Single-slit diffraction

Point a laser at a razor-thin slit cut into a metal sheet and something odd happens on the screen behind it. Ray optics — the physics of shadows and straight-edged beams — predicts a bright, sharply-edged strip exactly the width of the slit, and darkness everywhere else. What you actually see is nothing like that: a wide, brilliant central band flanked on both sides by a row of much fainter, narrower bands, with genuinely dark gaps between them. Make the slit narrower and, counter to every shadow-drawing instinct, the bright central band gets wider, not narrower. Light has bent around the edges of the opening and spread into the geometric shadow region — a phenomenon ray optics cannot explain at all. This is single-slit diffraction, and it is one of the cleanest demonstrations that light is a wave. The explanation comes straight from Huygens' principle, the same idea CBSE Class 12 uses to prove the laws of reflection and refraction: every point on a wavefront is itself a source of tiny secondary wavelets. In Young's double-slit experiment you treated each of the two slits as effectively a single point source and asked how two such wavelet trains interfere. Here you zoom into a single slit and ask a subtler question — how the (many) wavelets launched from different points across the same opening interfere with each other. The surprising answer is that a single slit, all on its own, produces its own bright-dark interference structure, and the width of its brightest band obeys a strikingly simple rule tying together the slit width $a$ , the wavelength $\lambda$ , and the screen distance $D$ . The CBSE syllabus keeps this topic deliberately light — no intensity-distribution calculus, no derivation of the minima condition from first principles — but it still expects you to use the minima formula confidently, reason correctly about how the pattern widens or narrows, and never confuse this single-slit dark-fringe condition with the double-slit bright-fringe condition from Young's experiment. That confusion is, reliably, the single most common way marks are lost on this topic in board exams.

Key takeaways

Single-slit diffraction is the spreading and self-interference of light passing through one narrow opening of width $a$ , producing a bright central maximum flanked by much fainter secondary maxima, separated by dark minima.
The condition for the minima (dark fringes) is $a\sin\theta_n = n\lambda$ , for $n = 1, 2, 3,\dots$ — this gives dark bands, unlike the double-slit condition which gives bright bands.
The central maximum is twice as wide (in angle) as any secondary maximum: it spans from $\theta = -\theta_1$ to $\theta = +\theta_1$ , while each secondary band occupies only $\lambda/a$ in angle.
The angular width of the central maximum is $2\theta_1 \approx 2\lambda/a$ (small-angle approximation); the corresponding linear width on a screen at distance $D$ is $w \approx 2\lambda D/a$ .
A narrower slit spreads light more (wider central maximum); a wider slit ( $a \gg \lambda$ ) narrows the pattern towards the sharp geometric shadow of ray optics.
CBSE 2025-26 keeps this qualitative: you must reason about and use the minima condition and the width formula, but you are not required to derive the intensity distribution across the secondary maxima.
Key idea: a single slit acts as a continuum of Huygens wavelet sources whose mutual interference produces dark minima at $a\sin\theta_n = n\lambda$ , so the resulting central bright band has angular width $2\lambda/a$ and grows wider as the slit is made narrower.

The core insight

Every point across a single slit of width $a$ radiates a Huygens wavelet, and the wavelets cancel in pairs to give dark minima at $a\sin\theta_n=n\lambda$ , leaving a central bright band of angular width $2\lambda/a$ that widens as the slit narrows.

Where it fits in the syllabus

Single-slit diffraction sits in the Wave Optics chapter, right after Huygens' principle and Young's double-slit experiment, and it builds on both. CBSE's 2025-26 prescription is explicit and narrow here: "diffraction due to a single slit" with the width of the central maximum treated qualitatively only — the full intensity-distribution derivation (the calculus-heavy argument giving the sinc-squared pattern across all the secondary maxima) is not examinable and does not belong in this article. What is expected is that you can state and apply the minima condition $a\sin\theta_n = n\lambda$ , reason qualitatively about why the pattern looks the way it does, and predict how the central maximum's width changes with $a$ , $\lambda$ , and $D$ — in the same spirit as the double-slit fringe-width formula $\beta = \lambda D/d$ , which CBSE also asks you to use without deriving. Diffraction gratings and the resolving power of optical instruments are explicitly out of scope; keep this article's numbers and reasoning to the single-slit case only.

The physical picture

Huygens' principle says a wavefront advances because every point on it behaves as a fresh source of spherical (here, cylindrical, since the slit is a long narrow line) secondary wavelets, and the new wavefront a moment later is the envelope of all these wavelets. Apply this inside a single slit of width $a$ illuminated by a plane wave arriving straight on. The slit itself becomes a continuous line of secondary sources, all starting in phase (since the incoming wavefront is flat when it reaches the slit). Consider the direction straight ahead, at angle $\theta = 0$ measured from the slit's normal. Every wavelet from every point of the slit travels the same distance to a far-away point on the screen there, so they all arrive in phase and add constructively — this builds the central maximum, the brightest and by far the widest feature of the pattern. Now consider light heading off at some angle $\theta$ to that normal. Wavelets from different points across the slit now travel slightly different path lengths to the screen, so they arrive with a spread of phases. For most angles this spread produces partial cancellation and a dim, complicated sum. But at certain special angles the cancellation becomes total: imagine splitting the slit into two equal halves. If the path difference between a wavelet from the very top of the slit and the wavelet from the point exactly halfway down is $\lambda/2$ , then every wavelet in the top half is exactly cancelled by a matching wavelet a distance $a/2$ below it, and the whole slit sums to zero. Reasoning this way (matching pairs across ever finer sub-divisions of the slit) shows that total cancellation — a minimum — occurs whenever the path difference between the two edges of the slit is a whole number of wavelengths:

a\sin\theta_n = n\lambda, \qquad n = 1, 2, 3, \dots

where: $a$ is the slit width, $\theta_n$ is the angle (measured from the slit's normal, i.e. from the straight-ahead direction) at which the $n$ th dark fringe is observed, $\lambda$ is the wavelength of the light used, and $n$ is a positive integer labelling which minimum ( $n=1$ is the first dark fringe on either side, $n=2$ the second, and so on). Notice this looks structurally like the double-slit condition $d\sin\theta = n\lambda$ from Young's experiment, but here it marks dark fringes, not bright ones — the single most important thing to get right about this topic. Between two consecutive minima lies a secondary maximum — a band of light, but much fainter than the central one, because only a fraction of the slit contributes constructively there rather than the whole aperture as at $\theta=0$ . Each secondary maximum occupies an angular width of only $\lambda/a$ (the gap between consecutive $n$ and $n+1$ minima), while the central maximum spans from $\theta=-\theta_1$ to $\theta=+\theta_1$ — twice that angular width, and by far the brightest part of the whole pattern, carrying most of the transmitted light energy. Using the small-angle approximation $\sin\theta \approx \theta$ (valid because $\lambda \ll a$ in typical lab setups, so $\theta_1$ is a small angle in radians), the first minimum sits at

\theta_1 \approx \frac{\lambda}{a},

so the angular width of the central maximum, from $-\theta_1$ to $+\theta_1$ , is

2\theta_1 \approx \frac{2\lambda}{a}.

On a screen placed a distance $D$ from the slit (with $D \gg a$ , the usual lab geometry), this angular spread converts to a linear width on the screen:

w \approx \frac{2\lambda D}{a}.

where: $w$ is the linear width of the central bright band on the screen (metres), $D$ is the slit-to-screen distance (metres), $\lambda$ the wavelength (metres), and $a$ the slit width (metres). The physical reading of this formula is the heart of the qualitative treatment CBSE wants: $w$ is inversely proportional to $a$ , so squeezing the slit narrower makes the pattern spread wider — diffraction gets stronger as the aperture shrinks relative to the wavelength. Widen the slit enough that $a \gg \lambda$ , and $w \to 0$ : the pattern collapses onto a sharp geometric image of the slit, recovering the ray-optics prediction as a limiting case.

Figure

A single slit of width a is struck by parallel light rays from the left, producing wavelets that spread out and form a bright wide central band straight ahead on a screen at the right, flanked by two narrower and much fainter bands on either side, with dark gaps marking the minima between them

Derivation AThe angular and linear width of the central maximum from the minima condition

The goal is to move from the single trusted starting point of this syllabus — the first-minimum condition $a\sin\theta_1=\lambda$ , obtained by the qualitative pairwise-cancellation argument across the slit — to the two working results CBSE actually expects you to use in numericals: the angular width and the linear width of the central bright band on a screen. No intensity calculus is used; only the small-angle approximation and simple geometry, exactly matching the "qualitative treatment" the syllabus calls for.

Step 1. Start from the first-minimum condition. Splitting the slit of width $a$ into matching pairs of strips shows total destructive interference first occurs when the path difference between wavelets from the two edges of the slit equals one full wavelength: $a\sin\theta_1 = \lambda$ . This fixes the angle of the first dark fringe on one side of the central maximum.

Step 2. Apply the small-angle approximation. In essentially every board-level setup the slit width $a$ is a fraction of a millimetre while $\lambda$ is a few hundred nanometres, so $\lambda/a$ is of order $10^{-3}$ — a genuinely small angle. For small $\theta$ , $\sin\theta \approx \theta$ (in radians), so Step 1 becomes $\theta_1 \approx \lambda/a$ .

Step 3. Use the pattern's symmetry to get the angular width. The slit is symmetric about its centre line, so the first minimum on the other side of the central direction occurs at the mirror-image angle $-\theta_1$ . Everything between $\theta=-\theta_1$ and $\theta=+\theta_1$ is the unbroken central maximum (no minimum occurs strictly inside this range, since $n=1$ is the smallest allowed integer). The full angular width of the central maximum is therefore twice the single-sided angle: $2\theta_1 \approx \dfrac{2\lambda}{a}$ .

Step 4. Convert the angular width to a linear width on a real screen. On a screen placed a distance $D$ from the slit, with $D$ much larger than the pattern size (so that the small-angle geometry of a right triangle applies), an angular spread $2\theta_1$ subtends a linear distance on the screen equal to $D \times (2\theta_1)$ . Substituting Step 3, $w \approx D\left(\dfrac{2\lambda}{a}\right) = \dfrac{2\lambda D}{a}$ . This is the usable, examinable result: the linear width of the central bright band, growing as $\lambda$ or $D$ increase and shrinking as the slit $a$ is widened. Result: $\theta_1\approx\lambda/a$ (angle of the first dark fringe), $2\theta_1\approx 2\lambda/a$ (angular width of the central maximum), and $w\approx 2\lambda D/a$ (its linear width on a distant screen) — the three usable quantities the syllabus expects you to compute with, all resting on the single quoted minima condition $a\sin\theta_n=n\lambda$ .

Worked examples

Board level - width of the central maximum

Monochromatic light of wavelength $\lambda = 6000$ angstrom ( $6\times10^{-7}\ \text{m}$ ) is incident normally on a single slit of width $a = 0.3\ \text{mm}$ . Find (a) the angle at which the first minimum occurs, and (b) the width of the central maximum on a screen placed $D = 1.5\ \text{m}$ away.

Step 1. Convert everything to consistent SI units: $\lambda = 6\times10^{-7}\ \text{m}$ , $a = 0.3\times10^{-3}\ \text{m} = 3\times10^{-4}\ \text{m}$ , $D = 1.5\ \text{m}$ .

Step 2. Apply the first-minimum condition $a\sin\theta_1=\lambda$ to find the angle. $\sin\theta_1 = \dfrac{\lambda}{a} = \dfrac{6\times10^{-7}}{3\times10^{-4}} = 2\times10^{-3}$ . Since this is small, $\theta_1 \approx 2\times10^{-3}\ \text{rad}$ (about $0.115$ degree).

Step 3. Apply the linear-width formula $w = \dfrac{2\lambda D}{a}$ . $w = \dfrac{2\times(6\times10^{-7})\times1.5}{3\times10^{-4}} = \dfrac{1.8\times10^{-6}}{3\times10^{-4}} = 6\times10^{-3}\ \text{m}$ .

Final answers: the first minimum occurs at $\theta_1 \approx 2\times10^{-3}$ rad, and the central maximum is $6\times10^{-3}\ \text{m} = 6\ \text{mm}$ wide on the screen — a comfortably visible bright band roughly the width of a pencil lead, from a slit only three-tenths of a millimetre across.

Competitive level - doubling the slit width, then recovering the wavelength

In a single-slit diffraction setup with slit width $a = 0.2\ \text{mm}$ , the central maximum on a screen has measured width $4\ \text{mm}$ . (a) If the slit width is doubled to $a=0.4\ \text{mm}$ , keeping $\lambda$ and $D$ fixed, find the new width of the central maximum. (b) If the screen distance in the original setup was $D = 1\ \text{m}$ , find the wavelength $\lambda$ used.

Step 1. Use the proportionality hidden in the formula rather than recomputing from scratch. Since $w = 2\lambda D/a$ and only $a$ changes, $w \propto 1/a$ for fixed $\lambda, D$ . Doubling $a$ therefore halves $w$ : $w_{\text{new}} = w_{\text{old}}/2 = 4\ \text{mm}/2 = 2\ \text{mm}$ .

Step 2. Solve part (b) by rearranging the linear-width formula for $\lambda$ , using the original numbers ( $a=0.2\ \text{mm}=2\times10^{-4}\ \text{m}$ , $w=4\ \text{mm}=4\times10^{-3}\ \text{m}$ , $D=1\ \text{m}$ ): $\lambda = \dfrac{wa}{2D}$ .

Step 3. Substitute and compute. $\lambda = \dfrac{(4\times10^{-3})(2\times10^{-4})}{2\times1} = \dfrac{8\times10^{-7}}{2} = 4\times10^{-7}\ \text{m} = 4000$ angstrom.

Final answers: doubling the slit width halves the central maximum to $\mathbf{2\ mm}$ , and the wavelength used in the original setup was $\boldsymbol{\lambda = 4\times10^{-7}\ \text{m} = 4000}$ angstrom — violet light, consistent with a narrower slit producing a comparatively modest $4\ \text{mm}$ band at $1\ \text{m}$ . Timing note: Part (a) should take under 20 seconds once you recognise $w\propto 1/a$ — never redo the full substitution for a pure proportionality question; reserve full substitution, as in part (b), for when an unknown quantity (here $\lambda$ ) actually needs to be extracted, and budget no more than 90 seconds total for a two-part question of this shape.

Common traps

Confusing the single-slit minima condition $a\sin\theta_n=n\lambda$ with the double-slit maxima condition from Young's experiment ( $d\sin\theta=n\lambda$ for bright fringes) — in single-slit diffraction this relation gives dark fringes, not bright ones; writing "bright fringe" here is an automatic mark loss.
Believing a narrower slit gives a sharper, narrower pattern — it is the reverse: a smaller $a$ makes the central maximum wider, because diffraction strengthens as the opening shrinks relative to $\lambda$ .
Forgetting the factor of 2: the central maximum's angular width is $2\lambda/a$ (from $-\theta_1$ to $+\theta_1$ ), not $\lambda/a$ , which is only the half-width or the width of a single secondary maximum.
Mixing up angular width (dimensionless, in radians, $2\lambda/a$ ) with linear width on the screen (in metres, $2\lambda D/a$ ) — always check what the question is actually asking for before substituting.
Assuming the secondary maxima are roughly as bright as the central maximum — they are markedly fainter, and the pattern's overwhelming majority of transmitted energy sits in the central band, a fact only qualitative reasoning (not a formula) is needed to state on an exam.
Trying to derive or reproduce the intensity-distribution formula across all the secondary maxima — CBSE 2025-26 explicitly keeps this qualitative; the examinable content is the minima condition and the resulting central-maximum width, nothing more.

You should be able to…

Explain, using Huygens' principle, why a single slit of finite width produces its own bright-dark diffraction pattern rather than a sharp geometric shadow.
State the minima (dark fringe) condition $a\sin\theta_n=n\lambda$ and correctly distinguish it from the double-slit bright-fringe condition.
Compute the angular width ( $2\lambda/a$ ) and linear width on a screen ( $2\lambda D/a$ ) of the central maximum, and rearrange either formula to solve for any one unknown quantity.
Predict, without recomputation, how the pattern broadens or narrows when the slit width, wavelength, or screen distance is changed (proportionality reasoning).
Justify qualitatively why the central maximum is both twice as wide and much brighter than any secondary maximum.
Recognise when a question is asking about single-slit diffraction versus double-slit (Young's) interference from the wording and the physical setup described.

Ask the tutor: Want to work through a mixed numerical that combines this with Young's double-slit fringe-width formula, so you can drill the dark-versus-bright distinction until it's automatic?

Why does a single slit even produce a pattern — shouldn't light from one opening just go straight through?

Ray optics treats the slit as simply blocking some light and letting the rest through in a straight beam, but light is a wave, and Huygens' principle says every point across the opening becomes its own little wave source. Those many wavelets, all launched from slightly different positions, travel slightly different distances to any given point on the screen and interfere with each other — reinforcing straight ahead, cancelling at certain angles. A "single opening" is not one source; it is a whole continuum of sources packed into a width $a$ , and that is exactly what makes the pattern possible.

Is this the same thing as Young's double-slit experiment, just with one slit instead of two?

No, and this is the trap CBSE tests most often. In Young's experiment you have two narrow slits acting as two essentially point-like sources, and their two wavelet trains interfere to give a rapid pattern of bright and dark fringes where $d\sin\theta=n\lambda$ gives bright fringes. Here you have one slit, but it is not narrow enough to act as a single point — its finite width $a$ means wavelets from many different points within it interfere with each other, and that self-interference gives $a\sin\theta_n=n\lambda$ for dark fringes instead. Same-looking formula, opposite physical meaning.

Why does the central maximum get wider when I make the slit narrower — shouldn't a smaller opening make a smaller, tighter beam?

That is exactly the ray-optics intuition, and wave optics overturns it. A narrower slit forces the wavelets across it to have path differences that build up to a full wavelength ( $a\sin\theta_1=\lambda$ ) at a larger angle $\theta_1$ , because $a$ is smaller — the wave has to spread further before the cancellation condition is met. Physically, squeezing light through a smaller aperture bends it more sharply around the edges. Only once $a$ becomes much larger than $\lambda$ does this bending become negligible and you recover the sharp, ray-optics shadow of the slit.

Do I need to memorise the full intensity formula across all the fringes for my board exam?

No. CBSE 2025-26 explicitly keeps single-slit diffraction at a qualitative level for anything beyond the central maximum's width. You should be comfortable using $a\sin\theta_n=n\lambda$ for the minima and $w\approx 2\lambda D/a$ for the central maximum's width, and you should be able to reason qualitatively about how the pattern changes with $a$ , $\lambda$ , or $D$ — but reproducing the calculus-based derivation of intensity across the secondary maxima is university-level content and outside your syllabus.