Newton's rings — derivation: Worked examples — Hinglish deep dive

2.5.13 · D3 · Physics › Optics › Newton's rings — derivation

Ye Newton's rings ka drill page hai. Parent note mein physics build ki thi; yahan hum har type ke question ko hammer karte hain jo ek problem mein aa sakta hai. Har example se pehle tum Forecast karoge answer (pehle guess karo — isi se formula ki shape intuition ban jaati hai), phir step by step kaam karo, phir Verify karo back-plug karke aur units check karke.

Sab kuch in earned facts pe tika hai parent se:

Recall Jo facts hum use karte hain (dark AND bright geometry)

Air film mein total optical path difference: $Δ = 2 μ t + \frac{λ}{2}$ . 2 isliye hai kyunki ray B gap ki thickness $t$ se neeche aur upar jaati hai; $\frac{λ}{2}$ ek reflection pe Stokes flip hai.
Dark ring: $2 μ t = mλ$ . Bright ring: $2 μ t = (m - \frac{1}{2}) λ$ .
Geometry: $t = \frac{r ^{2}}{2 R}$ .
Dark-ring radius/diameter: $r_{m} = \frac{mλ R}{μ}$ , $D_{m}^{2} = \frac{4 mλ R}{μ}$ .
Bright-ring radius/diameter: $r_{m} = \frac{( 2 m - 1 ) λ R}{2 μ}$ , $D_{m}^{2} = \frac{4 ( m - \frac{1}{2} ) λ R}{μ} = \frac{2 ( 2 m - 1 ) λ R}{μ}$ .

Yahan $r$ ring ki radius hai (metres mein), $D = 2 r$ diameter hai, $m$ integer ring order hai, $λ$ light ki wavelength hai, $R$ lens ka radius of curvature hai, aur $μ$ jo bhi gap mein bhara ho uska refractive index hai.

Related tools jo tum open rakhna chahoge: Thin film interference, Stokes' relations & phase change on reflection, Refractive index measurement, Wedge fringes / air wedge.

Scenario matrix

Har Newton's-rings problem in cells mein se ek hai. Neeche ke examples un cells ke saath label kiye gaye hain jo ve hit karte hain, taaki milke poori table cover ho jaye.

Cell	Kya vary karta hai	Trap / skill jo test hoti hai
A. Direct radius	$r_{m}$ ya $D_{m}$ nikalna $m, λ, R$ se	kaun sa formula (dark vs bright), factor 2
B. Degenerate centre	$m = 0$ , $t = 0$	dark hai ya bright? Stokes flip decide karta hai
C. Solve for $λ$	measured $D_{n}, D_{m}$	difference method centre cancel karta hai
D. Solve for $R$	measured diameters, known $λ$	same difference formula rearrange karo
E. Liquid in gap	$μ \neq = 1$	rings shrink hoti hain $1/ μ$ se; $μ$ measure karo
F. Bright vs dark counting	consecutive rings, spacing	$r \propto m$ , crowding, spacing shrinks
G. Transmitted light	neeche se dekha	pattern invert hota hai: centre bright
H. Word / exam twist	glass plate replace, ya $t$ nikalo	physics padhna, sirf plug-in nahi

Example 1 — Cell A: dark ring ki direct radius

Worked example 10th dark ring

Monochromatic light $λ = 600 nm$ , lens $R = 0.80 m$ , air film ( $μ = 1$ ). 10th dark ring ki radius aur diameter nikalo.

Forecast: Kyunki $r_{m} = mλ R$ hai, roughly $10 \times 6 \times 1 0^{- 7} \times 0.8 \approx 5 \times 1 0^{- 6} \sim 2$ mm expect karo. Guess ~2 mm.

Step 1. $r_{m} = mλ R / μ$ choose karo. Ye step kyun? Air mein dark ring, toh condition $2 μ t = mλ$ aur $t = r^{2} /2 R$ milake exactly yahi deta hai. Step 2. Plug karo: $r_{10}^{2} = \frac{10 \times 600 \times 1 0 ^{- 9} \times 0.80}{1} = 4.8 \times 1 0^{- 6} m^{2}$ . Ye step kyun? Har quantity SI mein (metres), toh answer $m^{2}$ mein aata hai. Step 3. $r_{10} = 4.8 \times 1 0^{- 6} = 2.19 \times 1 0^{- 3} m = 2.19 mm$ . Ye step kyun? Step 2 se $r^{2}$ ka square root lo taaki actual radius mile; $m^{2} = m$ . Step 4. Diameter $D_{10} = 2 r_{10} = 4.38 mm$ . Ye step kyun? Measurable quantity diameter hoti hai (edge to edge), jo sirf radius ki double hai.

Verify: Units: $m \cdot m = m$ ✓. ~2 mm forecast se match ✓. Back-check thickness: $t = r^{2} /2 R = 4.8 \times 1 0^{- 6} /1.6 = 3.0 μ m$ , aur $2 t = 6.0 μ m = 10 \times 600 nm = 10 λ$ ✓.

Example 2 — Cell B: centre dark kyun hota hai (degenerate $m = 0$ )

Worked example Contact point

Exactly contact point pe air gap $t = 0$ hai. Reflected light mein centre bright hai ya dark? Path difference se dikhao.

Forecast: Zero gap lagta hai "in phase → bright." Bright guess karo — phir dekho kaise flip hota hai.

Step 1. Total path difference likho: $Δ = 2 μ t + \frac{λ}{2}$ . Ye step kyun? Stokes flip include karna zaroori hai; sirf geometric term ek trap hai. Step 2. $t = 0$ set karo: $Δ = 0 + \frac{λ}{2} = \frac{λ}{2}$ . Ye step kyun? $t = 0$ geometric part hata deta hai; sirf flip bachta hai. Step 3. Exactly aadhi wavelength ka path difference matlab dono waves out of phase pahunchti hain → destructive → dark. Ye step kyun? Aadhi wavelength ka path ek wave ka crest doosre ke trough pe shift kar deta hai, toh ve cancel hoti hain — yahi destructive interference ki definition hai.

Verify: Dark condition $2 μ t = mλ$ se consistency: $m = 0$ rakhne pe $t = 0$ milta hai, jo centre hai. Toh $m = 0$ hai ek dark ring — centre zeroth dark ring hai ✓. Guess galat tha; flip hi poori kahani hai.

Example 3 — Cell C: difference method se $λ$ measure karna

Worked example Do dark rings se wavelength

Dark-ring diameters $D_{20} = 4.60 mm$ aur $D_{4} = 1.80 mm$ measured; $R = 1.00 m$ , air. $λ$ nikalo.

Forecast: Visible light 400–700 nm hoti hai, toh kuch hundred nm expect karo.

Step 1. $λ = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) R}$ use karo. Ye step kyun? Har dark $D^{2}$ mein uncertain, deformed contact point se same additive error aata hai (tum kabhi exactly nahi jaante centre kahan hai). Dono subtract karne se wo common offset cancel ho jaata hai, ek clean linear relation bacha rehta hai — Stokes $\frac{λ}{2}$ already dark condition ke integer $m$ -indexing mein absorb ho jaata hai, toh wahan cancel karne ke liye kuch extra nahi bachta. Step 2. $D_{20}^{2} = (4.60 \times 1 0^{- 3})^{2} = 2.116 \times 1 0^{- 5} m^{2}$ ; $D_{4}^{2} = (1.80 \times 1 0^{- 3})^{2} = 3.24 \times 1 0^{- 6} m^{2}$ . Ye step kyun? $m^{2}$ mein kaam karo taaki final $λ$ metres mein mile. Step 3. Numerator $= 2.116 \times 1 0^{- 5} - 3.24 \times 1 0^{- 6} = 1.792 \times 1 0^{- 5} m^{2}$ . Ye step kyun? Ye difference physically meaningful, centre-independent quantity hai jo $4 (n - m) λ R$ ke barabar hai. Step 4. Denominator $= 4 (20 - 4) (1.00) = 64$ . Ye step kyun? $4 (n - m) R$ assemble karo — wo known constants jo difference ko scale karte hain. Step 5. $λ = \frac{1.792 \times 1 0 ^{- 5}}{64} = 2.80 \times 1 0^{- 7} m = 280 nm$ . Ye step kyun? Measured numerator ( $m^{2}$ ) ko constant ( $m$ ) se divide karo taaki $λ$ metres mein isolate ho.

Verify: Units: $m^{2} / m = m$ ✓. Numerically chhota hai lekin mechanics exact hain; real lab mein itne close diameters tumhe accuracy ke liye zyada rings apart use karne par majboor karenge. Sanity: $(n - m)$ badhane se answer chhota hota hai, toh wide ring separations tight $λ$ estimates dete hain ✓.

Example 4 — Cell D: lens ka $R$ measure karna

Worked example Radius of curvature

Sodium light $λ = 589 nm$ , air film ke saath: $D_{15}^{2} - D_{5}^{2} = 1.20 \times 1 0^{- 5} m^{2}$ . $R$ nikalo.

Forecast: Lab lenses ka $R$ order of 1 m hota hai. Guess ~1 m.

Step 1. Same difference law rearrange karo: $R = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) λ}$ . Ye step kyun? Ek equation, ek unknown — jo bhi symbol question hide kare uske liye solve karo. Step 2. $n - m = 15 - 5 = 10$ ; $4 (n - m) λ = 4 \times 10 \times 589 \times 1 0^{- 9} = 2.356 \times 1 0^{- 5} m$ . Ye step kyun? Denominator poora SI mein assemble karo taaki ratio metres de. Step 3. $R = \frac{1.20 \times 1 0 ^{- 5}}{2.356 \times 1 0 ^{- 5}} = 0.509 m$ . Ye step kyun? Measured $D^{2}$ -difference ( $m^{2}$ ) ko assembled constant ( $m$ ) se divide karo taaki $R$ isolate ho.

Verify: Units: $m^{2} / m = m$ ✓. Order ~0.5 m, lens ke liye sensible ✓. Cross-check: is $R$ se, $D_{5}^{2}$ aur $D_{15}^{2}$ ka difference $4 \times 10 \times 589 nm \times 0.509 = 1.20 \times 1 0^{- 5} m^{2}$ hoga — exactly input jaisa ✓.

Example 5 — Cell E: gap mein liquid hai ( $μ$ measure karo)

Worked example Liquid se rings shrink hoti hain

Same setup air mein dark-ring diameter $D_{10}^{air} = 4.00 mm$ deta hai. Ek transparent liquid daali jaati hai aur ab $D_{10}^{liq} = 3.20 mm$ hai. Liquid ka $μ$ nikalo.

Forecast: Rings chhhoti ho gayi → $μ > 1$ . Kyunki $D \propto 1/ μ$ hai aur ratio $3.2/4.0 = 0.8$ hai, expect karo $μ = 1/0. 8^{2} \approx 1.56$ .

Step 1. Kisi given ring $m$ ke liye: $D_{m}^{2} = \frac{4 mλ R}{μ}$ . Ye step kyun? $μ$ denominator mein hai, isliye denser medium fill karne pe same order ke liye kam thickness chahiye → chhota ring. Step 2. Dono cases mein same $m, λ, R$ hain, toh divide karo: $\frac{D _{air}^{2}}{D _{liq}^{2}} = \frac{μ _{liq}}{μ _{air}} = μ_{liq}$ (air $μ \approx 1$ ). Ye step kyun? Ratio lene se har wo quantity jo hum nahi jaante cancel ho jaati hai ( $m, λ, R$ ), sirf $μ$ bachta hai. Step 3. $μ = \frac{( 4.00 ) ^{2}}{( 3.20 ) ^{2}} = \frac{16.0}{10.24} = 1.5625$ . Ye step kyun? Do measured diameters substitute karo; units (mm) ratio mein cancel ho jaate hain, isliye $μ$ sahi se dimensionless hai.

Verify: Ratio $D_{liq} / D_{air} = 3.2/4.0 = 0.8 = 1/ 1.5625$ kyunki $1.5625 = 1.25$ ✓. Rings $μ > 1$ ke liye predict ki tarah shrink hui ✓. Ye hi refractive-index measurement method hai — dekho Refractive index measurement.

Example 6 — Cell F: spacing & crowding (bright rings, geometry)

Figure: bright-ring radius $r$ (vertical axis, proportional units) ring order $m$ (horizontal axis) ke against plot kiya. Blue curve $r \propto 2 m - 1$ follow karta hai. Pink double-arrow rings 1 aur 2 ke beech bada gap mark karta hai; yellow double-arrow rings 9 aur 10 ke beech bahut chhota gap mark karta hai — square-root curve bahar jaake flat hoti hai, isliye $m$ mein equal steps $r$ mein shrinking steps dete hain. Yahi Newton's rings ka crowding hai.

Worked example Gaps bahar ki taraf kyun shrink hote hain

Air film, $λ = 500 nm$ , $R = 1.00 m$ . 1st aur 2nd bright rings ki radius nikalo, phir 9th aur 10th bright rings ke beech ka gap. Dono gaps compare karo.

Forecast: $r \propto m$ , isliye $1 \to 2$ jump bada hona chahiye $9 \to 10$ jump se. Guess: outer gap chhota hai.

Step 1. Bright ring radius: $r_{m} = \frac{( 2 m - 1 ) λ R}{2 μ}$ , air $μ = 1$ . Ye step kyun? Bright $2 μ t = (m - \frac{1}{2}) λ$ use karta hai, yaani odd-multiple half condition; $t = r^{2} /2 R$ substitute karne pe ye radius milti hai. Step 2. $r_{1} = \frac{( 1 ) ( 500 \times 1 0 ^{- 9} ) ( 1 )}{2} = 2.5 \times 1 0^{- 7} = 5.00 \times 1 0^{- 4} m = 0.500 mm$ . Ye step kyun? $m = 1$ ke liye, $2 m - 1 = 1$ ; SI mein plug karo taaki root metres de. Step 3. $r_{2} = \frac{( 3 ) ( 500 \times 1 0 ^{- 9} ) ( 1 )}{2} = 7.5 \times 1 0^{- 7} = 8.66 \times 1 0^{- 4} m = 0.866 mm$ . Inner gap $= r_{2} - r_{1} = 0.366 mm$ . Ye step kyun? $m = 2$ ke liye, $2 m - 1 = 3$ ; consecutive radii ka difference hi un do rings ke beech visible spacing hai. Step 4. $r_{9} = \frac{17 \times 500 \times 1 0 ^{- 9}}{2} = 4.25 \times 1 0^{- 6} = 2.062 mm$ ; $r_{10} = \frac{19 \times 500 \times 1 0 ^{- 9}}{2} = 4.75 \times 1 0^{- 6} = 2.179 mm$ . Outer gap $= r_{10} - r_{9} = 0.117 mm$ . Ye step kyun? $m = 9, 10$ ke liye odd factors 17 aur 19 hain; is gap ko inner gap se compare karna crowding claim test karta hai. Figure dekho: curve centre ke paas steep hai, bahar flat.

Verify: Har jagah units: $m \cdot m = m$ har radius ke liye ✓, aur har gap metres ka difference hai, isliye bhi metres mein ✓. Outer gap $0.117 mm$ < inner gap $0.366 mm$ ✓ — rings bahar crowd karti hain, $r \propto m$ se match ✓. Guess confirm hua.

Example 7 — Cell G: transmitted light sab kuch invert karta hai

Worked example Neeche se dekhna

Reflected light ki jagah pattern transmitted light mein dekha jaata hai. Ab centre kya hai, aur dark-ring condition kya hai?

Forecast: Transmitted aur reflected patterns complementary hote hain (energy conservation), isliye centre bright hona chahiye.

Step 1. Transmission mein do interfering rays dono transmit hoti hain; koi bhi wo air→glass external-reflection $λ /2$ flip experience nahi karta jo reflected ray B ne ki thi. Ye step kyun? Net Stokes half-wave nahi hai, isliye geometric path difference $2 μ t$ akele khada rehta hai. Step 2. Bright (transmitted) jab $2 μ t = mλ$ ; dark jab $2 μ t = (m - \frac{1}{2}) λ$ . Ye step kyun? $\frac{λ}{2}$ hatane se reflected conditions swap ho jaati hain — bright aur dark jagah badal lete hain. Step 3. $t = 0$ pe: $2 μ t = 0 = 0 \cdot λ$ → bright centre. Ye step kyun? Koi flip nahi, toh zero path difference matlab truly in-phase → constructive → bright.

Verify: Reflected centre dark tha (Example 2); transmitted centre bright hai — complementary, jaisa energy conservation demand karta hai ✓. Dekho Interference — path difference & coherence ye jaanne ke liye ki dono views ko incident intensity tak kyun add karna chahiye.

Example 8 — Cell H: exam twist, air-gap thickness nikalo

Worked example Named ring pe thickness

Air film, $λ = 650 nm$ . 8th dark ring pe gap ki thickness $t_{8}$ hai. $t_{8}$ nikalo, aur ye batao ki ye geometric path ke kitne wavelengths correspond karta hai.

Forecast: $2 t = mλ$ toh $t = mλ /2 = 8 \times 650/2 nm = 2600 nm \approx 2.6 μ m$ .

Step 1. Dark condition, air: $2 μ t = mλ$ with $μ = 1$ → $t = \frac{mλ}{2}$ . Ye step kyun? Koi radius nahi chahiye — thickness directly order se follow hoti hai. Step 2. $t_{8} = \frac{8 \times 650 \times 1 0 ^{- 9}}{2} = 2.60 \times 1 0^{- 6} m = 2.60 μ m$ . Ye step kyun? SI mein $m = 8$ substitute karo; denominator mein 2 down-and-up round trip undo karta hai. Step 3. Geometric down-and-up path $= 2 t_{8} = 5.20 \times 1 0^{- 6} m = 8 \times 650 nm = 8 λ$ . Ye step kyun? Physical meaning confirm karta hai: round trip exactly 8 poori wavelengths hai, aur extra $\frac{λ}{2}$ flip total $8.5 λ$ banata hai → destructive ✓.

Verify: $2 t_{8} / λ = 5.20 \times 1 0^{- 6} /650 \times 1 0^{- 9} = 8.0$ exactly ✓. $t$ ki units: $nm$ / $μ m$ ✓. Micron-scale gap, optical films ke liye expected ✓.

Quick self-test

Kaun sa centre dark hai?

Reflected light — single Stokes

λ /2

flip

t = 0

Δ = λ /2

banata hai.

Rings shrink hoti hain ya grow karti hain jab liquid (

μ > 1

) gap fill kare?

Shrink, factor

1/ μ

se.

Radii ki jagah diameters kyun measure karte hain?

Offset/deformed contact point

D_{n}^{2} - D_{m}^{2}

mein cancel ho jaata hai.

Measurements se

R

nikalne ka formula?

R = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) λ}