2.5.15 · D3Diffraction grating — condition for maxima
Intuition Yeh page kis liye haiParent note Diffraction grating — condition for maxima ne tumhe wo ek rule diya jo sab kuch chalata hai:
d sin θ m = mλ . θ m θ m m kis bright spot ki taraf point kar rahe hain (m = 1 m = 2 θ
Yahan hum naya physics nahi seekhte — hum har tarah ke questions dhundte hain jo us rule se nikl sakte hain, aur ek-ek ka solve karte hain. Agar tum yeh sab kar sako, toh koi bhi grating problem tumhe surprise nahi kar sakti.
Shuru karne se pehle, aao har symbol ko dobara samjhein taaki koi bhi reader seedha aa ke follow kar sake:
d do neighbouring slits ke beech ki centre-to-centre distance , metres mein measure ki gayi. Ek fence mein do tiny gaps socho; d θ angle jo outgoing ray straight-ahead direction se banati hai ("normal", yaani grating ke perpendicular line). Seedha aage θ = 0 θ = 9 0 ∘ θ m m λ wavelength , light ki ek poori crest-to-crest ripple ki length. Dekho Wavelength and the visible spectrum .m order , ek poora counting number 0 , ± 1 , ± 2 , … kitne poore wavelengths ka extra path neighbouring rays ko alag karta hai.sin θ tan cos θ opposite hoti hai, d
Intuition Signs, negative orders aur arcsine ke do answersHar worked example neeche positive θ left–right symmetric hai, isliye signs ke baare mein honest rehna zaroori hai:
Negative orders. Kyunki m 0 , ± 1 , ± 2 , … m = + 1 + θ m = − 1 − θ doosri taraf centre se. Isliye counting problems (Cell D) mein hamesha ek odd total 2 m m a x + 1 Arcsine ke do answers. Jab tum θ = sin − 1 ( x ) sin usi value par θ 18 0 ∘ − θ − 9 0 ∘ ≤ θ ≤ + 9 0 ∘ The ceiling. sin θ 1 − 1 − 9 0 ∘ ≤ θ ≤ + 9 0 ∘ poora physically allowed range hai. Koi bhi m ∣ mλ / d ∣ > 1 koi real angle nahi hota — woh order simply exist hi nahi karta.
Yeh dhyan mein rakho: jab bhi hum likhein "θ 1 = 19. 3 ∘ m = − 1 − 19. 3 ∘
Har grating question in cells mein se ek (ya blend) hoti hai. Har worked example neeche us cell ke saath tagged hai jise woh solve karta hai.
Cell
Kya ise alag banata hai
Example
A. Forward — angle nikalo d , λ , m θ Ex 1
B. Backward — d θ , λ , m Ex 2
C. Backward — λ θ , d , m Ex 3
D. Orders count karo Kitne maxima fit hote hain; sin θ ≤ 1 ± m
Ex 4
E. Degenerate: m = 0 Zero order — sab colours overlap karte hain, koi dispersion nahi
Ex 5
F. Limiting: order exactly 9 0 ∘ sin θ = 1 Ex 6
G. Do wavelengths / dispersion Ek hi order mein do colours ka angular separation
Ex 7
H. Real-world word problem Starlight spectrometer, numbers extract karni padegi
Ex 8
I. Exam twist — overlapping orders Ek colour ka order m m ′
Ex 9
Cells covered: A B C D E F G H I — sab ke sab.
Worked example Example 1 (Cell A)Green light, λ = 550 nm 600 lines/mm ki grating par padti hai. First-order (m = 1
Forecast: Jab d
Step 1 — d
d = 600 mm − 1 1 = 600 1 mm = 1.667 × 1 0 − 6 m . Yeh step kyun? Equation ko physical spacing chahiye, na ki "lines per mm". Hum invert karte hain, aur mm→m convert karte hain taaki d λ
Step 2 — grating rule ko sin θ
sin θ 1 = d mλ = 1.667 × 1 0 − 6 1 ⋅ 550 × 1 0 − 9 = 0.330. Yeh step kyun? θ sin θ mλ / d
Step 3 — sine undo karo.
θ 1 = sin − 1 ( 0.330 ) = 19. 3 ∘ . Yeh step kyun? sin − 1 [ − 9 0 ∘ , + 9 0 ∘ ] m = − 1 − 19. 3 ∘
Verify: Plug back karo: d sin θ = 1.667 × 1 0 − 6 × sin ( 19. 3 ∘ ) = 1.667 × 1 0 − 6 × 0.330 = 5.5 × 1 0 − 7 m = 550 nm = 1 ⋅ λ
Worked example Example 2 (Cell B)Ek red laser, λ = 633 nm third-order maximum θ = 35. 0 ∘ d
Forecast: Modest angle par higher order ka matlab hai ki slits relatively... wide hain ya narrow? Guess karo.
Step 1 — rule ko d
d = s i n θ mλ = s i n 3 5 ∘ 3 ⋅ 633 × 1 0 − 9 . Yeh step kyun? Ab d sin θ
Step 2 — evaluate karo.
sin 3 5 ∘ = 0.5736 , d = 0.5736 1.899 × 1 0 − 6 = 3.31 × 1 0 − 6 m . Yeh step kyun? Symbolic answer ko number mein badalna hai: calculator se sin 3 5 ∘ mλ = 3 × 633 nm
Step 3 — lines per mm mein convert karo.
lines/mm = d 1 = 3.31 × 1 0 − 3 mm 1 ≈ 302 lines/mm . Yeh step kyun? Lines-per-mm spacing ka reciprocal hota hai, mm mein expressed.
Verify: d sin θ = 3.31 × 1 0 − 6 × 0.5736 = 1.90 × 1 0 − 6 m mλ = 3 × 633 × 1 0 − 9 = 1.90 × 1 0 − 6 m
Worked example Example 3 (Cell C)Ek unknown spectral line second order (m = 2 θ = 28. 0 ∘ 400 lines/mm ki grating par. Wavelength kya hai, aur kya yeh visible hai?
Forecast: Visible light ~400–700 nm tak hoti hai. Kya tumhara answer us band ke andar land karega?
Step 1 — d
d = 400 mm − 1 1 = 2.5 × 1 0 − 6 m . Yeh step kyun? Equation physical spacing d d λ
Step 2 — rule ko λ
λ = m d s i n θ = 2 2.5 × 1 0 − 6 × s i n 2 8 ∘ . Yeh step kyun? λ d sin θ m
Step 3 — evaluate karo.
sin 2 8 ∘ = 0.4695 , λ = 2 2.5 × 1 0 − 6 × 0.4695 = 5.87 × 1 0 − 7 m = 587 nm .
Verify: 587 nm visible spectrum ke yellow mein hai — yeh famous sodium line hai. Plug back: d sin θ = 2.5 × 1 0 − 6 × 0.4695 = 1.174 × 1 0 − 6 mλ = 2 × 5.87 × 1 0 − 7 = 1.174 × 1 0 − 6
Worked example Example 4 (Cell D)Blue light λ = 450 nm 800 lines/mm ki grating par. Total kitne bright maxima appear hote hain?
Forecast: Compute karne se pehle total guess karo. (Yaad rakho maxima dono sides par appear hote hain!)
Step 1 — d
d = 800 mm − 1 1 = 1.25 × 1 0 − 6 m . Yeh step kyun? Counting formula physical spacing d d / λ
Step 2 — sabse bada possible order nikalo.
λ d = 450 × 1 0 − 9 1.25 × 1 0 − 6 = 2.78. Yeh step kyun? sin θ 1 m m a x = ⌊ d / λ ⌋ m sin θ > 1
Step 3 — floor lo aur dono sides count karo.
m m a x = ⌊ 2.78 ⌋ = 2. m = − 2 , − 1 , 0 , + 1 , + 2 5 maxima .
Yeh step kyun? Pattern centre ke baare mein symmetric hai: har positive order + m − m m = 0 2 m m a x + 1
Verify: Check karo order 2 real hai: sin θ 2 = 2 × 450 × 1 0 − 9 /1.25 × 1 0 − 6 = 0.72 ≤ 1 3 × 450/1250 = 1.08 > 1 = 2 × 2 + 1 = 5
Worked example Example 5 (Cell E)White light (sab wavelengths 400–700 nm) kisi bhi grating par padti hai. Zero-order (m = 0
Forecast: Kya zero-order angle colour par depend karta hai? Haan ya nahi guess karo.
Step 1 — m = 0
d sin θ 0 = 0 ⋅ λ = 0 ⇒ sin θ 0 = 0 ⇒ θ 0 = 0 ∘ . Yeh step kyun? Zero extra path difference ke saath, har wavelength seedha aage constructively interfere karti hai.
Step 2 — colour ke baare mein soco.
Kyunki θ 0 = 0 λ sab colours ek hi jagah land karte hain aur recombine ho jaate hain.
Yeh step kyun? Koi wavelength bend nahi hoti, isliye centre par koi dispersion nahi hoti.
Figure mein kya dekhna hai: seedha aage wala red arrow m = 0 single arrow hai jo har colour ko ek saath carry karta hai, kyunki sab wavelengths θ = 0 m = ± 1 m ≥ 1 λ centre white hai; fans coloured hain.
Verify: Equation θ 0 = 0 λ = 400 λ = 700 0 ∘ white central maximum. ✓ (Sirf m ≥ 1
Worked example Example 6 (Cell F)Ek grating ka spacing d = 1200 nm λ = 600 nm second-order maximum exist karta hai? Woh kahan appear hota hai, aur kya hoga agar tum λ
Forecast: d / λ = 2
Step 1 — sin θ 2
sin θ 2 = d mλ = 1200 2 × 600 = 1.000. Yeh step kyun? Hum order ko ceiling sin θ ≤ 1
Step 2 — sin θ = 1
θ 2 = sin − 1 ( 1 ) = 9 0 ∘ . Yeh step kyun? θ = 9 0 ∘ ke saath saath skim karti hai — mathematically exist karti hai lekin infinitely spread hai aur aage screen tak kabhi nahi pahunchti. Yeh visibility ki exact boundary hai.
Step 3 — wavelength nudge karo.
λ = 610 sin θ 2 = 2 × 610/1200 = 1.017 > 1 Koi real angle nahi — order 2 gayab ho jaata hai.
Yeh step kyun? Yeh dikhata hai ki boundary ek genuine cliff hai: d / λ = m
Verify: λ = 600 m m a x = ⌊ 1200/600 ⌋ = 2 λ = 610 ⌊ 1200/610 ⌋ = ⌊ 1.97 ⌋ = 1
Worked example Example 7 (Cell G)500 lines/mm ki grating par do lines se light padti hai: λ 1 = 486 nm λ 2 = 656 nm First order mein in do lines ka angular separation nikalo.
Forecast: Kaun sa colour zyada bend hoga — shorter blue ya longer red? Compute karne se pehle guess karo.
Step 1 — d
d = 500 mm − 1 1 = 2.0 × 1 0 − 6 m .
Step 2 — har colour ka angle (m = 1
sin θ 1 , blue = 2.0 × 1 0 − 6 486 × 1 0 − 9 = 0.243 ⇒ θ = 14.0 6 ∘ . sin θ 1 , red = 2.0 × 1 0 − 6 656 × 1 0 − 9 = 0.328 ⇒ θ = 19.1 4 ∘ . Yeh step kyun? Same equation do baar apply ki; sirf λ
Step 3 — subtract karo.
Δ θ = 19.1 4 ∘ − 14.0 6 ∘ = 5.0 8 ∘ . Yeh step kyun? "Angular separation" literally do directions ke beech ka gap hai, isliye hum do coloured spots kitne door hain yeh nikalne ke liye chhote angle ko bade se subtract karte hain.
Verify: Red (656 486 sin θ ∝ λ ≈ 5. 1 ∘ Resolving power and spectrometers ka basis hai. ✓
Worked example Example 8 (Cell H)Ek astronomer starlight ko ek spectrometer mein feed karta hai jiskin grating mein 1200 lines/mm hai. Ek calcium absorption line first order mein measured angle θ = 28. 4 ∘ K line ≈ 393.4 nm
Forecast: 1200 lines/mm ek fine grating hai (small d
Step 1 — numbers extract karo. Maano N ℓ = 1200 lines/mm m = 1 θ = 28. 4 ∘ Yeh step kyun? Word problems variables ko prose mein chhupaate hain — pehle unhe name karo. Note karo N ℓ lines per length hai; spacing d
Step 2 — d
d = N ℓ 1 = 1200 mm − 1 1 = 8.333 × 1 0 − 7 m .
Step 3 — λ
λ = m d s i n θ = 1 8.333 × 1 0 − 7 × s i n 28. 4 ∘ . sin 28. 4 ∘ = 0.4756 , λ = 8.333 × 1 0 − 7 × 0.4756 = 3.96 × 1 0 − 7 m = 396 nm .
Verify: 396 nm H line ≈ 396.8 nm K line 393.4 nm ≈ 3 nm d sin θ = 8.333 × 1 0 − 7 × 0.4756 = 3.96 × 1 0 − 7 m = 1 ⋅ λ
Worked example Example 9 (Cell I)Ek grating par, red light λ R = 700 nm second order mein ek shorter wavelength λ x third order ke saath overlap karti hai (same angle). λ x
Forecast: Same angle ke liye, kya higher order ko longer ya shorter wavelength chahiye? Guess karo.
Step 1 — overlap condition likho.
Same d θ d sin θ d sin θ = m R λ R and d sin θ = m x λ x , m R λ R = m x λ x ⇒ 2 × 700 = 3 × λ x . Yeh step kyun? "Same angle" ka matlab hai d sin θ mλ d sin θ
Step 2 — solve karo.
λ x = 3 2 × 700 = 466.7 nm . Yeh step kyun? Seedha algebra: known product m R λ R m x = 3
Step 3 — identify karo.
466.7 nm visible band mein blue hai.
Yeh step kyun? Bare number ko physical colour mein convert karte hain taaki answer meaningful ho, sirf arithmetic nahi.
Verify: Equal path difference check karo — koi bhi grating lo, maano d = 2000 sin θ = 2 × 700/2000 = 0.700 sin θ = 3 × 466.7/2000 = 0.700 shorter wavelength chahiye — jaise forecast mein check kiya.
Common mistake Overlap problems:d Kyun sahi lagta hai: Har doosra grating problem d Fix: Jab do maxima same angle share karte hain, quantity d sin θ d sin θ = m 1 λ 1 d sin θ = m 2 λ 2 m 1 λ 1 = m 2 λ 2 d sin θ spacing jaane bina solve karte ho — jaisa Example 9 dikhata hai.
Common mistake Count karte waqt negative orders bhoolnaKyun sahi lagta hai: Tum ek angle ke liye solve karte ho, ek bright spot dekhte ho, aur wahi count report karte ho. Fix: Har order + m − m − θ 2 m m a x + 1 m = 0
Recall Poora matrix self-test karoYeh kaunse cell honge? "Measured angle se lines/mm nikalo" ::: Cell B (backward for d ± m sin θ ≤ 1
Har cell same equation hai ek alag letter ke liye solve ki gayi: "d sin θ = mλ Angle → sin − 1 ( mλ / d ) mλ / sin θ d sin θ / m 2 ⌊ d / λ ⌋ + 1