Single-slit diffraction

Shine light through a single narrow slit and something strange happens: instead of a sharp-edged shadow, the screen shows a broad, bright central band flanked by faint, evenly spaced side bands. This is single-slit diffraction — light bending around and spreading past an obstacle, the clearest everyday proof that light behaves as a wave. It sits at the end of the Class XII wave optics unit, right after Young's double-slit experiment (YDSE), and the two are constantly compared and confused in exams. Here every point across the slit's width acts as a fresh wave source (Huygens' principle), and the pattern you see is what happens when countless such wavelets interfere with each other. The goal of this article is modest but exam-critical: know the condition for the dark fringes (minima), know how wide the central bright band is, and never mix this pattern up with YDSE's evenly-spaced, evenly-bright fringes.

The core insight

Every point across a single slit of width $a$ sends out Huygens wavelets that interfere; pairwise cancellation across the slit gives minima at $a\sin\theta_n=n\lambda$ ( $n\neq0$ ), producing one wide, bright central maximum flanked by much fainter secondary bands.

The core idea

Setup. A plane wavefront (parallel rays — the Fraunhofer condition, achieved with a distant source/screen or a lens) falls on a narrow slit of width $a$ . By Huygens' principle, every point across that width radiates secondary wavelets in all forward directions. On a distant screen these wavelets superpose, and the resulting intensity pattern is the single-slit diffraction pattern.

Why there is a bright centre. Straight ahead, at $\theta=0$ , every wavelet has travelled the same path length from the slit to the screen, so they all arrive in phase and add constructively. This is the central maximum — broad and by far the most intense part of the pattern.

Why there are dark fringes (minima). Off to one side, at some angle $\theta$ , the extreme wavelets (from the top and bottom edges of the slit) differ in path length by $a\sin\theta$ . Split the slit into two equal halves. Wavelets from corresponding points in the two halves (each pair a distance $a/2$ apart) then differ in path by $\dfrac{a}{2}\sin\theta$ . When this equals $\lambda/2$ , every wavelet from the top half is exactly cancelled by its partner in the bottom half — total darkness at that angle. Extending this halves-within-halves cancellation argument (dividing the slit into $2n$ strips) gives the general condition for the $n$ -th minimum:

a\sin\theta_n = n\lambda \qquad (n=\pm1,\pm2,\pm3,\ldots)

where: $a$ is the slit width, $\theta_n$ is the angle of the $n$ -th dark fringe from the centre, $\lambda$ is the wavelength of light, and $n$ is a non-zero integer. Note carefully that $n=0$ is deliberately excluded — that direction is the centre of the bright central maximum, not a minimum.

Width of the central maximum. The central bright band runs from the first minimum on one side ( $n=-1$ ) to the first minimum on the other ( $n=+1$ ). For small angles, $\sin\theta\approx\theta$ , so the first minimum sits at $\theta_1\approx\lambda/a$ . If the screen is a distance $D$ from the slit, the linear half-width on the screen is $D\theta_1$ , so the total width of the central maximum is:

w=\frac{2\lambda D}{a}

where: $w$ is the linear width of the central bright band on the screen, $\lambda$ is the wavelength, $D$ is the slit-to-screen distance, and $a$ is the slit width. This central band is twice as wide as each secondary maximum flanking it, and carries most of the diffracted energy — the secondary maxima fall off rapidly in brightness further out.

Figure

Intensity pattern of single slit diffraction showing a tall central maximum and much smaller side maxima at the minima positions

Derivation of the width of the central maximum

Derivation ACentral maximum width from the minima condition

The goal is to show, using only the small-angle approximation and the minima condition $a\sin\theta_n=n\lambda$ already established, that the linear width of the bright central band on a screen at distance $D$ works out to $w=2\lambda D/a$ — a single formula students must be able to reproduce and apply directly in numericals without re-deriving it each time.

Step 1. Locate the first minimum. Setting $n=1$ in $a\sin\theta_n=n\lambda$ gives $a\sin\theta_1=\lambda$ , so $\sin\theta_1=\lambda/a$ . Since $\lambda\ll a$ is not always true here, but the angle itself stays small in typical setups, apply the small-angle approximation $\sin\theta_1\approx\theta_1$ , giving $\theta_1=\lambda/a$ radians.

Step 2. Convert the angle to a screen distance. A ray leaving the slit at angle $\theta_1$ strikes the screen, placed a distance $D$ away, at a linear distance $y_1=D\tan\theta_1\approx D\theta_1$ from the centre (again using the small-angle form, this time for tangent). Substituting $\theta_1=\lambda/a$ gives $y_1=D\lambda/a$ — the half-width of the central maximum, measured from the centre to the first dark fringe on one side.

Step 3. Double it for the full width. The central maximum is symmetric, stretching from the first minimum at $n=-1$ to the first minimum at $n=+1$ , so the total width is twice the half-width found in Step 2:

w=2y_1=\frac{2\lambda D}{a}

Step 4. Sanity-check the trend. Larger $\lambda$ (redder light) or larger $D$ (farther screen) widens the central band; a larger slit width $a$ narrows it — a wider opening diffracts light less, which matches the everyday observation that diffraction effects are most visible when the opening is comparable in size to the wavelength.

Worked example

Light of wavelength 500 nm falls on a single slit of width 0.2 mm, with a screen 2 m away. Find the angular position of the first minimum, the width of the central maximum, and the slit width needed to place the second minimum at 0.006 rad if a different light of wavelength 600 nm is used instead.

Step 1. Convert to SI: $\lambda=500\ \text{nm}=5\times10^{-7}\ \text{m}$ , $a=0.2\ \text{mm}=2\times10^{-4}\ \text{m}$ , $D=2\ \text{m}$ .

Step 2. First minimum angle, using $\theta_1\approx\lambda/a$ : $\theta_1=\dfrac{5\times10^{-7}}{2\times10^{-4}}=2.5\times10^{-3}\ \text{rad}$ .

Step 3. Width of the central maximum: $w=\dfrac{2\lambda D}{a}=2\times(2.5\times10^{-3})\times2=1\times10^{-2}\ \text{m}=1\ \text{cm}$ .

Step 4 (extension). For the second scenario, $n=2$ , $\lambda=600\ \text{nm}=6\times10^{-7}\ \text{m}$ , and the given angle $\theta_2=0.006\ \text{rad}$ . Rearranging $a\theta_n=n\lambda$ (small-angle form): $a=\dfrac{n\lambda}{\theta_2}=\dfrac{2\times6\times10^{-7}}{0.006}=\dfrac{1.2\times10^{-6}}{6\times10^{-3}}=2\times10^{-4}\ \text{m}$ .

Answer: The first minimum is at $2.5\times10^{-3}$ rad, the central maximum is 1 cm wide, and the second scenario needs a slit width of $2\times10^{-4}\ \text{m}=0.2\ \text{mm}$ .

Competitive stretch

If both experiments above used the same slit width $a=2\times10^{-4}\ \text{m}$ , the numbers agreeing exactly is not a coincidence — it shows the minima condition is fully reversible: given any three of $a$ , $\lambda$ , $\theta_n$ , $n$ , the fourth follows directly from $a\sin\theta_n=n\lambda$ . Competitive numericals often chain this with the central-maximum-width formula, asking you to find $a$ from one condition and then $w$ from another — always identify which quantities are shared across the two parts before setting up equations.

Common traps

Confusing single-slit minima with YDSE bright fringes. Single-slit minima follow $a\sin\theta_n=n\lambda$ ( $n\neq0$ ); YDSE bright fringes follow $\Delta x=n\lambda$ (constructive, from two point sources). Same-looking formula, opposite physical meaning — a frequent exam mix-up.
Treating the central maximum as just another fringe. It is twice as wide as each secondary maximum and far brighter; do not assume uniform fringe width or intensity as in YDSE.
Forgetting $n=0$ is excluded. $\theta=0$ is the centre of the bright central maximum, not a minimum — the minima start at $n=\pm1$ .
Applying the two-point-source path-difference method here. Single-slit diffraction needs the pairwise-cancellation argument across the continuous slit width, not the simple two-source geometry used for YDSE.

Ask the tutor: Want the pairwise-cancellation argument walked through step by step with a diagram, or a numerical comparing how far red versus violet light spreads through the same slit?

Why does a narrower slit spread the light out more, not less?

It seems backwards at first, but look at the formula $\theta_1\approx\lambda/a$ : as the slit width $a$ shrinks, the angle to the first minimum grows, so the central bright band widens on the screen. Physically, the path difference across the slit is $a\sin\theta$ , and for a smaller $a$ you need a larger angle $\theta$ before that path difference builds up to $\lambda$ , so the first dark fringe (and hence the whole central band) is pushed further out. The rule of thumb: diffraction becomes obvious once the slit width is comparable to the wavelength.

How is this different from Young's double-slit experiment if both use path differences and integer n?

The formulas look structurally similar, but the physics is opposite. In YDSE, two narrow, coherent point sources interfere, and $\Delta x=n\lambda$ gives bright fringes of roughly equal width and intensity. In single-slit diffraction, a continuous, extended slit produces infinitely many wavelets across its width, and $a\sin\theta_n=n\lambda$ gives dark fringes (minima), with one dominant bright band in the centre and rapidly weakening side bands. Same-looking condition, different phenomenon — always check whether the setup is two slits or one before applying a formula.

Why is n = 0 excluded from the minima formula but included in YDSE's bright fringe formula?

In YDSE, $n=0$ in $\Delta x=n\lambda$ correctly gives the central bright fringe, because zero path difference between the two sources means constructive interference there. In single-slit diffraction, the minima condition $a\sin\theta_n=n\lambda$ comes from a different argument — pairwise cancellation across the slit — and at $\theta=0$ there is no path difference to cancel anything, so all wavelets add constructively instead. That is exactly the central maximum, not a minimum, which is why $n=0$ is explicitly excluded from the minima list.