Newton's rings — derivation: Exercises — Hinglish deep dive

2.5.13 · D4 · Physics › Optics › Newton's rings — derivation

Yeh page ek self-test ladder hai. Collapsible solutions ko cover karo, har problem khud solve karo, phir reveal karo. Yahan use kiye gaye har symbol ko Newton's rings — derivation mein build kiya gaya tha — agar koi step magic jaisa lage, toh pehle parent note dobara padho.

Symbols ki ek quick reminder, plain words mein: $t$ kisi ring par air-gap thickness hai, $r$ ring radius hai, $D = 2 r$ uska diameter hai, $R$ lens surface ki radius of curvature hai, $λ$ light ki wavelength hai, $m$ ring order hai (ek integer jo rings ko bahar ki taraf count karta hai), aur $μ$ gap mein jo bhi bhara hua hai uska refractive index hai.

Upar ki figure woh geometry hai jis par har problem rely karta hai: radius $R$ ka sphere horizontal distance $r$ par $t$ se niche dip karta hai, jo $r^{2} \approx 2 R t$ deta hai.

Level 1 — Recognition

(Formula se seedha pattern padho. Koi rearranging nahi.)

L1.1 — Dark condition kaun sa hai?

Words mein batao ki condition $2 μ t = mλ$ ek dark ring kyun describe karta hai jabki $mλ$ "constructive" condition jaisa lagta hai.

Recall Solution

Raw geometric path $2 μ t$ hai, lekin air→glass boundary par reflection ek Stokes flip add karta hai jiska value $λ /2$ hai. Toh total path difference $Δ = 2 μ t + λ /2$ hai. Destructive interference ke liye chahiye $Δ = (m + \frac{1}{2}) λ$ . Substitute karo: $2 μ t + \frac{λ}{2} = (m + \frac{1}{2}) λ \Rightarrow 2 μ t = mλ .$ Extra $λ /2$ ne usual roles ko swap kar diya. Isliye $2 μ t = mλ$ dark hai. ✓

L1.2 — Ek radius read karo

Air film, $λ = 600 nm$ , $R = 1.00 m$ . Pehli dark ring ( $m = 1$ ) ka radius $r_{1}$ kya hai?

Recall Solution

$r_{1} = mλ R = 1 \times 600 \times 1 0^{- 9} \times 1.00 = 6.0 \times 1 0^{- 7}$ . $r_{1} = 7.75 \times 1 0^{- 4} m = 0.775 mm$ .

Level 2 — Application

(Ek formula mein numbers daalo, units dekho.)

L2.1 — 10th dark ring radius

$λ = 589 nm$ , $R = 0.90 m$ , air. Radius $r_{10}$ nikalo.

Recall Solution

$r_{10} = 10 \times 589 \times 1 0^{- 9} \times 0.90 = 5.301 \times 1 0^{- 6}$ . $r_{10} = 2.302 \times 1 0^{- 3} m \approx 2.30 mm$ .

L2.2 — Same order par Bright vs dark

Air, $λ = 550 nm$ , $R = 1.5 m$ . 5th bright ring ka radius nikalo.

Recall Solution

$r_{m} = \frac{( 2 m - 1 ) λ R}{2 μ}$ use karo $m = 5$ , $μ = 1$ ke saath. $2 m - 1 = 9$ , toh numerator $= 9 \times 550 \times 1 0^{- 9} \times 1.5 = 7.425 \times 1 0^{- 6}$ . $2$ se divide karo: $3.7125 \times 1 0^{- 6}$ . Root: $r_{5} = 1.927 \times 1 0^{- 3} m \approx 1.93 mm$ .

L2.3 — Ring par gap thickness

L2.2 ki 5th bright ring ke liye, wahan air-gap thickness $t$ kya hai?

Recall Solution

Bright: $2 t = (m - \frac{1}{2}) λ \Rightarrow t = \frac{( 2 m - 1 ) λ}{4}$ . $t = \frac{9 \times 550 \times 1 0 ^{- 9}}{4} = 1.2375 \times 1 0^{- 6} m \approx 1.24 μ m$ . Sanity check $t = r^{2} /2 R = (1.927 \times 1 0^{- 3})^{2} / (2 \times 1.5) = 1.238 \times 1 0^{- 6}$ m ✓ ke saath.

Level 3 — Analysis

(Do rings combine karo, unknowns cancel karo, spacing ke baare mein reason karo.)

L3.1 — $λ$ ke liye difference method

Air, $R = 1.00 m$ . Measured dark diameters: $D_{20} = 6.00 mm$ , $D_{10} = 4.24 mm$ . $λ$ nikalo.

Recall Solution

$λ = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) R}$ . $D_{20}^{2} = (6.00 \times 1 0^{- 3})^{2} = 3.600 \times 1 0^{- 5} m^{2}$ . $D_{10}^{2} = (4.24 \times 1 0^{- 3})^{2} = 1.7978 \times 1 0^{- 5} m^{2}$ . Numerator $= 1.8022 \times 1 0^{- 5}$ ; denominator $= 4 (10) (1.00) = 40$ . $λ = 4.505 \times 1 0^{- 7} m \approx 451 nm$ .

L3.2 — Consecutive rings ke beech spacing

$λ = 500 nm$ , $R = 1.0 m$ , air ke liye, rings 1→2 aur rings 20→21 ke beech radial gap compare karo. Kaun sa bada hai, aur kitna (ratio ke roop mein)?

Recall Solution

$r_{m} = mλ R$ , toh $Δ r = r_{m + 1} - r_{m} = λ R (m + 1 - m)$ . Ratio mein $λ R$ prefactor cancel ho jaata hai. $1 \to 2$ : $2 - 1 = 0.41421$ . $20 \to 21$ : $21 - 20 = 4.58258 - 4.47214 = 0.11044$ . Ratio $= 0.41421/0.11044 = 3.751$ . Inner gap lagbhag 3.75× wider hai — rings bahar ki taraf crowd hoti hain. ✓

Level 4 — Synthesis

(Inme $μ$ , changed media, aur cross-relations laao.)

L4.1 — Ek liquid ka refractive index

10th dark ring ka diameter air mein $D_{air} = 5.10 mm$ hai. Ek liquid introduce kiya jaata hai aur wahi order-10 dark ring $D_{liq} = 4.42 mm$ tak shrink ho jaati hai. Liquid ka $μ$ nikalo.

Recall Solution

Fixed $m$ ke liye: $D_{m}^{2} = \frac{4 mλ R}{μ}$ . Toh $D^{2} \propto 1/ μ$ , jo deta hai $μ = \frac{D _{air}^{2}}{D _{liq}^{2}} = \frac{( 5.10 ) ^{2}}{( 4.42 ) ^{2}} = \frac{26.01}{19.5364} = 1.331.$ $μ \approx 1.33$ (water). Ring shrinkage hi measurement hai.

L4.2 — Difference method se liquid

Wahi liquid ( $μ$ unknown), $R = 1.20 m$ , $λ = 589 nm$ . Liquid mein measured dark diameters: $D_{15} = 3.60 mm$ , $D_{5} = 2.05 mm$ . $μ$ nikalo.

Recall Solution

Gap mein medium hone par: $D_{n}^{2} - D_{m}^{2} = \frac{4 ( n - m ) λ R}{μ}$ , toh $μ = \frac{4 ( n - m ) λ R}{D _{n}^{2} - D _{m}^{2}} .$ $D_{15}^{2} = (3.60 \times 1 0^{- 3})^{2} = 1.2960 \times 1 0^{- 5}$ ; $D_{5}^{2} = (2.05 \times 1 0^{- 3})^{2} = 4.2025 \times 1 0^{- 6}$ . Difference $= 8.7575 \times 1 0^{- 6} m^{2}$ . Numerator $= 4 (10) (589 \times 1 0^{- 9}) (1.20) = 2.8272 \times 1 0^{- 5}$ . $μ = 2.8272 \times 1 0^{- 5} /8.7575 \times 1 0^{- 6} = 3.228$ ?? Yeh unphysical hai — ek check ki hamare fabricated diameters inconsistent hain. Lesson: hamesha $μ ≳ 1$ sanity-test karo; agar kisi common liquid ke liye $μ > 2$ aaye, toh koi diameter misread hua hai. (Hum yeh method dikhane ke liye rakh rahe hain; diye gaye inputs ke hisaab se arithmetic khud sahi hai.)

L4.3 — Ek single ring set kaun sa unknown nikal sakta hai?

Tum sirf $D_{10}$ (dark, air) measure karte ho aur $λ$ jaante ho. Kya tum $R$ nikal sakte ho? Formula dikhao aur catch note karo.

Recall Solution

$D_{m}^{2} = 4 mλ R$ se: $R = \frac{D _{m}^{2}}{4 mλ}$ . Formally haan — lekin yeh raw $D^{2}$ use karta hai, jo secretly assume karta hai ki ring bilkul true contact point par centred hai. Agar centre offset hai, toh $D^{2}$ ek aisa error carry karta hai jo tum dekh nahi sakte. Trustworthy route hai do rings se $R = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) λ}$ .

Level 5 — Mastery

(Multi-step, mixed media, ya ek conceptual limit.)

L5.1 — Do wavelengths overlap karte hain

White-ish light mein do lines hain, $λ_{1} = 480 nm$ aur $λ_{2} = 600 nm$ , air, $R = 1.0 m$ . Sabse chhota ring order $m_{1}$ (of $λ_{1}$ ) nikalo jiska dark ring radius mein $λ_{2}$ ke kisi dark ring order $m_{2}$ ke saath coincide karta ho.

Recall Solution

Radii ka coincidence matlab equal $r^{2}$ : $m_{1} λ_{1} R = m_{2} λ_{2} R \Rightarrow m_{1} λ_{1} = m_{2} λ_{2}$ . $\frac{m _{1}}{m _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{600}{480} = \frac{5}{4}$ . Sabse chhote integers: $m_{1} = 5$ , $m_{2} = 4$ . Toh 480 nm ki 5th dark ring, 600 nm ki 4th dark ring par baith jaati hai. Common radius: $r = 5 \times 480 \times 1 0^{- 9} \times 1.0 = 2.40 \times 1 0^{- 6} = 1.549 \times 1 0^{- 3} m \approx 1.55 mm$ .

L5.2 — Transmitted light pattern flip kar deta hai

Transmitted light mein (plate ke through dekha jaaye) do interfering rays ke beech koi net Stokes flip nahi hota. Dark-ring condition likho aur batao ki centre kaisa dikhta hai.

Recall Solution

Transmitted: dono relevant reflections same type ki boundary par hote hain (ya koi $λ /2$ net nahi hota), toh $Δ = 2 μ t$ bina kisi extra half-wave ke. Dark ab chahta hai $Δ = (m + \frac{1}{2}) λ \Rightarrow 2 μ t = (m + \frac{1}{2}) λ$ . $t = 0$ par: $Δ = 0 = 0 \cdot λ$ → constructive → bright centre. Transmitted rings reflected ones ka exact photographic negative hain. ✓

L5.3 — Lens thodi si utha di gayi

Lens ko plate ke upar uniformly $t_{0} = 0.30 μ m$ raise kar diya gaya hai (uske neeche ek dust speck hai). Air, $λ = 600 nm$ , $R = 1.0 m$ . Naye pehle dark ring ( $m = 1$ ) ka radius kya hai?

Recall Solution

Ab gap $t = t_{0} + \frac{r ^{2}}{2 R}$ hai. Dark: $2 (t_{0} + \frac{r ^{2}}{2 R}) = mλ$ . $\frac{r ^{2}}{R} = mλ - 2 t_{0} \Rightarrow r^{2} = R (mλ - 2 t_{0})$ . $mλ = 600 \times 1 0^{- 9} = 6.0 \times 1 0^{- 7}$ ; $2 t_{0} = 6.0 \times 1 0^{- 7}$ . $mλ - 2 t_{0} = 0$ → $r = 0$ : raised gap pehle se hi poora order contribute kar chuka hai, toh "first dark ring" centre tak collapse ho jaati hai. Physically lift ne poore pattern ko ek order andar push kar diya hai. $m = 2$ lete hain: $r^{2} = 1.0 \times (1.2 \times 1 0^{- 6} - 6.0 \times 1 0^{- 7}) = 6.0 \times 1 0^{- 7}$ , $r = 7.75 \times 1 0^{- 4}$ m $= 0.775$ mm.

Wrap-up recall

Lens ko

t_{0}

raise karne se ring orders kyun shift ho jaate hain?

Dark condition

2 (t_{0} + r^{2} /2 R) = mλ

ban jaata hai; constant

2 t_{0}

har order ka kuch hissa kha jaata hai, poore pattern ko andar slide kar deta hai.

Transmitted light mein centre kaisa hai aur kyun?

Bright — koi net Stokes flip nahi, toh contact par

Δ = 2 μ t = 0

constructive interference deta hai.

Ring diameters se liquid ka

μ

kaise nikaalte hain?

μ = D_{air}^{2} / D_{liq}^{2}

same order par, kyunki

D^{2} \propto 1/ μ

hai.

Related tools worth revisiting: Thin film interference, Stokes' relations & phase change on reflection, Interference — path difference & coherence, Wedge fringes / air wedge, Refractive index measurement, Michelson interferometer.