Fraunhofer diffraction

Shine a laser through a narrow slit and the spot on a distant screen is not a sharp line but a bright central band flanked by faint, evenly-spaced side bands — light bending into the geometric shadow. This is diffraction, and its cleanest case is the single slit viewed in the far field, called Fraunhofer diffraction. Every point across the slit width acts as a Huygens source, and adding up their wavelets at each viewing angle produces a precise sinc-squared intensity pattern with sharply located dark fringes. This article derives that pattern, shows how to locate its minima, and contrasts the far-field (Fraunhofer) regime with the near-field (Fresnel) regime where the simple formula breaks down.

The core insight

A single slit of width $a$ produces a far-field intensity pattern $I(\theta)=I_0(\sin\alpha/\alpha)^2$ with $\alpha=\pi a\sin\theta/\lambda$ , whose dark fringes sit at $a\sin\theta=m\lambda$ for integer $m\neq0$ .

The core idea

Picture a slit of width $a$ illuminated by a plane wave of wavelength $\lambda$ . In the Fraunhofer limit, the screen is effectively at infinity — either it really is very far away, or a converging lens brings the parallel rays leaving at angle $\theta$ to a single focal point — so every ray direction $\theta$ corresponds to one clean interference calculation, not a smeared-out near-field mess. Treat every point across the slit as a Huygens secondary source. A ray leaving the slit at angle $\theta$ from a point a distance $x$ from the slit's edge travels a path that is shorter or longer than its neighbours by an amount proportional to $x\sin\theta$ , so the wavelets arrive at the screen with a phase that varies linearly across the slit width. Summing (integrating) this continuous fan of phasors gives a resultant amplitude proportional to $\sin\alpha/\alpha$ , where

\alpha=\frac{\pi a\sin\theta}{\lambda}.

where: $a$ is the slit width, $\theta$ is the angle from the slit normal to the observation direction, and $\lambda$ is the wavelength of the light. Squaring the amplitude gives the observed intensity,

I(\theta)=I_0\left(\frac{\sin\alpha}{\alpha}\right)^2,

where: $I_0$ is the intensity straight ahead at $\theta=0$ , where every wavelet arrives in phase and the phasor sum is maximal. As $\theta$ grows, $I(\theta)$ collapses to zero, climbs back to a much weaker secondary maximum, falls to zero again, and repeats — the classic single-slit diffraction envelope. The dark fringes (minima) occur wherever $\sin\alpha=0$ but $\alpha\neq0$ , i.e. at $\alpha=m\pi$ for integer $m\neq0$ . Substituting the definition of $\alpha$ turns this into the compact, exam-ready condition

a\sin\theta=m\lambda,\qquad m=\pm1,\pm2,\pm3,\dots

Physically, this is exactly the condition under which the slit can be sliced into $2m$ equal strips whose wavelets cancel pairwise, strip by strip. The central maximum is twice as angularly wide as any side lobe, and the secondary maxima die off fast — the first one carries only about $4.7\%$ of $I_0$ , because the phasor sum never closes into a full circle away from $\theta=0$ . Fraunhofer diffraction is valid when the wavefronts at the slit and at the screen are effectively flat, which holds once the screen distance $L$ satisfies $L \gg a^2/\lambda$ , or once a lens images the far field onto a nearby screen. Fresnel (near-field) diffraction is the general case where the wavefronts curve appreciably across the aperture, the phase is no longer a simple linear function of position, and the resulting patterns (a straight edge, a circular aperture at finite distance) need Fresnel-integral or zone-construction methods instead of the clean sinc-squared result above. The two regimes are best remembered as the far-field/plane-wave and near-field/curved-wavefront limits of the same underlying Huygens–Fresnel superposition.

Figure

Single slit diffraction intensity pattern showing a tall central peak and much smaller side lobes at the minima positions

Derivation ALocating the single-slit diffraction minima

The goal of this derivation is to show, using only the phasor picture of wave addition and no calculus beyond recognising $\sin\alpha/\alpha$ as the amplitude sum across the slit, exactly why the dark fringes of a single-slit Fraunhofer pattern sit at the angles given by $a\sin\theta=m\lambda$ for nonzero integer $m$ , and why $\theta=0$ is never one of them.

Step 1. Split the slit of width $a$ into many thin strips, each acting as a Huygens source. At an observation angle $\theta$ , a strip at position $x$ (measured from one edge) sends out a ray whose extra path length, relative to the strip at $x=0$ , is $x\sin\theta$ . That path difference becomes a phase difference $\delta=\frac{2\pi}{\lambda}x\sin\theta$ , so the total phase spread across the full slit width is $\phi=\frac{2\pi a\sin\theta}{\lambda}=2\alpha$ .

Step 2. Add up the wavelets as phasors. Because the phase increases linearly and smoothly across the slit, the phasor diagram is a circular arc of total length proportional to $a$ and total angular sweep $\phi=2\alpha$ . The resultant amplitude is the chord of that arc, which standard geometry gives as proportional to $\sin\alpha/\alpha$ — this is the origin of the sinc function, and squaring it gives $I(\theta)=I_0(\sin\alpha/\alpha)^2$ .

Step 3. Find where the resultant amplitude vanishes. The chord length is zero exactly when the arc closes into a complete loop (or an integer number of complete loops), i.e. when the total sweep $\phi=2\alpha$ is an integer multiple of $2\pi$ other than zero. This means $\alpha=m\pi$ for $m=\pm1,\pm2,\pm3,\dots$ — the case $m=0$ is excluded because $\alpha=0$ is the removable singularity where $\sin\alpha/\alpha\to1$ , the central maximum, not a closed loop.

Step 4. Translate back into slit width and wavelength. Substituting $\alpha=\pi a\sin\theta/\lambda=m\pi$ and cancelling $\pi$ gives the minima condition $a\sin\theta=m\lambda$ . An equivalent, more physical way to see this: at each such angle, the slit can be divided into $2m$ equal strips such that each strip's wavelet is exactly cancelled by the wavelet from the corresponding strip one path-difference-of- $\lambda/2$ away, so the net amplitude vanishes strip pair by strip pair.

Worked example

A single slit of width $a=0.20\ \text{mm}$ is illuminated by light of wavelength $\lambda=600\ \text{nm}$ , and the pattern is viewed on a screen $L=1.5\ \text{m}$ away.

Step 1. Find the first minimum using $a\sin\theta_1=m\lambda$ with $m=1$ :

\sin\theta_1=\frac{\lambda}{a}=\frac{600\times10^{-9}}{0.20\times10^{-3}}=3.0\times10^{-3}.

Since this is small, $\theta_1\approx3.0\times10^{-3}\ \text{rad}=0.172^\circ$ .

Step 2. Find the second minimum with $m=2$ : $\sin\theta_2=2\lambda/a=6.0\times10^{-3}$ , so $\theta_2\approx6.0\times10^{-3}\ \text{rad}=0.344^\circ$ .

Step 3. Convert the first minimum to a linear position on the screen. For small angles, $y\approx L\tan\theta\approx L\sin\theta$ , so

y_1=L\sin\theta_1=1.5\times3.0\times10^{-3}=4.5\times10^{-3}\ \text{m}=4.5\ \text{mm}.

The full central maximum, spanning from the $m=-1$ minimum to the $m=+1$ minimum, therefore has width $2y_1=9.0\ \text{mm}$ — the bright envelope dominating the screen.

Competitive stretch

Now suppose the slit width is instead $a=0.15\ \text{mm}$ and a second-order minimum ( $m=2$ ) is observed at $\theta=0.50^\circ$ ; find the wavelength. Rearranging the minima condition gives $\lambda=a\sin\theta/m$ . Converting $\theta=0.50^\circ=0.00873\ \text{rad}$ , so $\sin\theta\approx0.00873$ , and

\lambda=\frac{(0.15\times10^{-3})(0.00873)}{2}=6.5\times10^{-7}\ \text{m}=650\ \text{nm}.

This lies in the visible red band ( $\sim620$ – $750\ \text{nm}$ ), consistent with a red laser or filtered source.

Common traps

Confusing the single-slit diffraction minima condition $a\sin\theta=m\lambda$ with the double-slit interference maxima condition $d\sin\theta=m\lambda$ — the two formulas look identical but describe opposite effects (dark vs bright) in different setups; keep slit width $a$ and slit separation $d$ conceptually distinct.
Treating $m=0$ as a minimum — $\alpha=0$ at $\theta=0$ is the central maximum (the removable singularity of $\sin\alpha/\alpha\to1$ ), not a dark fringe.
Assuming secondary maxima are evenly spaced or comparably bright to the central peak — they sit roughly midway between successive minima and fall off rapidly, with the first one at only about $4.7\%$ of $I_0$ .
Using the Fraunhofer sinc-squared formula at short screen distances without checking $L\gg a^2/\lambda$ — at short range the pattern is Fresnel diffraction and the simple formula no longer applies.

Ask the tutor: Want to see the phasor-arc picture worked through step by step, or try a fresh minima-location problem with different numbers?

Why does the central maximum look twice as wide as the side lobes?

The minima sit at $\alpha=m\pi$ for $m=\pm1,\pm2,\dots$ , so the two minima bounding the central maximum are at $m=-1$ and $m=+1$ , one full unit of $\alpha=\pi$ on each side of zero. Every other bright lobe, by contrast, sits between two consecutive minima that are both nonzero integers apart by only one unit of $\pi$ — so it spans the same $\alpha$ -width as one side of the central peak, making the full central maximum exactly twice as wide in $\theta$ (or in $\alpha$ ) as any secondary maximum.

How is single-slit diffraction different from double-slit interference if the formulas look so similar?

They come from different physics even though $a\sin\theta=m\lambda$ and $d\sin\theta=m\lambda$ look alike. Double-slit interference adds contributions from two point-like sources separated by $d$ , giving bright fringes at $d\sin\theta=m\lambda$ . Single-slit diffraction instead adds a continuum of sources spread across one aperture of width $a$ , giving dark fringes (not bright ones) at $a\sin\theta=m\lambda$ . In a real double slit with finite slit width, both effects combine: an interference pattern from the two slits sits inside a diffraction envelope from each individual slit's width.

When does the simple Fraunhofer formula stop being valid?

It stops being valid once the far-field condition $L\gg a^2/\lambda$ fails — for example, if the screen is placed too close to the slit, or the slit is unusually wide relative to the wavelength. In that near-field regime the wavefronts curve noticeably across the aperture, the phase is no longer a simple linear function of position across the slit, and you must use Fresnel-diffraction methods (Fresnel integrals or zone constructions) instead of the sinc-squared intensity formula used here.

Fraunhofer diffraction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads