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Concept 04 / 07Wave Optics

Diffraction at a Single Slit: Minima, the Central Maximum, and Why Diffraction and Interference Are the Same Physics

Deriving a sin(theta) = n lambda by pairing sub-wavelets, the phasor result I = I0 (sin b / b) squared, why the central maximum is twice as wide, and how diffraction limits resolution.

Figure 1
Diffraction is the bending of waves around obstacles and into the geometrical shadow, and it is unavoidable whenever a wave is restricted by an aperture comparable in size to its wavelength. The reason is already in Huygens' principle: the moment you block part of a wavefront, the surviving part still acts as a collection of secondary sources, and those wavelets spread sideways. The sharp-edged shadow of ray optics is only the limit in which λ\lambda is vanishingly small compared with the aperture. Two regimes exist. In Fresnel (near-field) diffraction the screen is close and the wavefronts reaching it are curved; in Fraunhofer (far-field) diffraction the source and screen are effectively at infinity, achieved in practice with a collimating lens and a focusing lens, so that the interfering rays are parallel. The JEE syllabus deals with the Fraunhofer case, which is the one with clean formulas. Deriving the minima by pairing. Let a plane wave fall normally on a slit of width aa and consider light leaving the slit at an angle θ\theta. Two wavelets from points separated by a distance ss across the slit have a path difference ssin⁡θs\sin\theta; across the whole slit, from one edge to the other, the path difference is asin⁡θa\sin\theta. Now split the slit into two equal halves. Pair the point at the top of the upper half with the point at the top of the lower half, the second point of the upper half with the second point of the lower half, and so on. Every such pair is separated by exactly a/2a/2, so every pair has the same path difference a2sin⁡θ\dfrac{a}{2}\sin\theta. If we choose θ\theta so that this equals λ/2\lambda/2, every single pair cancels exactly, and since the pairing covers the slit completely, the total amplitude at that angle is zero. The condition is a2sin⁡θ=λ2\dfrac{a}{2}\sin\theta = \dfrac{\lambda}{2}, that is asin⁡θ=λa\sin\theta = \lambda: the first minimum. Repeat with four equal parts: pairing the first part against the second and the third against the fourth cancels everything if a4sin⁡θ=λ2\dfrac{a}{4}\sin\theta = \dfrac{\lambda}{2}, that is asin⁡θ=2λa\sin\theta = 2\lambda. In general, dividing into 2n2n equal parts gives asin⁡θ=nλa\sin\theta = n\lambda, with n=±1,±2,±3,…n = \pm1, \pm2, \pm3, \ldots The exclusion n≠0n \neq 0 is not a technicality, it is the whole point: θ=0\theta = 0 makes every path difference zero, so all the wavelets add in phase and give the central maximum, not a minimum. Students who write "minima at asin⁡θ=nλa\sin\theta = n\lambda, n=0,1,2…n = 0, 1, 2\ldots" have deleted the brightest part of the pattern. The full intensity formula by phasors. Divide the slit into a very large number of equal strips, each sending a wavelet of the same small amplitude, with a constant phase step between neighbours. Adding such phasors head to tail produces a circular arc whose total length equals A0A_0, the amplitude when everything is in phase. If the total phase spread from one edge of the slit to the other is 2β2\beta, where β=πasin⁡θλ\beta = \dfrac{\pi a\sin\theta}{\lambda}, then the arc of length A0A_0 subtends an angle 2β2\beta at its centre, so its radius is R=A02βR = \dfrac{A_0}{2\beta} and the resultant, which is the chord, is A=2Rsin⁡β=A0sin⁡ββA = 2R\sin\beta = A_0\dfrac{\sin\beta}{\beta}. Hence I=I0(sin⁡ββ)2I = I_0\left(\dfrac{\sin\beta}{\beta}\right)^2. This single formula contains everything. It equals I0I_0 at β→0\beta \to 0 (the central maximum, using sin⁡β/β→1\sin\beta/\beta \to 1), and it vanishes whenever β=nπ\beta = n\pi, that is asin⁡θ=nλa\sin\theta = n\lambda, reproducing the pairing result. Secondary maxima lie roughly midway between minima, at asin⁡θ≈(2n+1)λ2a\sin\theta \approx (2n+1)\dfrac{\lambda}{2}; solving dIdβ=0\dfrac{dI}{d\beta} = 0 exactly gives tan⁡β=β\tan\beta = \beta, whose roots are β=1.4303π, 2.4590π, 3.4707π\beta = 1.4303\pi,\ 2.4590\pi,\ 3.4707\pi, giving relative intensities 0.04720.0472, 0.01650.0165 and 0.00830.0083. The secondary maxima are thus feeble: the first carries under 5%5\% of the central peak, which is why the pattern looks like one bright band with faint ripples. Width of the central maximum, and why it is twice the rest. The central maximum runs from the first minimum on one side to the first minimum on the other, that is from sin⁡θ=−λ/a\sin\theta = -\lambda/a to sin⁡θ=+λ/a\sin\theta = +\lambda/a: its angular width is 2θ1=2λa2\theta_1 = \dfrac{2\lambda}{a} and its linear width on a screen at distance DD is 2λDa\dfrac{2\lambda D}{a} (or 2fλa\dfrac{2f\lambda}{a} with a focusing lens of focal length ff). Every other maximum, however, is squeezed between two consecutive minima on the same side, say nn and n+1n+1, which are separated in sin⁡θ\sin\theta by only λ/a\lambda/a. So each secondary maximum has half the width of the central one. The reason is simply that the central peak is the only one that straddles the axis and therefore owns a minimum-to-minimum interval on both sides, as the graph above shows. The roles of aa and λ\lambda. The pattern width scales as λ/a\lambda/a. Make the slit wider and the diffraction pattern narrows into the geometrical image, which is why we get away with ray optics for everyday apertures. Make it narrower and the light fans out more. If a<λa < \lambda, then sin⁡θ=λ/a>1\sin\theta = \lambda/a > 1 has no solution and there are no minima at all: the light simply spreads over the whole forward hemisphere. Note also the energy bookkeeping when aa is doubled: the amplitude at the centre doubles so the peak intensity quadruples, while the width halves, so the total power through the slit only doubles, exactly as it must since twice as much light is admitted. How far can ray optics be trusted? A beam of width aa spreads by an angle λ/a\lambda/a, so after travelling a distance zz its diffraction spread is zλ/az\lambda/a. This equals the original width when z=a2λz = \dfrac{a^2}{\lambda}, the Fresnel distance zFz_F. For distances much smaller than zFz_F the beam keeps its shape and ray optics is safe; well beyond zFz_F, diffraction dominates. Interference versus diffraction: the same physics, different counting. Both are superposition. In interference we superpose waves from a small number of discrete coherent sources (two slits), and the result is a set of equally spaced fringes of equal intensity with perfectly dark minima. In diffraction we superpose waves from a continuous infinity of secondary sources spread across one wavefront, and the result is a set of unequal maxima falling off rapidly, with a central band twice as wide as the rest. Increase the number of slits from two to many and the interference pattern sharpens into the sharp lines of a grating; let the slits merge into one continuous aperture and you have diffraction. Nature draws no line between them, only we do. In a real double-slit experiment both act together. Each slit of width aa produces its own diffraction envelope, and the two slits separated by dd produce interference within it: I=4I0cos⁡2(πdsin⁡θλ)(sin⁡ββ)2I = 4I_0\cos^2\left(\dfrac{\pi d\sin\theta}{\lambda}\right)\left(\dfrac{\sin\beta}{\beta}\right)^2. If an interference maximum dsin⁡θ=nλd\sin\theta = n\lambda happens to fall exactly on a diffraction minimum asin⁡θ=mλa\sin\theta = m\lambda, that fringe is wiped out. Dividing, this happens when nm=da\dfrac{n}{m} = \dfrac{d}{a}, and those orders nn are called missing orders. Resolving power. Because every aperture diffracts, the image of a point source through a circular lens of diameter DD is not a point but an Airy disc of angular radius 1.22λ/D1.22\lambda/D. Rayleigh's criterion says two point sources are just resolved when the central maximum of one falls on the first minimum of the other, so the smallest resolvable angular separation is θmin=1.22λD\theta_{min} = \dfrac{1.22\lambda}{D}. Bigger telescopes resolve better, and blue light resolves better than red. For a microscope the corresponding limit on the smallest resolvable separation is 1.22λ2μsin⁡θ=0.61λμsin⁡θ\dfrac{1.22\lambda}{2\mu\sin\theta} = \dfrac{0.61\lambda}{\mu\sin\theta}, where μsin⁡θ\mu\sin\theta is the numerical aperture; this is why oil-immersion objectives (raising μ\mu) and short-wavelength illumination are used, and ultimately why electron microscopes, with their tiny de Broglie wavelength, see so much more than light microscopes. Key Takeaways and Formulae: Minima: asin⁡θ=nλa\sin\theta = n\lambda, n=±1,±2,…n = \pm1, \pm2,\ldots (never n=0n = 0). Secondary maxima: approximately asin⁡θ=(2n+1)λ2a\sin\theta = (2n+1)\dfrac{\lambda}{2}; relative intensities 1:0.047:0.017:0.0081 : 0.047 : 0.017 : 0.008. Intensity: I=I0(sin⁡ββ)2I = I_0\left(\dfrac{\sin\beta}{\beta}\right)^2 with β=πasin⁡θλ\beta = \dfrac{\pi a\sin\theta}{\lambda}. Central maximum: angular width 2λa\dfrac{2\lambda}{a}, linear width 2λDa\dfrac{2\lambda D}{a}; all other maxima are half as wide. Fresnel distance zF=a2λz_F = \dfrac{a^2}{\lambda}; ray optics is valid for z≪zFz \ll z_F. Double slit with finite slit width: I=4I0cos⁡2(πdsin⁡θλ)(sin⁡ββ)2I = 4I_0\cos^2\left(\dfrac{\pi d\sin\theta}{\lambda}\right)\left(\dfrac{\sin\beta}{\beta}\right)^2; missing orders at nm=da\dfrac{n}{m} = \dfrac{d}{a}. Rayleigh: θmin=1.22λD\theta_{min} = \dfrac{1.22\lambda}{D} for a circular aperture; microscope limit 0.61λμsin⁡θ\dfrac{0.61\lambda}{\mu\sin\theta}. Common mistake: mixing up dd (slit separation, interference) with aa (slit width, diffraction). In the combined formula, d>ad > a always, so the interference fringes are always finer than the diffraction envelope. Common mistake: treating the single-slit central bright band as a "fringe" of the same width as the others. Brain Teaser 1: Explain from first principles why the central diffraction maximum of a single slit is exactly twice as wide as every other maximum. Then predict what happens to the width, the peak intensity and the total transmitted power if the slit width is doubled, and check that your three answers are mutually consistent. Solution: Width. The minima lie at sin⁡θ=±λa,±2λa,±3λa,…\sin\theta = \pm\dfrac{\lambda}{a}, \pm\dfrac{2\lambda}{a}, \pm\dfrac{3\lambda}{a},\ldots and are therefore equally spaced in sin⁡θ\sin\theta with a step of λ/a\lambda/a. A maximum occupies the interval between two consecutive minima. Every secondary maximum sits between, say, nλ/an\lambda/a and (n+1)λ/a(n+1)\lambda/a, an interval of λ/a\lambda/a. The central maximum, on the other hand, is not bounded by two minima on one side; it stretches from −λ/a-\lambda/a to +λ/a+\lambda/a because there is no minimum at θ=0\theta = 0. Its interval is therefore 2λ/a2\lambda/a, exactly twice as large. The absence of an n=0n = 0 minimum is the sole cause. Doubling aa. First, the width: it scales as λ/a\lambda/a, so the pattern shrinks to half its angular size. Second, the peak intensity: at θ=0\theta = 0 every wavelet from the slit arrives in phase, so the amplitude is proportional to the number of secondary sources, that is to aa. Doubling aa doubles the amplitude at the centre and so multiplies the peak intensity by 44. Third, the power: the slit is twice as wide, so it admits exactly twice as much light. Consistency: the total power is roughly (peak intensity) ×\times (width of the pattern), which scales as a2×1a=aa^2 \times \dfrac{1}{a} = a. So quadrupling the peak while halving the width gives a net doubling of the power, matching the amount of light admitted. All three statements agree, which is the check that the reasoning is sound. The same argument, run backwards, explains why a very narrow slit gives a very broad but very faint pattern. Brain Teaser 2: In a double slit experiment the slit separation is d=4ad = 4a, where aa is the width of each slit. Which interference orders are missing, and how many bright fringes are actually seen within the central diffraction maximum? Solution: Interference maxima occur at dsin⁡θ=nλd\sin\theta = n\lambda and diffraction minima at asin⁡θ=mλa\sin\theta = m\lambda. An interference maximum is wiped out if the two conditions hold at the same angle, so dividing them, nm=da=4\dfrac{n}{m} = \dfrac{d}{a} = 4. With m=1,2,3,…m = 1, 2, 3,\ldots the missing orders are n=±4,±8,±12,…n = \pm4, \pm8, \pm12, \ldots Now count the fringes inside the central envelope. The central diffraction maximum extends over ∣sin⁡θ∣<λa|\sin\theta| < \dfrac{\lambda}{a}. Interference maxima inside it satisfy ∣nλd∣<λa\left|\dfrac{n\lambda}{d}\right| < \dfrac{\lambda}{a}, that is ∣n∣<da=4|n| < \dfrac{d}{a} = 4. So n=0,±1,±2,±3n = 0, \pm1, \pm2, \pm3 are inside, which is 77 fringes. The orders n=±4n = \pm 4 would sit exactly on the edge of the envelope, where the diffraction factor is zero, so they are the missing ones and do not appear. Answer: 77 bright fringes inside the central maximum, and every fourth order is absent throughout the pattern. A quick general rule worth remembering: the number of fringes in the central envelope is 2da−12\dfrac{d}{a} - 1, here 2(4)−1=72(4) - 1 = 7. Brain Teaser 3: Light of wavelength 600600 nm falls normally on a slit of width 0.300.30 mm and the pattern is observed on a screen 2.02.0 m away. Find the width of the central maximum and the distance of the first secondary maximum from the centre. Also verify that the far-field (Fraunhofer) treatment is justified at this distance. Solution: Width of the central maximum =2λDa=2×600×10−9×2.00.30×10−3=2.4×10−63.0×10−4=8.0×10−3= \dfrac{2\lambda D}{a} = \dfrac{2 \times 600\times10^{-9}\times 2.0}{0.30\times10^{-3}} = \dfrac{2.4\times10^{-6}}{3.0\times10^{-4}} = 8.0\times10^{-3} m =8.0= 8.0 mm. The first minimum is therefore 4.04.0 mm from the centre, and the second minimum 8.08.0 mm from the centre. The first secondary maximum lies near asin⁡θ=3λ2a\sin\theta = \dfrac{3\lambda}{2}, that is at y≈1.5λDa=1.5×4.0=6.0y \approx \dfrac{1.5\lambda D}{a} = 1.5 \times 4.0 = 6.0 mm. (Using the exact root β=1.4303π\beta = 1.4303\pi gives y=1.4303×4.0=5.72y = 1.4303 \times 4.0 = 5.72 mm, so the "midway" estimate is good to about 5%5\% and is always slightly too far out.) Far-field check: the Fresnel distance is zF=a2λ=(0.30×10−3)2600×10−9=9.0×10−86.0×10−7=0.15z_F = \dfrac{a^2}{\lambda} = \dfrac{(0.30\times10^{-3})^2}{600\times10^{-9}} = \dfrac{9.0\times10^{-8}}{6.0\times10^{-7}} = 0.15 m. Since the screen at 2.02.0 m is more than ten times zFz_F, the beam has spread far beyond its original width and we are comfortably in the Fraunhofer regime, so using asin⁡θ=nλa\sin\theta = n\lambda with sin⁡θ≈y/D\sin\theta \approx y/D is justified. Had the screen been at, say, 55 cm, the pattern would have been a Fresnel pattern and these formulas would not apply.

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Diffraction at a Single Slit: Minima, the Central Maximum, and Why Diffraction and Interference Are the Same Physics — DPP

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