When light of wavelength λ \lambda λ a a a not a single source — it is an infinite collection of tiny Huygens wavelets spread across its width. Each wavelet travels to a point on a far screen, but because they start at different positions across the slit, they arrive with different path lengths , hence different phases . The screen brightness is the result of all these wavelets interfering . Where they add up → bright. Where they cancel in a closed loop → dark.
WHY does a single slit make a pattern at all? Because a slit of finite width is really many sources. Two-slit interference is the special case of "two point sources"; single-slit diffraction is "a continuous strip of sources."
Find the intensity I ( θ ) I(\theta) I ( θ ) θ \theta θ = = a = = ==a== == a == λ \lambda λ I ( θ ) = I 0 ( sin β β ) 2 , β = π a sin θ λ . I(\theta) = I_0 \left( \frac{\sin\beta}{\beta} \right)^2, \qquad \beta = \frac{\pi a \sin\theta}{\lambda}. I ( θ ) = I 0 ( β s i n β ) 2 , β = λ π a s i n θ .
Intuition WHY phase mattersTwo wavelets that travel slightly different distances arrive "out of step." If wavelet B travels an extra distance Δ \Delta Δ 2 π λ Δ \dfrac{2\pi}{\lambda}\Delta λ 2 π Δ 2 π 2\pi 2 π
Divide the slit into N N N Δ y = a / N \Delta y = a/N Δ y = a / N y y y 0 0 0 a a a
Path difference of a strip at position y y y relative to the top strip, heading toward angle θ \theta θ Δ ( y ) = y sin θ . \Delta(y) = y\sin\theta. Δ ( y ) = y sin θ .
Worked example Why this step?Drop a perpendicular from the top of the slit onto the ray from a strip at depth y y y y sin θ y\sin\theta y sin θ same geometry as the two-slit path difference d sin θ d\sin\theta d sin θ y y y
So the phase of the strip at y y y ϕ ( y ) = 2 π λ y sin θ . \phi(y) = \frac{2\pi}{\lambda}\, y\sin\theta. ϕ ( y ) = λ 2 π y sin θ .
The total phase difference between the top (y = 0 y=0 y = 0 bottom (y = a y=a y = a Φ = 2 π λ a sin θ = 2 β , β ≡ π a sin θ λ . \boxed{\;\Phi = \frac{2\pi}{\lambda}\,a\sin\theta \;=\; 2\beta\;}, \qquad \beta \equiv \frac{\pi a \sin\theta}{\lambda}. Φ = λ 2 π a sin θ = 2 β , β ≡ λ π a s i n θ .
β \beta β 2 β 2\beta 2 β β \beta β half of it — it will turn out to be the natural variable.
Each strip contributes a tiny field. With N N N A 0 / N A_0/N A 0 / N A 0 A_0 A 0 θ = 0 \theta=0 θ = 0
E = A 0 N ∑ n = 0 N − 1 e i ϕ n , ϕ n = 2 π λ sin θ ⋅ n a N . E = \frac{A_0}{N}\sum_{n=0}^{N-1} e^{i\phi_n}, \qquad \phi_n = \frac{2\pi}{\lambda}\sin\theta\cdot \frac{na}{N}. E = N A 0 ∑ n = 0 N − 1 e i ϕ n , ϕ n = λ 2 π sin θ ⋅ N na .
Take N → ∞ N\to\infty N → ∞ E = A 0 a ∫ 0 a exp ( i 2 π λ y sin θ ) d y . E = \frac{A_0}{a}\int_0^{a} \exp\!\left(i\,\frac{2\pi}{\lambda}y\sin\theta\right) dy. E = a A 0 ∫ 0 a exp ( i λ 2 π y sin θ ) d y .
a a a θ = 0 \theta=0 θ = 0 A 0 A_0 A 0
Let k ′ = 2 π λ sin θ k' = \dfrac{2\pi}{\lambda}\sin\theta k ′ = λ 2 π sin θ E = A 0 a ∫ 0 a e i k ′ y d y = A 0 a ⋅ e i k ′ a − 1 i k ′ . E = \frac{A_0}{a}\int_0^a e^{ik'y}\,dy = \frac{A_0}{a}\cdot \frac{e^{ik'a}-1}{ik'}. E = a A 0 ∫ 0 a e i k ′ y d y = a A 0 ⋅ i k ′ e i k ′ a − 1 .
HOW to simplify: factor out the half-angle (the standard "phasor chord" trick):
e i k ′ a − 1 = e i k ′ a / 2 ( e i k ′ a / 2 − e − i k ′ a / 2 ) = e i k ′ a / 2 ⋅ 2 i sin ( k ′ a 2 ) . e^{ik'a}-1 = e^{ik'a/2}\left(e^{ik'a/2}-e^{-ik'a/2}\right) = e^{ik'a/2}\cdot 2i\sin\!\left(\frac{k'a}{2}\right). e i k ′ a − 1 = e i k ′ a /2 ( e i k ′ a /2 − e − i k ′ a /2 ) = e i k ′ a /2 ⋅ 2 i sin ( 2 k ′ a ) .
Now k ′ a 2 = π a sin θ λ = β \dfrac{k'a}{2} = \dfrac{\pi a\sin\theta}{\lambda} = \beta 2 k ′ a = λ π a sin θ = β E = A 0 a ⋅ e i β 2 i sin β i k ′ = A 0 e i β sin β β E = \frac{A_0}{a}\cdot \frac{e^{i\beta}\,2i\sin\beta}{i k'} = A_0\, e^{i\beta}\,\frac{\sin\beta}{\beta} E = a A 0 ⋅ i k ′ e i β 2 i s i n β = A 0 e i β β s i n β k ′ a / 2 = β ⇒ k ′ = 2 β / a k'a/2=\beta \Rightarrow k' = 2\beta/a k ′ a /2 = β ⇒ k ′ = 2 β / a 1 a ⋅ 2 k ′ = 1 a ⋅ a β = 1 β \dfrac{1}{a}\cdot\dfrac{2}{k'} = \dfrac{1}{a}\cdot\dfrac{a}{\beta}=\dfrac{1}{\beta} a 1 ⋅ k ′ 2 = a 1 ⋅ β a = β 1
Worked example Why this step?The phasor chord = full circle. Summing many small phasors of equal length but uniformly increasing phase traces an arc; the resultant is the chord, and the chord length of an arc subtending total angle 2 β 2\beta 2 β ∝ sin β / β \propto \sin\beta/\beta ∝ sin β / β
Intensity = (amplitude)². The phase factor e i β e^{i\beta} e i β I ( θ ) = I 0 ( sin β β ) 2 , β = π a sin θ λ , I 0 = A 0 2 . \boxed{\;I(\theta) = I_0\left(\frac{\sin\beta}{\beta}\right)^2, \qquad \beta = \frac{\pi a\sin\theta}{\lambda}, \quad I_0 = A_0^2.\;} I ( θ ) = I 0 ( β sin β ) 2 , β = λ π a sin θ , I 0 = A 0 2 .
Intuition Central maximum (WHY it's brightest)As θ → 0 \theta\to 0 θ → 0 β → 0 \beta\to 0 β → 0 sin β β → 1 \dfrac{\sin\beta}{\beta}\to 1 β sin β → 1 I → I 0 I\to I_0 I → I 0
Minima (dark fringes): I = 0 I=0 I = 0 sin β = 0 \sin\beta=0 sin β = 0 β ≠ 0 \beta\neq 0 β = 0 β = m π \beta = m\pi β = mπ m = ± 1 , ± 2 , … m=\pm1,\pm2,\dots m = ± 1 , ± 2 , … π a sin θ λ = m π ⇒ a sin θ = m λ , m = ± 1 , ± 2 , … \frac{\pi a\sin\theta}{\lambda}=m\pi \;\Rightarrow\; \boxed{a\sin\theta = m\lambda}, \quad m=\pm1,\pm2,\dots λ π a s i n θ = mπ ⇒ a sin θ = mλ , m = ± 1 , ± 2 , …
a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ m m m whole-slit path difference is one full λ \lambda λ λ / 2 \lambda/2 λ /2 d sin θ = m λ d\sin\theta=m\lambda d sin θ = mλ bright .
Secondary maxima: approximately where tan β = β \tan\beta=\beta tan β = β β ≈ 1.43 π , 2.46 π , … \beta\approx 1.43\pi,\,2.46\pi,\dots β ≈ 1.43 π , 2.46 π , … I I 0 = ( sin ( 1.43 π ) 1.43 π ) 2 ≈ 1 ( 1.5 π ) 2 ≈ 0.045 = 4.5 % . \frac{I}{I_0}=\left(\frac{\sin(1.43\pi)}{1.43\pi}\right)^2 \approx \frac{1}{(1.5\pi)^2}\approx 0.045 = 4.5\%. I 0 I = ( 1.43 π s i n ( 1.43 π ) ) 2 ≈ ( 1.5 π ) 2 1 ≈ 0.045 = 4.5%.
Angular width of central maximum: between m = ± 1 m=\pm1 m = ± 1 sin θ = ± λ a ⇒ half-width θ 1 ≈ λ a ( small θ ) . \sin\theta = \pm\frac{\lambda}{a}\;\Rightarrow\; \text{half-width } \theta_1\approx \frac{\lambda}{a}\ \ (\text{small }\theta). sin θ = ± a λ ⇒ half-width θ 1 ≈ a λ ( small θ ) .
Intuition WHY narrow slit ⇒ wide patternθ 1 ≈ λ / a \theta_1 \approx \lambda/a θ 1 ≈ λ / a a a a spread out more — a hallmark of wave behaviour (and a baby version of the uncertainty principle: confining position Δ y ∼ a \Delta y\sim a Δ y ∼ a
Worked example Example 1 — Where is the first dark fringe?λ = 600 \lambda=600 λ = 600 a = 0.10 a=0.10 a = 0.10 Why: first minimum uses m = 1 m=1 m = 1 a sin θ = λ a\sin\theta=\lambda a sin θ = λ sin θ = λ a = 600 × 10 − 9 0.10 × 10 − 3 = 6 × 10 − 3 . \sin\theta=\frac{\lambda}{a}=\frac{600\times10^{-9}}{0.10\times10^{-3}}=6\times10^{-3}. sin θ = a λ = 0.10 × 1 0 − 3 600 × 1 0 − 9 = 6 × 1 0 − 3 . θ ≈ 0.34 ∘ \theta\approx 0.34^\circ θ ≈ 0.3 4 ∘ Why small? Because a ≫ λ a\gg\lambda a ≫ λ
Worked example Example 2 — Linear width of central max on a screenSame slit, screen distance L = 2.0 L=2.0 L = 2.0 m = − 1 m=-1 m = − 1 m = + 1 m=+1 m = + 1 Why: position y ≈ L tan θ ≈ L θ y\approx L\tan\theta\approx L\theta y ≈ L tan θ ≈ L θ y 1 = L λ / a y_1 = L\lambda/a y 1 = L λ / a y 1 = 2.0 × 600 × 10 − 9 0.10 × 10 − 3 = 0.012 m = 12 mm . y_1 = \frac{2.0\times 600\times10^{-9}}{0.10\times10^{-3}}=0.012\text{ m}=12\text{ mm}. y 1 = 0.10 × 1 0 − 3 2.0 × 600 × 1 0 − 9 = 0.012 m = 12 mm . = 2 y 1 = 24 =2y_1=24 = 2 y 1 = 24 Why double? Symmetric on both sides.
Worked example Example 3 — Intensity at a given angleFind I / I 0 I/I_0 I / I 0 θ \theta θ a sin θ = λ / 2 a\sin\theta=\lambda/2 a sin θ = λ /2 Why: compute β = π a sin θ λ = π 2 \beta=\dfrac{\pi a\sin\theta}{\lambda}=\dfrac{\pi}{2} β = λ π a sin θ = 2 π I I 0 = ( sin ( π / 2 ) π / 2 ) 2 = ( 1 π / 2 ) 2 = 4 π 2 ≈ 0.405. \frac{I}{I_0}=\left(\frac{\sin(\pi/2)}{\pi/2}\right)^2=\left(\frac{1}{\pi/2}\right)^2=\frac{4}{\pi^2}\approx 0.405. I 0 I = ( π /2 s i n ( π /2 ) ) 2 = ( π /2 1 ) 2 = π 2 4 ≈ 0.405. a sin θ = λ a\sin\theta=\lambda a sin θ = λ
Common mistake "Bright fringes are ata sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ Why it feels right: In Young's double slit, d sin θ = m λ d\sin\theta=m\lambda d sin θ = mλ bright , so students transfer the rule.
Fix: For a single slit, a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ DARK fringes. Reason: a slit is a continuous set of sources that cancel in pairs when the full-width path difference is a whole number of wavelengths. Different physics → opposite rule.
m = 0 m=0 m = 0 Why it feels right: the formula a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ m = 0 m=0 m = 0 Fix: m = 0 m=0 m = 0 β = 0 \beta=0 β = 0 sin β / β → 1 \sin\beta/\beta\to 1 sin β / β → 1 central bright maximum , not a minimum. Minima require m = ± 1 , ± 2 , … m=\pm1,\pm2,\dots m = ± 1 , ± 2 , …
Common mistake "Secondary maxima sit exactly halfway between minima atβ = ( m + 1 2 ) π \beta=(m+\tfrac12)\pi β = ( m + 2 1 ) π Why it feels right: sounds symmetric and tidy.
Fix: True maxima satisfy tan β = β \tan\beta=\beta tan β = β β ≈ 1.43 π \beta\approx1.43\pi β ≈ 1.43 π 1.5 π 1.5\pi 1.5 π slightly shifted toward the central peak because the 1 / β 1/\beta 1/ β
a a a d d d Fix: a a a diffraction envelope (single slit). d d d interference fringes (two slits). In a real double-slit experiment both appear together — diffraction modulates the interference.
#flashcards/physics
What is the single-slit intensity formula? I = I 0 ( sin β β ) 2 I=I_0\left(\dfrac{\sin\beta}{\beta}\right)^2 I = I 0 ( β sin β ) 2 with
β = π a sin θ λ \beta=\dfrac{\pi a\sin\theta}{\lambda} β = λ π a sin θ What does β \beta β Half of the total phase difference between the top and bottom edges of the slit.
Condition for dark fringes (minima)? a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ , with
m = ± 1 , ± 2 , … m=\pm1,\pm2,\dots m = ± 1 , ± 2 , … (NOT
m = 0 m=0 m = 0 ).
Why is a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ Wavelets pair up across the slit and cancel exactly (
λ / 2 \lambda/2 λ /2 apart in pairs).
What happens at θ = 0 \theta=0 θ = 0 β → 0 \beta\to0 β → 0 ,
sin β / β → 1 \sin\beta/\beta\to1 sin β / β → 1 , intensity
= I 0 =I_0 = I 0 : central maximum (brightest).
Angular half-width of central maximum? θ 1 ≈ λ / a \theta_1\approx\lambda/a θ 1 ≈ λ / a (small angle).
Effect of decreasing slit width a a a Pattern spreads wider (
θ 1 ∝ 1 / a \theta_1\propto1/a θ 1 ∝ 1/ a ); more diffraction.
Relative intensity of first secondary maximum? About
4.5 % 4.5\% 4.5% of
I 0 I_0 I 0 (at
β ≈ 1.43 π \beta\approx1.43\pi β ≈ 1.43 π ).
Exact condition for secondary maxima? tan β = β \tan\beta=\beta tan β = β (not
β = ( m + 1 2 ) π \beta=(m+\tfrac12)\pi β = ( m + 2 1 ) π ).
Intensity at a sin θ = λ / 2 a\sin\theta=\lambda/2 a sin θ = λ /2 ( 2 / π ) 2 ≈ 0.405 I 0 (2/\pi)^2\approx0.405\,I_0 ( 2/ π ) 2 ≈ 0.405 I 0 .
Difference between a a a d d d a a a =slit width → diffraction envelope;
d d d =slit separation → interference fringes.
Recall Feynman: explain to a 12-year-oldImagine a wide doorway and a crowd of tiny torch-people standing shoulder to shoulder across it, all flashing light at the same instant toward a far wall. If you stand straight ahead, every torch's light arrives together and it's super bright. But if you walk sideways, the light from one edge of the doorway has to travel a bit farther than the light from the other edge. When that extra distance is exactly one full "wave step," the torches on the left half line up perfectly opposite to the torches on the right half — left says "up," right says "down," and they cancel to zero : a dark stripe. Squeeze the doorway narrow and the bright patch on the wall gets wider , because narrow gaps make waves fan out more. That spreading is what diffraction is.
Mnemonic Remember the rule flip"Single slit: same formula, opposite result."
a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ DARK (single slit) vs d sin θ = m λ d\sin\theta=m\lambda d sin θ = mλ BRIGHT (double slit).
And "Beta zero, brightest hero" — β = 0 \beta=0 β = 0
Young's Double Slit Experiment — interference; single-slit is the continuous-source generalization.Huygens Principle — provides the wavelets we summed.Phasor Addition of Waves — the chord-of-an-arc method used in Step 2.Diffraction Grating — many slits; envelope here multiplies the grating's sharp peaks.Rayleigh Criterion and Resolution — uses θ ≈ 1.22 λ / D \theta\approx1.22\lambda/D θ ≈ 1.22 λ / D λ / a \lambda/a λ / a Fraunhofer vs Fresnel Diffraction — our far-screen (parallel ray) assumption is Fraunhofer.Heisenberg Uncertainty Principle — narrow slit ⇒ wide angular spread is its optical analogue.
Continuous strip of Huygens wavelets
Path difference y sin theta
Total phase spread 2 beta
Beta = pi a sin theta over lambda
I0 times sin beta over beta squared
Intuition Hinglish mein samjhoDekho, single slit ka matlab hai ek chhoti si gap jisme se light guzarti hai. Galti yeh hoti hai ki hum slit ko ek single source samajh lete hain — par actually woh ek continuous patti hai jisme infinite chhote-chhote Huygens sources bhare hue hain. Har source se light screen tak alag-alag distance travel karti hai, isliye unka phase alag hota hai, aur ye saare wavelets aapas mein interfere karte hain. Jahan sab milke add hote hain — bright; jahan aapas mein cancel ho jaate hain — dark.
Derivation ka dil yeh hai: top edge aur bottom edge ke beech ka total path difference hota hai a sin θ a\sin\theta a sin θ 2 π λ a sin θ = 2 β \frac{2\pi}{\lambda}a\sin\theta = 2\beta λ 2 π a sin θ = 2 β A 0 sin β β A_0\frac{\sin\beta}{\beta} A 0 β s i n β I = I 0 ( sin β / β ) 2 I=I_0(\sin\beta/\beta)^2 I = I 0 ( sin β / β ) 2
Sabse important baat (aur sabse common galti): single slit mein a sin θ = m λ a\sin\theta=m\lambda a sin θ = mλ DARK fringe milti hai, bright nahi! Kyunki slit ko aadha-aadha baant do — upar wale half ka har wavelet neeche wale half ke apne partner se exactly λ / 2 \lambda/2 λ /2 d sin θ = m λ d\sin\theta=m\lambda d sin θ = mλ
Practical importance: agar slit a a a θ 1 ≈ λ / a \theta_1 \approx \lambda/a θ 1 ≈ λ / a β \beta β
