Young's double slit — fringe width derivation: Worked examples

2.5.11 · D3 · Physics › Optics › Young's double slit — fringe width derivation

Yeh page fringe-width derivation ka exhaustive problem-drill child hai. Parent ne do master tools diye the:

$Δ = \frac{y d}{D} and β = \frac{λ D}{d} .$

Yahan hum har tarah ke numbers ko hit karte hain jo yeh formulas throw kar sakti hain: centre ke upar wala point, centre ke neeche wala point, bilkul centre, tiny-input aur huge-input limits, medium ka change, ek real-world word problem, aur ek exam twist. Har example batata hai ki woh matrix ka kaunsa cell fill karta hai, aur — sabse zaroori — woh tumhe answer forecast karne par majboor karta hai pehle dikhane se.

Apparatus — har symbol, words mein aur ek picture mein

Koi bhi formula se pehle, physical setup se milo. Neeche ki figure woh schematic hai jo har example silently assume karta hai; ise ek baar dekho aur letters $d$ , $D$ , $S_{1}$ , $S_{2}$ , $P$ , $O$ , $y$ mysterious rehna band ho jayenge.

Figure — Young's double slit — fringe width derivation

Notation jo hum use karenge — ek baar define karke

Do aur symbols counting ka kaam karte hain, toh inhe pin down karte hain pehle kabhi bhi use karne se.

Hum dark-fringe positions bhi chahiye honge. Ek dark band do bright ones ke exactly beech mein baith ti hai, toh uski order half-integer hai:

Scenario matrix

Examples karne se pehle, har case-class list karte hain jo ek YDSE problem mein ho sakti hai. Agar koi cell cover nahi hai toh tumhe ek aisa scenario milega jo tumne kabhi nahi dekha — exactly wahi hum forbid kar rahe hain.

#	Case class	Kya weird hai / kya dhyaan rakhna hai	Example jo isse cover karta hai
A	Point centre ke upar, $y > 0$	Seedha $Δ = + y d / D$	Ex 1
B	Point centre ke neeche, $y < 0$	$Δ$ negative hai → same fringe by symmetry	Ex 2
C	Point centre par, $y = 0$	$Δ = 0$ → zeroth (central) bright fringe	Ex 2
D	$λ$ find karo ek known fringe se	$y_{n}$ rearrange karo, $β$ nahi	Ex 3
E	Bright ya dark? ek given $y$ classify karo	$Δ/ λ$ compute karo, integer vs half-integer	Ex 4
F	Medium change (water/oil)	$λ \to λ / n$ se $β$ shrink hoti hai	Ex 5
G	Real-world word problem	Lab ruler padhna, $N$ fringes count karna	Ex 6
H	Exam twist — screen distance shift karo	Dhyan se padhna ki kaunsa variable badla	Ex 7
I	Degenerate / limiting input ( $d \to 0$ , $D \to \infty$ , $λ \to 0$ )	Kya formula sane rehta hai?	Ex 8

Ab har cell fill karte hain.

Example 1 — Cell A (point above centre)

Step 1 — SI mein convert karo. $λ = 500 \times 1 0^{- 9} m$ , $d = 0.20 \times 1 0^{- 3} m$ , $D = 1.0 m$ . Yeh step kyun? Formula $β = λ D / d$ tabhi sahi hai jab sabhi lengths same ruler use karein; nm aur mm mix karne par nonsense aata hai.

Step 2 — Pehle fringe width. $β = \frac{λ D}{d} = \frac{( 500 \times 1 0 ^{- 9} ) ( 1.0 )}{0.20 \times 1 0 ^{- 3}} = 2.5 \times 1 0^{- 3} m = 2.5 mm .$ Yeh step kyun? $β$ "unit gap" hai; ek baar mile toh $n$ -th fringe sirf $n$ baar woh gap hai.

Step 3 — 3rd bright fringe ki position. $y_{3} = 3 β = 3 (2.5 mm) = 7.5 mm .$ Yeh step kyun? $y_{n} = nλ D / d = n β$ , toh fringes count karna literally $β$ 's count karna hai.

Neeche ki figure is arithmetic ko ek picture mein badle ti hai: blue bars bright bands hain $y_{0} = 0, y_{1} = 2.5, y_{2} = 5.0, \dots$ mm par; notice karo ki har ek exactly ek green arrow ( $β$ ) pichle se upar hai, aur orange bar hamara answer hai $y_{3} = 7.5$ mm — ladder ki teesri seedi.

Verify: Directly, $y_{3} = 3 λ D / d = 3 (500 \times 1 0^{- 9}) (1) / (0.2 \times 1 0^{- 3}) = 7.5 \times 1 0^{- 3} m$ ✓. Millimetres, jaise forecast kiya tha — lab bench ke liye sensible.

Example 2 — Cells B & C (below centre, aur centre khud)

Step 1 — Centre ke neeche matlab $y < 0$ . $O$ ke $2 β$ neeche wale point ke liye, $y_{- 2} = - 2 β = - 5.0 mm$ . Yeh step kyun? $y$ $O$ se measure ki gayi signed height hai; centre ke neeche negative hoti hai. Path difference ban jaata hai $Δ = y d / D < 0$ , matlab ab $S_{1} P$ (slit $S_{1}$ se point tak ki distance) zyada lamba path hai instead of $S_{2} P$ .

Step 2 — Negative $Δ$ bhi valid bright condition hai. $Δ = - 2 λ$ satisfy karta hai $Δ = nλ$ with $n = - 2$ . Wavelengths ki poori number — positive ya negative — phir bhi constructive hai. Yeh step kyun? Pattern $O$ ke baare mein symmetric hai; $2 β$ neeche wali fringe $2 β$ upar wali ke identical dikhti hai. Isliye hum $n = 0, \pm 1, \pm 2, \dots$ allow karte hain.

Figure mirror symmetry ko explicit banati hai: left par blue slits $S_{1}, S_{2}$ teen screen points tak dashed rays bhejte hain. Green dot $O$ hai (screen par $n = 0$ central bright fringe at $y = 0$ ); orange dot upar ka ek point hai ( $y > 0$ , $n = + 2$ ) aur red dot uska mirror image neeche ( $y < 0$ , $n = - 2$ ). Do double-headed arrows length mein equal hain — woh equality hi symmetry hai.

Step 3 — Centre. $O$ par, $y = 0 \Rightarrow Δ = 0 = 0 \cdot λ$ . Yeh $n = 0$ hai: ek bright fringe. Yeh step kyun? Zero path difference matlab dono waves perfectly in step arrive karti hain — maximum brightness. Centre ek symmetric YDSE mein hamesha bright hota hai.

Verify: $∣ y_{- 2} ∣ = 2 β = 5.0 mm$ symmetry se $y_{+ 2} = + 5.0 mm$ se match karta hai ✓. Central fringe $Δ = 0$ → constructive → bright ✓ (ek common trap: log kehte hain centre dark hai).

Example 3 — Cell D (wavelength find karo)

Step 1 — Sahi formula choose karo. Hum ek specific fringe ki position jaante hain, toh $y_{n} = nλ D / d$ use karo, sirf $β$ nahi. Yeh step kyun? $β$ spacing batata hai; yahan humein fringe number $n = 5$ ki absolute position di gayi hai, toh $y_{n}$ form direct fit hai.

Step 2 — $λ$ ke liye rearrange karo. $λ = \frac{y _{n} d}{n D} = \frac{( 6.0 \times 1 0 ^{- 3} ) ( 0.40 \times 1 0 ^{- 3} )}{( 5 ) ( 0.80 )} .$ Yeh step kyun? Numbers plug karne se pehle algebraically unknown ke liye solve karo — kam arithmetic slips.

Step 3 — Compute karo. $λ = \frac{2.4 \times 1 0 ^{- 6}}{4.0} = 6.0 \times 1 0^{- 7} m = 600 nm .$

Verify: $600 nm$ orange-red visible light hai ✓ — 400–700 nm ke andar jaise forecast kiya tha. Back-check: $y_{5} = 5 (600 e- 9) (0.8) / (0.4 e- 3) = 6.0 mm$ ✓.

Yeh upar derive ki gayi small-angle result $y_{n} = nλ D / d$ use karta hai.

Example 4 — Cell E (bright ya dark? ek point classify karo)

Step 1 — Path difference compute karo. $Δ = \frac{y d}{D} = \frac{( 4.0 \times 1 0 ^{- 3} ) ( 0.30 \times 1 0 ^{- 3} )}{1.2} = 1.0 \times 1 0^{- 6} m .$ Yeh step kyun? $Δ$ master quantity hai — sab kuch (bright/dark) $Δ$ ko $λ$ se compare karke decide hota hai. Dekho Path difference and phase difference.

Step 2 — $λ$ se divide karo wavelengths count karne ke liye. $\frac{Δ}{λ} = \frac{1.0 \times 1 0 ^{- 6}}{600 \times 1 0 ^{- 9}} = \frac{5}{3} \approx 1.667.$ Yeh step kyun? Do exact conditions ke against compare karo equations ki form mein: $bright ⟺ \frac{Δ}{λ} = n \in {0, \pm 1, \pm 2, \dots}, dark ⟺ \frac{Δ}{λ} = m + \frac{1}{2} \in {\pm \frac{1}{2}, \pm \frac{3}{2}, \dots} .$

Step 3 — Classify karo. $\frac{5}{3} = 1.667$ na integer hai na half-integer. Yeh $\frac{3}{2}$ (dark) aur $2$ (bright) ke beech mein hai, toh point partially lit hai — minimum se zyada bright, maximum se kam bright. Yeh step kyun? Real detectors intensity smoothly read karte hain; sirf exact integer/half-integer points pure max/min hain.

Verify: $Δ/ λ = 5/3 \approx 1.667$ ✓; ${0, 1, 2, \dots}$ mein nahi aur na ${0.5, 1.5, \dots}$ mein, toh "neither" sahi hai ✓.

Example 5 — Cell F (medium mein submerge karo)

Step 1 — Water mein kya badlega? Frequency fixed hai (source se set hoti hai); light $n_{w}$ factor se slow hoti hai, toh $λ^{'} = λ / n_{w}$ . Yeh step kyun? $β$ in-medium wavelength par depend karta hai, aur sirf $λ$ badalta hai — $d$ aur $D$ physical distances hain jo medium ki parwah nahi karte.

Step 2 — Fringe width scale karo. $β^{'} = \frac{λ ^{'} D}{d} = \frac{λ D}{n _{w} d} = \frac{β}{n _{w}} = \frac{1.8 mm}{1.33} .$ Yeh step kyun? $β \propto λ$ hai, toh $λ$ ko $n_{w}$ se divide karne par $β$ bhi usi $n_{w}$ se divide ho jaata hai.

Step 3 — Compute karo. $β^{'} = 1.35 mm (to 3 s.f.) .$

Verify: $1.8/1.33 = 1.353 mm$ ✓ — tange, jaise forecast kiya tha (zyada index pattern squeeze karta hai).

Example 6 — Cell G (real-world word problem)

Step 1 — Gaps count karo, fringes nahi. Fringe number 1 se fringe number 10 tak $10 - 1 = 9$ gaps hain (har gap ek $β$ hai). Nine fence-panels ke beech ten fence-posts ki picture: ten posts, lekin sirf nine panels. Toh $9 β = 16.2 mm ⟹ β = \frac{16.2 mm}{9} = 1.80 mm .$ Yeh step kyun? Yeh single most common lab error hai. Fringe $a$ aur fringe $b$ ke beech ki distance $(b - a) β$ hai, kyunki $y_{b} - y_{a} = (b - a) β$ from $y_{n} = n β$ . Fringes count karna (ten) instead of gaps (nine) spacing inflate karta hai.

Step 2 — $β = λ D / d$ ko $λ$ ke liye solve karo. $λ = \frac{β d}{D} = \frac{( 1.80 \times 1 0 ^{- 3} ) ( 0.25 \times 1 0 ^{- 3} )}{1.5} .$ Yeh step kyun? Baaki sab measured hai; sirf $λ$ unknown hai, toh isolate karo.

Step 3 — Compute karo. $λ = \frac{4.5 \times 1 0 ^{- 7}}{1.5} = 3.0 \times 1 0^{- 7} m = 300 nm .$

Verify: $λ = 300 nm$ ✓ arithmetic — lekin yeh ultraviolet hai, aankhon ko invisible. Real lab mein yeh ek flag hai: ya toh light sach mein UV hai (film par detect hoti hai, aankhon se nahi), ya student ne galat count kiya aur 9 ki jagah 10 gaps use kiye. Agar unhone galti se $10 β = 16.2$ mm use kiya toh unhe $β = 1.62$ mm aur $λ = 270 nm$ milta — phir bhi UV, aur kisi bhi visible value se zyada door. Lesson: gaps count karo = (last − first), phir colour band sanity-check karo.

Example 7 — Cell H (exam twist)

Step 1 — $β_{1}$ compute karo. $β_{1} = \frac{λ D _{1}}{d} = \frac{( 650 \times 1 0 ^{- 9} ) ( 1.0 )}{0.50 \times 1 0 ^{- 3}} = 1.30 \times 1 0^{- 3} m = 1.30 mm .$ Yeh step kyun? "Before" value anchor karo.

Step 2 — Proportionality use karo, scratch se recompute mat karo. $β \propto D$ hai, toh $β_{2} = β_{1} \cdot (D_{2} / D_{1}) = 1.30 mm \times 2 = 2.60 mm$ . Yeh step kyun? $β \propto D$ recognize karna faster hai aur exam-safe hai: sirf $D$ badla (slits $S_{1}, S_{2}$ aur unka spacing $d$ untouched hain), toh scale karo.

Step 3 — Shift. $Δ β = β_{2} - β_{1} = 2.60 - 1.30 = 1.30 mm .$ Yeh step kyun? Exams aksar change poochte hain, test karte hue ki tumhe linear dependence pata hai.

Verify: $β_{2} = 650 e- 9 \cdot 2.0/0.5 e- 3 = 2.6 mm$ ✓ ; $D$ double karne par $β$ double hua ✓ jaise forecast kiya tha.

Example 8 — Cell I (degenerate & limiting inputs)

Step 1 — $d \to 0$ (slits merge hote hain). $β = λ D / d \to \infty$ . Yeh step kyun? Jaise do slits $S_{1}, S_{2}$ ek mein merge hoti hain, fringes infinitely far apart spread ho jaate hain — matlab poori screen ek single bright blur ban jaati hai (koi distinguishable pattern nahi). Yeh two-source pattern wapas ek source par collapse ho raha hai. Physically: ek single opening se interference nahi ho sakti (woh single-slit diffraction hai instead).

Step 2 — $D \to \infty$ (screen bahut door). $β \to \infty$ bhi, lekin angular width $θ = λ / d$ fixed aur finite rehti hai. Yeh step kyun? Linear spacing infinitely badh jaata hai, phir bhi fringes ke beech ka angle $D$ se independent hai — isliye parent note ne angular form $θ = λ / d$ ko prize diya.

Step 3 — $λ \to 0$ (imagine light with no wavelength). $β \to 0$ : fringes infinitely close crowd ho jaate hain, uniform brightness mein wash ho jaate hain — koi visible pattern nahi. Yeh step kyun? Interference ek wave effect hai; zero-wavelength wave ek straight ray ki tarah behave karta hai (geometric optics), aur rays interfere nahi karte. Isliye hum kabhI kabhi, say, cricket ball ke saath interference nahi dekhte.

Figure $β$ ko $d$ ke against plot karta hai fixed $λ, D$ ke liye: dekho blue curve sky ki taraf rocket karta hai jaise $d$ left side par sihrta hai ( $d \to 0$ blow-up, red dashed line), aur right side par small $β$ ki taraf flatten hota hai jaise $d$ barhta hai (fringes crowding, green arrow). Yeh $β \propto 1/ d$ ki shape visible hai.

Verify: Teeno limits $β = λ D / d$ ke consistent hain: numerator $\to 0$ deta hai $β \to 0$ ; denominator $\to 0$ deta hai $β \to \infty$ ; $D \to \infty$ deta hai $β \to \infty$ jabki $β / D = λ / d$ finite rehta hai ✓.

Matrix ka Recap

Recall Kis example ne har cell fill kiya?

A→Ex1, B→Ex2, C→Ex2, D→Ex3, E→Ex4, F→Ex5, G→Ex6, H→Ex7, I→Ex8. Har cell covered.

Recall Har problem ke peeche woh ek sawaal

$Δ = y d / D$ compute karo, $λ$ se divide karo: integer = bright, half-integer = dark, warna in-between. ::: Haan — woh ek comparison hi nau cells drive karta hai.

Recall "N fringes mein distance" ko

$β$ mein kaise convert karte ho? Gaps count karo = (last fringe number − first fringe number), phir measured distance ko un gaps ki number se divide karo. ::: Fringe 1 se fringe 10 tak 9 gaps hain, toh $β = distance /9$ .

Connections

Young's double slit — fringe width derivation — parent derivation jin drills ko yeh exercise karte hain.
Path difference and phase difference — woh $Δ$ jo hum har example mein compute karte hain.
Small angle approximation — isliye $sin θ \approx tan θ = y / D$ throughout.
Refractive index — Example 5 ka medium change.
Coherence and coherent sources — fringes exist karne ke liye chahiye (Ex 8a).
Diffraction grating — $d \to 0$ single-opening limit.
Interference of light — umbrella phenomenon.