}}{\underbrace{\sin \theta }}=\underset{\text{one wavelength}}{\underbrace{\lambda }}$

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Step 4 — Ise small angle mein convert karo

WHAT. $a\sin \theta =\lambda$ ko angle ke liye solve karo:

$\sin {\theta }_{\min }=\frac{\lambda }{a}.$

WHY approximate. Light ke liye, $\lambda$ (hundreds of nanometres) $a$ (millimetres) ke compared tiny hai, isliye ${\theta }_{\min }$ ek tiny angle hai. Tiny angles ke liye, $\sin \theta \approx \theta$ (Small angle approximation). Isse hum $\sin$ drop kar sakte hain aur angle directly likh sakte hain:

${\theta }_{\min }\approx \frac{\lambda }{a}(\text{slit}).$

PICTURE. $\sin \theta$ ka graph origin ke paas seedhi line $\theta$ se chipka hua — woh indistinguishable hain jahan hamare angles hain (red band). Yahi haara license hai ek ko doosre se swap karne ka.

Yahan har symbol:

• $\lambda$ — light ki wavelength (numerator: lambi wave ⇒ bada blur).
• $a$ — slit width (denominator: chauda slit ⇒ chhota blur).

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Step 5 — ROUND hole ke liye fix karo: 1.22

WHAT. Real lenses aur pupils circular hote hain, slit-shaped nahi. Ek strip ki jagah ek disc par ray-pairing dobara karne se sirf ek cheez badlti hai: ek numerical factor.

WHY 1.22. Slit ke liye pehla zero $\lambda /a$ par tha. Disc ke liye, light rings mein spread hoti hai aur pehla dark ring thoda aur door push ho jaata hai. Exact location woh hai jahan ek special function — Bessel function $J_1$ — pehli baar zero hit karta hai, aur yeh slit value ka $1.22$ times par hota hai:

${\theta }_{\min }=1.22\frac{\lambda }{D}(\text{circular aperture}).$

Yahan $D$ aperture size ke roop mein $a$ ki jagah leta hai (disc ka diameter).

PICTURE. Side by side: slit ki dark line vs disc ki dark ring. Red ring thoda bade angle par baith gayi hai — woh "thoda" hi poora $1.22$ hai.

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Step 6 — Ab DOOSRA source laate hain

WHAT. Hamare paas ek blurry disc hai. Ek doosra nearby source ek doosra identical disc banata hai, jo dono sources ke beech angle $\theta$ se shift hota hai. Resolution ka sawaal hai: dono discs kitni door honi chahiye jab main bataa sakoon ki do hain?

WHY a rule is needed. Koi ek physical instant nahi hai jahan "ek blob" "do" ban jaaye. Jaise discs alag hote hain unke beech ki dip dheere-dheere gehri hoti hai. Isliye hume ek agreed convention chahiye — ek fair line in the sand.

PICTURE. Teen panels: discs bahut close (ek hump), discs bas alag ho rahe (ek shallow red dip dikhta hai), discs kaafi alag (do clean peaks). Beech wala panel woh borderline hai jo hum abhi name karne wale hain.

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Step 7 — Rayleigh ki line in the sand batao

WHAT. Rayleigh ka rule: do sources just resolved hote hain jab ek ka central peak exactly doosre ki first dark ring par baith jaaye.

WHY here. Is exact spacing par dono patterns ka sum peaks se lagbhag $26\%$ neeche dip karta hai — chhota, lekin aankhon se detect hone wala. Aur paas aao aur dip gayab ho jaata hai; aur door ho aur yeh obvious hai. Ise isliye choose kiya gaya kyunki compute karna clean hai: hum pehle hi Steps 4–5 mein first-minimum angle nikal chuke hain.

Toh sabse chhoti resolvable separation us first-minimum angle ke barabar hai:

$\boxed{{\theta }_R=1.22\frac{\lambda }{D}}(\text{circular}),{\theta }_R=\frac{\lambda }{a}(\text{slit}).$

Ek ek term:

• ${\theta }_R$ — do sources ke beech sabse chhota angle jo abhi bhi alag bataya ja sake (radians).
• $1.22$ — round-hole Bessel factor.
• $\lambda$ — wavelength; bada ⇒ bura (bada ${\theta }_R$).
• $D$ — aperture diameter; bada ⇒ better (chhota ${\theta }_R$).

PICTURE. Peak-on-first-minimum, drawn: source A ka peak (black) precisely source B ke red first zero par land karta hua, aur resulting summed curve apni shallow dip ke saath.

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Step 8 — Edge aur degenerate cases (inhe kabhi skip mat karo)

WHAT. Formula ko uski extremes tak push karo aur check karo ki woh abhi bhi sensible hai.

PICTURE. Char labelled limits, ek panel each — red curve ${\theta }_R$ ko us variable ke against dikhata hai jo push ho raha hai.

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Step 9 — Microscope twist: $D$ se N.A. tak

WHAT. Ek paas ke object (microscope) ke liye hum sky mein angle nahi balki slide par do dots ke beech ki sabse chhoti distance $d$ ki parwah karte hain. Relevant aperture measure ban jaata hai half-angle $\beta$ of the light cone jo objective collect karta hai, times medium ka refractive index $n$:

$d_{\min }=\frac{1.22\lambda }{2n\sin \beta }=\frac{0.61\lambda }{\text{N.A.}},\text{N.A.}=n\sin \beta .$

WHY the cone, not the diameter. Ek paas ka object lens ko ek wide cone se subtend karta hai; woh lens us cone ka kitna part pakad sakta hai yahi detail ko limit karta hai. Gap ko oil se bharna ($n>1$) effective cone ko chaudha karta hai, isliye N.A. badhta hai aur $d_{\min }$ ghatta hai — finer detail. Numerical aperture dekho.

PICTURE. Slide se dekha gaya objective: red cone of half-angle $\beta$ lens mein ghusta hua, unke beech index $n$ ka medium.

Ek ek term:

• $\beta$ — collected cone ka half-angle.
• $n$ — medium ka refractive index (air $=1$, oil $\approx 1.5$).
• N.A. $=n\sin \beta$ — dono ko combine karta hai; bada ⇒ chhota $d_{\min }$.
• $0.61=1.22/2$ — wahi Bessel factor, two-sided cone geometry se half ho gaya.

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Ek-picture summary

Upar ki sari cheez, compressed: hole par fan-out → Airy pattern apni first dark ring ke saath → do overlapping patterns → peak-on-first-minimum → ${\theta }_R=1.22\lambda /D$.

Recall Feynman retelling — plain words mein vapas bolo

Light ek wave hai, aur har hole ki edges hoti hain, isliye light hamesha thodi si spread hoke nikalti hai — woh spread diffraction hai. Ek door ka dot isliye ek point ki jagah ek bright disc banata hai jiske aas-paas dark circles hoti hain (Airy pattern). Do dots do aisi discs banate hain. Inhe paas laao aur unke beech ki dip fill in hoti hai; ek point par tum yaqeen se nahi keh sakte ki do hain. Rayleigh ka fair rule kehta hai "just resolved" tab hai jab ek disc ka bright centre doosri disc ki first dark ring par seedha baith jaaye. Hum pehle se jaante the woh dark ring kahan hai: slit ke liye yeh $\lambda /a$ par hai, aur round hole ke liye Bessel maths ise $1.22\lambda /D$ tak nudge karta hai. Toh sabse chhota angle jo tum abhi bhi split kar sako woh hai ${\theta }_R=1.22\lambda /D$ — bada hole ya chhoti wave matlab chhota angle matlab teez aankhein. Microscope ke liye hum diameter ko collecting cone se swap karte hain, aur ${\theta }_R$ ban jaata hai $d_{\min }=0.61\lambda /\text{N.A.}$, isliye oil immersion, cone ko chaudha karke, tumhe finer cheezein dekhne deta hai.

Recall Quick self-check

Ek pattern ka peak doosre ki ...... par baith jaata hai. ::: first minimum (first dark ring) Slit first-minimum angle? ::: $\theta \approx \lambda /a$ Circular-aperture limit? ::: ${\theta }_R=1.22\lambda /D$ (radians) 1.22 kahan se aata hai? ::: Bessel function $J_1$ ka first zero Microscope limit N.A. mein? ::: $d_{\min }=0.61\lambda /\text{N.A.}$