Diffraction — single slit intensity pattern derivation: Question bank

2.5.14 · D5 · Physics › Optics › Diffraction — single slit intensity pattern derivation

Do anchors ka reminder jo tumhe baar baar chahiye honge: $I (θ) = I_{0} (\frac{s i n β}{β})^{2}, β = \frac{π a s i n θ}{λ}, minima: a sin θ = mλ (m = \pm 1, \pm 2, \dots) .$

Yahan $a$ slit width hai, $λ$ wavelength hai, $θ$ diffraction angle hai, aur $β$ slit ke across total phase spread ka half hai.

True or false — justify karo

Single-slit minimum ki condition

a sin θ = mλ

hai, to

m = 0

pehli dark fringe deta hai

False.

m = 0

ka matlab

β = 0

hai, jahan

sin β / β \to 1

— ye central bright maximum hai. Minima ke liye sirf

m = \pm 1, \pm 2, \dots

chahiye.

Formula

a sin θ = mλ

bright fringes mark karta hai, bilkul Young's double slit ki tarah

False. Single slit ke liye ye dark fringes mark karta hai: jab full-width path difference ek poora

λ

hota hai, to wavelets top-bottom pairs mein cancel ho jaate hain. Same dikhne wala formula, opposite physics.

Slit ko narrower banane se diffraction pattern narrower ho jaata hai

False. Half-width

θ_{1} \approx λ / a

hai, isliye chhota

a

zyada wide spread deta hai. Beam ko zyada confine karne se woh aur bhi zyada fan out karti hai.

Secondary maxima bilkul adjacent minima ke beech mein barabar doori par hote hain

False. Ye

tan β = β

satisfy karte hain, jisse

β \approx 1.43 π, 2.46 π, \dots

milta hai —

1/ β

envelope ke kaaran thoda centre ki taraf khinche hue, na ki saaf-suthre

1.5 π, 2.5 π

par.

Central maximum angle mein har side maximum se exactly do guna wide hota hai

Linear width mein True, roughly. Central peak

m = - 1

m = + 1

tak span karta hai (width

\sim 2 λ / a

), jabki ek side maximum sirf ek

π

-interval of

β

span karta hai (width

\sim λ / a

). To centre lagbhag do guna broad hai.

Saare secondary maxima ek dusre se same brightness ke hote hain

False. Unki heights roughly

1/ (m + \frac{1}{2})^{2} π^{2}

ke hisaab se fall off hoti hain: pehla

\approx 4.5%

I_{0}

hai, doosra

\approx 1.6%

, aur aage bhi. Envelope decay karta hai.

Agar tum

a

aur

λ

dono double kar do, to minima same angles par aate hain

True. Minima

sin θ = mλ / a

follow karte hain, jo sirf ratio

λ / a

par depend karta hai. Dono ko same factor se scale karne par har angle unchanged rehta hai.

Incoming light ki intensity badhaane se fringe positions baahir ki taraf shift ho jaati hain

False. Fringe positions sirf

λ

a

, aur geometry par depend karti hain. Zyada bright light

I_{0}

ko raise karta hai (saare peaks ek saath scale hote hain) lekin kuch move nahi karta.

Error dhundho

β = π

par

sin β = 0

hai, aur kyunki

sin β / β = 0/ π = 0

hai, pattern dark hai — ye pehla minimum hai." Flaw dhundho

Is mein koi flaw nahi hai — ye correct hai.

β = π

exactly

a sin θ = λ

hai, genuine pehla minimum. (Trap ye hai ki ek sahi statement par doubt karo.)

sin β / β

mein denominator mein

β

hai, isliye

β = 0

par intensity blow up ho jaati hai."

Galat. Jab

β \to 0

sin β \approx β

, to ratio

\to 1

hota hai, infinity nahi. Numerator bilkul utni hi fast vanish hota hai jitna denominator — peak

I_{0}

par finite hai.

"Pehla secondary maximum

β = 1.5 π

par hai, to uski intensity

(sin (1.5 π) /1.5 π)^{2} = 1/ (1.5 π)^{2}

hai."

Do slip hain. Asli peak

β \approx 1.43 π

par hai, aur

sin (1.5 π) = - 1

hai (sinc ka max nahi).

\approx 4.5%

value ek coincidentally-close estimate hai, exact peak nahi.

"Kyunki ek single slit 'do slits ko ek saath push karne jaisi' hai, double slit ka missing-order rule yahan bhi apply hota hai."

Single slit mein alag 'orders' hi nahi hote jo miss ho sakein; ye ek continuous source hai, do point sources nahi. Double-slit missing-order effect diffraction envelope se aata hai jo interference fringes ko modulate karta hai — ek do-scale story jiske liye

a

aur slit separation

d

dono chahiye.

"Slit ke across phase difference

β = π a sin θ / λ

hai."

Edge-to-edge total phase spread

2 β = 2 π a sin θ / λ

hai.

β

sirf uska aadha hai — natural half-angle variable jo chord formula

sin β / β

ko clean banata hai.

"Pehle minimum par slit ka top edge aur bottom edge

λ /2

out of phase hain, isliye wo cancel hote hain."

Edges pehle minimum par ek poore

λ

(yaani

2 π

) apart hain,

λ /2

nahi. Cancellation isliye hoti hai kyunki top half ki har strip bottom half ki ek strip ke saath pair karti hai jo

λ /2

away hoti hai — halves ke across pairwise cancellation.

Why questions

Pattern ka centre bright kyun hota hai, dark kyun nahi?

θ = 0

par har wavelet same distance travel karta hai, isliye saare phases zero hain aur sab perfectly in step add ho jaate hain (

sin β / β \to 1

). Maximum constructive interference centre par rehti hai.

Single-slit rule darkness kyun deta hai jahan double-slit rule brightness deta hai same

a sin θ = mλ

ke liye?

Double slit mein do point sources hain jo

mλ

par in step add hote hain; single slit ek continuum hai jiske wavelets

mλ

par pairs mein cancel ho jaate hain. Source ki distributed nature outcome ko flip kar deti hai.

Amplitude mein factor

e^{i β}

intensity ko affect kyun nahi karta?

Iska magnitude 1 hai (ek pure phase), aur intensity

∣ E ∣^{2}

hai. Unit-magnitude phase se multiply karne par complex plane par direction badalta hai lekin length nahi — jo square hoti hai.

β

(phase spread ka half) "natural variable" kyun hai, poora spread

2 β

kyun nahi?

Total angle

2 β

mein fan out ho rahe bahut saare equal phasors ka resultant us arc ki chord hoti hai, aur chord ki length half-angle

sin β

involve karti hai. To

β

wahi hai jo geometry tumhe deta hai.

Minima ke beech bhi bade angles par jaane se intensity kyun fall off hoti hai?

1/ β

envelope: bada

β

matlab phasors ek tighter, smaller-radius arc mein curl hote hain, isliye best partial cancellation bhi kam resultant chhodti hai. Poora pattern ek shrinking

1/ β^{2}

ceiling ke neeche hai.

Ek screen point tak pahunchne wali saari rays ko parallel kyun treat kar sakte hain?

Screen slit width ki comparison mein bahut door hai (Fraunhofer condition, Fraunhofer vs Fresnel Diffraction mein contrast kiya gaya hai). Door point tak rays ki direction mein negligible amount ka farq hota hai, isliye path difference cleanly

y sin θ

hai.

Single-slit spreading ko uncertainty principle ka "baby version" kyun kehte hain?

Photon ki transverse position ko width

Δ y \sim a

tak confine karne par transverse momentum spread

\sim λ / a

in angle force hoti hai — chhota

a

, wider angular fan. Ye exactly Heisenberg Uncertainty Principle ka position-momentum trade-off hai.

Edge cases

a \to λ

par pattern kaisa dikhta hai (slit width ek wavelength tak shrink ho jaati hai)?

Pehle minimum ko

sin θ = λ / a = 1

, yaani

θ = 9 0^{\circ}

chahiye — ye edge tak slide ho jaata hai. Central maximum poore forward hemisphere ko fill kar leta hai; slit lagbhag ek single Huygens point source ki tarah behave karta hai.

a < λ

hone par kya hota hai?

Koi minimum exist nahi karta, kyunki

sin θ = mλ / a

sin θ > 1

chahiye hoga. Sirf central maximum bachta hai, jo saare forward angles mein spread ho jaata hai — essentially pure spherical-wavelet emission.

a \to \infty

(bahut wide slit) limit mein

I

kya hota hai?

Minima

θ = 0

ki taraf crowd ho jaate hain (

sin θ = mλ / a \to 0

), isliye central peak ek razor-sharp spike seedha aage tak collapse ho jaata hai. Ye ray-optics limit hai: wide aperture ek sharp geometric beam banata hai, koi visible diffraction nahi.

Exactly

θ = 9 0^{\circ}

par jab

a sin θ

kisi

mλ

ke barabar ho, kya wo direction physically reachable hai?

θ = 9 0^{\circ}

ka matlab hai light slit plane ke along travel kar rahi hai — ek grazing, physically limiting edge. Jinhe

sin θ > 1

chahiye hoga wo orders exist hi nahi karte;

sin θ = 1

hard ceiling hai.

Agar hum same slit mein se do wavelengths shine karein, to kya unke central maxima overlap karte hain?

Haan — dono central maxima

θ = 0

par hi baithte hain

λ

se regardless, isliye unke centres coincide karte hain. Unke pehle minima diverge karte hain (

θ_{1} \approx λ / a

), isliye longer wavelength ka pattern shared centre ke around wider hota hai.

Ek secondary maximum ke neighbouring minimum par intensity exactly kya hai, aur kya peak

I_{0}

tak pahunchti hai?

Kisi bhi minimum par

I = 0

exactly hota hai. Koi bhi secondary peak kabhi

I_{0}

tak nahi pahunchta; pehla sirf

\approx 4.5%

hai, kyunki

1/ β^{2}

envelope har baad ke peak ko centre se bahut neeche cap karta hai.

Recall Fast self-test

Single-slit $a sin θ = mλ$ ka matlab kya hai, aur kis $m$ ke liye? ::: DARK fringes, $m = \pm 1, \pm 2, \dots$ (kabhi bhi $m = 0$ nahi). Slit narrow hone par pattern kis taraf jaata hai? ::: Ye wider ho jaata hai ( $θ_{1} \approx λ / a$ ). True secondary maxima kahan hain? ::: $tan β = β$ par, yaani $β \approx 1.43 π, 2.46 π, \dots$ , $π$ ke half-integer multiples par nahi.

Related builds: Phasor Addition of Waves (chord $sin β / β$ kyun deta hai), Huygens Principle (slit many sources kyun hai), aur Rayleigh Criterion and Resolution (kyun $λ / a$ optics ki separation limit set karta hai).