Resolving power — Rayleigh criterion: Worked examples

2.5.16 · D3 · Physics › Optics › Resolving power — Rayleigh criterion

Yeh page Rayleigh criterion ki practice arena hai. Parent note ne formula build kiya; yahan hum har tarah ke question hit karte hain jo yeh generate kar sakta hai. Pehle ek map of all cases, phir worked examples — har ek us map cell ke saath labelled jo woh fill karta hai.

The scenario matrix

Is topic ka har problem in cells mein se ek hai. Kuch bhi skip mat karo.

#	Case class	Kya ise special banata hai	Example
A	Circular aperture, angle dhundho	Seedha $θ_{R} = 1.22 λ / D$	Ex 1
B	Angle → physical separation	Arc length $s = θ L$ use karo (small-angle)	Ex 2
C	Slit aperture (no 1.22)	Factor drop karo, width $a$ use karo	Ex 3
D	Microscope / N.A.	Near object → $d_{m i n} = 0.61 λ / N.A.$	Ex 4
E	Limiting / proportional reasoning	$λ$ aur $D$ badlo, koi numbers plug mat karo	Ex 5
F	Degenerate input ( $D \to \infty$ , $λ \to 0$ )	Resolution limit vanish ho jaati hai	Ex 6
G	Real-world word problem	Prose se $λ, D, L$ extract karo	Ex 7
H	Exam twist / trap	Diameter vs radius, degrees vs radians	Ex 8

Case A — Circular aperture, angle dhundho

Worked example Ex 1 (Cell A) — Ek satellite camera

Ek spy-satellite camera ka circular lens diameter $D = 0.30 m$ hai aur yeh green light $λ = 550 nm$ mein images leta hai. Do ground points ke beech sabse chhota angle kya hai jo yeh resolve kar sakta hai?

Forecast: Teri aankh (2 mm) se ~150× bada lens hai. Guess: kya $θ_{R}$ around $1 0^{- 6}$ rad hai? Usse bada ya chhota?

Step 1 — Formula choose karo. Circular aperture, distant objects ⇒ $θ_{R} = 1.22 λ / D$ . Yeh step kyun? Cell A plain angular-resolution case hai — koi slit nahi, koi N.A. nahi. Lens round hai, isliye $1.22$ ( $J_{1}$ ka pehla zero, dekho Airy disc and Bessel functions) zaroori hai.

Step 2 — Units consistent rakho. Dono lengths ko metres mein dalo: $λ = 550 \times 1 0^{- 9} m$ , $D = 0.30 m$ . Yeh step kyun? $λ / D$ dimensionless hona chahiye taaki answer radians mein aaye (angle ek pure number hai). nm ko m ke saath mix karna $1 0^{9}$ ka ek nonsense factor dega.

Step 3 — Plug in karo. $θ_{R} = 1.22 \times \frac{550 \times 1 0 ^{- 9}}{0.30} = 2.24 \times 1 0^{- 6} rad .$

Verify: Dimensionless ratio → radians ✓. Yeh teri aankh ke $3.4 \times 1 0^{- 4}$ rad se about $150 \times$ chhota hai, exactly diameter ratio $0.30/0.002 = 150$ . Consistent hai, kyunki $θ_{R} \propto 1/ D$ . ✓

Case B — Angle se ground par ek real distance tak

Worked example Ex 2 (Cell B) — Woh camera actually kya dekhta hai

Wahi camera ( $θ_{R} = 2.24 \times 1 0^{- 6} rad$ ) height $L = 250 km$ par orbit karta hai. Do objects ki sabse chhoti separation $s$ kya hai jo yeh ground par alag bata sakta hai?

Forecast: Metres? Centimetres? Compute karne se pehle guess karo.

Step 1 — Geometry draw karo. Ground par do points $s$ se separate hain, lens se $L$ door angle $θ_{R}$ subtend karte hain.

Step 2 — Small-angle arc relation use karo. Ek tiny angle ke liye, separation arc length hai: $s = θ_{R} L .$ Yeh step kyun? $θ_{R} \sim 1 0^{- 6}$ rad minuscule hai, isliye $\tan\theta\approx\sin\theta\approx\theta$ $1 0^{12}$ mein ek part se behtar hold karta hai. "Straight gap" $s$ aur "curved arc" $θ L$ yahan indistinguishable hain.

Step 3 — Plug in karo ( $L = 250 km = 2.5 \times 1 0^{5} m$ ): $s = 2.24 \times 1 0^{- 6} \times 2.5 \times 1 0^{5} = 0.559 m \approx 56 cm .$

Verify: rad $\times$ m $=$ m ✓ (radians dimensionless hain, isliye woh "disappear" ho jaate hain aur metres bacha lete hain — ek units sanity check ki tumne degrees nahi radians use kiye). ~0.5 m ground resolution aisi optics ke liye realistic hai. ✓

Case C — Circle nahi, slit (1.22 drop karo)

Worked example Ex 3 (Cell C) — Ek rectangular telescope aperture

Ek telescope ka aperture ek slit of width $a = 0.80 m$ tak masked hai (circle nahi). $λ = 600 nm$ observe karte hue, narrow direction ke along resolution limit dhundho.

Forecast: $1.22 λ / a$ jaisa? Ya alag? Kaunsa hai aur kyun?

Step 1 — Aperture shape pehchano. Ek slit ka pehla diffraction minimum $a sin θ = λ$ par baithta hai, jo $θ_{m i n} \approx λ / a$ deta hai — koi 1.22 nahi. Yeh step kyun? $1.22$ purely circular-geometry Bessel factor hai (dekho Single-slit diffraction). Ek slit circular nahi hai, isliye use apply karna classic trap hoga.

Step 2 — Plug in karo. $θ_{R} = \frac{λ}{a} = \frac{600 \times 1 0 ^{- 9}}{0.80} = 7.5 \times 1 0^{- 7} rad .$

Verify: Agar koi galti se $1.22$ use karta, toh woh $9.15 \times 1 0^{- 7}$ paata — ek 22% error. Slit ke liye bare $λ / a$ correct hai. ✓ Units: m/m → rad ✓.

Case D — Microscope, near object, numerical aperture

Worked example Ex 4 (Cell D) — Dry vs oil objective

Ek microscope objective ka half-angle $β = 6 4^{\circ}$ hai. Air mein ( $n = 1$ ), $λ = 500 nm$ . (i) Sabse chhoti resolvable detail $d_{m i n}$ dhundho. (ii) Ab immersion oil $n = 1.52$ add karo — recompute karo.

Forecast: Oil $d_{m i n}$ ko chhota (better) banana chahiye. Roughly $n$ values ke ratio se kitna — lagbhag utna hi?

Step 1 — Numerical aperture build karo. $N.A. = n sin β$ (dekho Numerical aperture). Yeh step kyun? Near object ke liye lens light ka ek cone collect karta hai; relevant "size" cone half-angle $β$ hai, diameter nahi. $sin β$ measure karta hai ki woh cone kitna wide khulta hai; $n$ ise scale karta hai kyunki oil zyada rays lens mein bend karta hai.

Step 2 — Air case. $sin 6 4^{\circ} = 0.8988$ , isliye $N.A. = 1 \times 0.8988 = 0.8988$ . $d_{m i n} = \frac{0.61 λ}{N.A.} = \frac{0.61 \times 500 \times 1 0 ^{- 9}}{0.8988} = 3.39 \times 1 0^{- 7} m \approx 339 nm .$

Step 3 — Oil case. $N.A. = 1.52 \times 0.8988 = 1.366$ . $d_{m i n} = \frac{0.61 \times 500 \times 1 0 ^{- 9}}{1.366} = 2.23 \times 1 0^{- 7} m \approx 223 nm .$

Verify: Do answers ka ratio $= 339/223 = 1.52 = n_{oil}$ ✓ — exactly $n$ factor, kyunki $β$ unchanged tha. Oil resolution improve karta hai. ✓

Case E — Proportional reasoning (koi numbers nahi)

Case F — Degenerate / limiting inputs

Worked example Ex 6 (Cell F) — Perfect-lens ka myth

Do limiting thought-experiments: (i) aperture ko bound ke bina badhne do, $D \to \infty$ ; (ii) wavelength ko zero tak shrink hone do, $λ \to 0$ . Har ek mein $θ_{R}$ kya approach karta hai?

Forecast: Kya blur kabhi fully vanish hota hai? Kaun sa limit physically reachable hai?

Step 1 — $D \to \infty$ lo. $θ_{R} = 1.22 λ / D \to 0$ . Yeh step kyun? Infinitely wide aperture wavefront se kuch bhi chop nahi karta, isliye diffract karne ke liye kuch nahi hai — parent note ka "chopping a wave = diffraction" logic reverse mein chalta hai. Perfect point, lekin $D = \infty$ unbuildable hai.

Step 2 — $λ \to 0$ lo. $θ_{R} = 1.22 λ / D \to 0$ . Yeh step kyun? Zero wavelength geometric-optics (ray) limit hai: waves jo wave nahi karti woh diffract nahi karti. Isliye electron microscopes ( $λ$ light se hazaaron times chhoti) optical ones ko beat karte hain.

Step 3 — Nonsense se bachao. $D = 0$ (koi aperture nahi) ya $λ = \infty$ deta hai $θ_{R} \to \infty$ : light sab angles par phailti hai, zero resolving power — tum kuch bhi image nahi karte.

Verify: Dono "good" limits $θ_{R} \to 0$ push karte hain (infinite resolving power $1/ θ_{R} \to \infty$ ), dono "bad" limits ise $\infty$ tak push karte hain. Formula har extreme par monotonically aur sensibly behave karta hai. ✓

Case G — Real-world word problem

Worked example Ex 7 (Cell G) — Drone se car number plate padhna

Ek camera-drone $L = 120 m$ par hover karta hai. Uski lens aperture $D = 8.0 mm$ hai, $λ = 550 nm$ . Number-plate characters ke strokes about $4 mm$ apart hote hain. Kya yeh plate padh sakta hai?

Forecast: Gut call — readable hai ya nahi?

Step 1 — Angular limit dhundho. $θ_{R} = 1.22 \times \frac{550 \times 1 0 ^{- 9}}{8.0 \times 1 0 ^{- 3}} = 8.39 \times 1 0^{- 5} rad .$ Yeh step kyun? "Kya yeh do features alag bata sakta hai?" exactly resolution question hai — Cell A finest angle deta hai.

Step 2 — Ground distance mein convert karo $L = 120 m$ par: $s_{m i n} = θ_{R} L = 8.39 \times 1 0^{- 5} \times 120 = 1.01 \times 1 0^{- 2} m \approx 10 mm .$ Yeh step kyun? Like ke saath like compare karo: smallest resolvable gap $s_{m i n}$ versus actual $4 mm$ stroke spacing.

Step 3 — Decide karo. $s_{m i n} \approx 10 mm > 4 mm$ ⇒ strokes resolution limit se closer hain ⇒ woh blur ho jaate hain ⇒ padh nahi sakte.

Verify: rad $\times$ m $=$ m ✓. 4 mm feature padhne ke liye tumhe $s_{m i n} \leq 4$ mm chahiye, yani $D \geq 8.0 \times (10/4) = 20 mm$ — ek bada lens. Physically sensible. ✓

Case H — Exam twist / classic trap

Worked example Ex 8 (Cell H) — Radius diya, degrees maanga

Ek telescope mirror ka radius $r = 1.25 m$ hai; $λ = 500 nm$ observe karo. Resolution limit arcseconds mein express karo. (Ek arcsecond ek degree ka $1/3600$ hota hai.)

Forecast: Yahan do traps chhupe hain — compute karne se pehle unhe spot karo.

Step 1 — Trap 1: radius nahi, diameter. Formula diameter $D = 2 r = 2.50 m$ use karta hai. Yeh step kyun? $θ_{R} = 1.22 λ / D$ — aperture ki full width wave ko diffract karti hai; ise halve karna (radius $r$ use karna) $θ_{R}$ ko double kar dega aur resolution halve kar dega.

Step 2 — Pehle radians mein compute karo. $θ_{R} = 1.22 \times \frac{500 \times 1 0 ^{- 9}}{2.50} = 2.44 \times 1 0^{- 7} rad .$ Pehle radians kyun? Formula hamesha radians output karta hai (yeh ek length ratio hai). Convert sirf end mein karo.

Step 3 — Trap 2: radians → arcseconds. $\frac{180}{π}$ se multiply karo (rad→deg) phir $3600$ se (deg→arcsec): $θ_{R} = 2.44 \times 1 0^{- 7} \times \frac{180}{π} \times 3600 = 0.0503 arcsec .$

Verify: $1 rad = 206265 arcsec$ , isliye $2.44 \times 1 0^{- 7} \times 206265 = 0.0503$ arcsec ✓ — match karta hai. 2.5 m mirror ke liye about $0.0 5^{''}$ sahi hai. ✓

Recall Kaun sa cell kaun sa hai? (self-test)

Camera resolution in radians ::: Cell A — plain $1.22 λ / D$ Ek angle se ground separation ::: Cell B — $s = θ_{R} L$ , small angle Slit-shaped aperture ::: Cell C — $λ / a$ , koi 1.22 nahi N.A. se microscope detail ::: Cell D — $0.61 λ / N.A.$ "Agar dono change ho jaayein toh?" ::: Cell E — ratio track karo, plug mat karo " $D \to \infty$ " ::: Cell F — $θ_{R} \to 0$ , perfect lekin unbuildable Radius diya / degrees maanga ::: Cell H — $D = 2 r$ use karo, last mein convert karo

Flashcards

Satellite lens D = 0.30 m, λ = 550 nm — angular resolution?

1.22 \times 550 \times 1 0^{- 9} /0.30 = 2.24 \times 1 0^{- 6}

rad.

250 km par, θ_R = 2.24×10⁻⁶ rad kya ground separation deta hai?

s = θ_{R} L = 2.24 \times 1 0^{- 6} \times 2.5 \times 1 0^{5} \approx 0.56

Slit aperture a = 0.80 m, λ = 600 nm — resolution?

λ / a = 7.5 \times 1 0^{- 7}

rad (koi 1.22 nahi).

Oil objective N.A. = 1.37, λ = 500 nm — sabse chhoti detail?

0.61 λ / N.A. \approx 223

nm.

λ aur D dono ek saath halve karo — θ_R par effect?

Unchanged; ratio λ/D preserve rehta hai.

2.5 m mirror, λ = 500 nm — resolution in arcsec?

≈ 0.050 arcsec (D = diameter use karo, last mein radians convert karo).