Diffraction

Hold a hair up to a bright streetlamp at night, squint, and you will see faint fringes of light bleeding into what should be a perfectly sharp dark shadow. That bleeding is diffraction — proof that light is a wave, caught in the act of bending around an edge too fine for straight-line rays to explain. Ordinary daylight objects never show this trick: a pillar, a book, even a coin all cast crisp shadows, because light's wavelength, roughly half a micrometre, is minuscule next to their size. But shrink the gap down to match that wavelength — a narrow slit, a strand of hair, the tiny aperture of your own pupil — and the wave nature of light stops hiding and starts painting bands. This article builds the concept from Huygens' principle, the same secondary-wavelet idea used earlier in the chapter to prove reflection and refraction. Every unblocked point of a wavefront crossing an opening becomes its own tiny source, and when the opening is narrow, those tiny sources spread wavelets sideways into the region a ray picture would insist stays dark. The result, for a single narrow slit, is not a sharp-edged bright strip but a broad, bright central band flanked by weaker, evenly spaced secondary bands — a diffraction pattern, qualitatively and geometrically different from the equally spaced fringes of Young's double-slit interference met earlier in this chapter. We will pin down exactly when diffraction becomes noticeable, walk through the simple path-difference argument (algebra only, no calculus) that fixes how wide that central band is, and work two numeric examples — one comparing sound and light at an everyday doorway, one tracking what happens to the pattern as a slit is narrowed — before turning to the traps CBSE examiners love to set around this exact idea.

Key takeaways

Diffraction is the bending of waves into the geometric shadow of an obstacle or through an aperture, governed by Huygens' principle: every point on an unblocked wavefront acts as a source of secondary wavelets.
It is only noticeable when the aperture or obstacle size $a$ is comparable to the wavelength $\lambda$ . For light ( $\lambda \sim 5\times10^{-7}\ \text{m}$ ) this means slits a fraction of a millimetre wide; everyday objects are millions of times larger, so light casts sharp shadows.
A single narrow slit produces a bright central maximum flanked by much dimmer secondary maxima, separated by dark minima — very different in shape and spacing from the equally-bright fringes of Young's double-slit interference.
The first dark fringe (first minimum) occurs at angle $\theta_1 \approx \lambda/a$ (small-angle, radians), obtained by splitting the slit into two halves and demanding their wavelets cancel in pairs — algebra, not calculus.
The central maximum has angular half-width $\lambda/a$ and total angular width $2\lambda/a$ — roughly twice as wide as each secondary maximum, and it narrows as the slit is widened, broadens as the slit is narrowed.
Diffraction and interference are the same underlying superposition principle applied to different numbers of coherent sources: two point sources for YDSE interference, infinitely many points across one aperture for single-slit diffraction.
Key idea: light bends noticeably into shadow only when the gap it passes through is comparable to its wavelength, and splitting a single slit into wavelet pairs that cancel gives the first diffraction minimum at θ₁ ≈ λ/a, so the bright central band has angular width 2λ/a.

The core insight

Diffraction is the spreading of a wave into the geometric shadow whenever the aperture or obstacle size is comparable to the wavelength, and for a single slit the simple two-half wavelet-cancellation argument gives the first minimum at $\theta_1 \approx \lambda/a$ , fixing the bright central band's angular width at $2\lambda/a$ .

Where it fits in the syllabus

Diffraction closes the Wave Optics chapter, arriving right after Young's double-slit interference and coherent sources. The 2025-26 CBSE prescription is explicit and narrow: diffraction due to a single slit, and the width of the central maximum, treated qualitatively — you are not asked to derive or apply the full Fraunhofer intensity distribution, and polarisation, diffraction gratings, and resolving power of instruments are all rationalised out of the current syllabus entirely. This article gives the conceptual foundation and the one simple, non-calculus geometric result (the angle to the first minimum) that the "qualitative treatment" comfortably allows; the detailed shape and consequences of the single-slit pattern are the dedicated focus of the next node in this chapter.

The physical picture

Start from Huygens' principle, already used earlier in the chapter to derive the laws of reflection and refraction from wavefronts. Every point on an unobstructed wavefront is itself a source of secondary spherical wavelets, and the new wavefront a little later is the envelope that touches all of them. When a wavefront meets a wide opening, wavelets from across the opening combine to reconstruct essentially the same plane wave moving straight ahead — no visible spreading. When the opening is narrow enough that only a few wavelengths' worth of wavefront gets through, the wavelets no longer reconstruct a simple plane wave; they spread sideways into the region a ray picture calls the shadow. That sideways spreading is diffraction, and its strength is governed by one comparison:

\frac{a}{\lambda}\ \text{small (comparable to 1)} \;\Rightarrow\; \text{strong diffraction}, \qquad \frac{a}{\lambda}\ \text{large} \;\Rightarrow\; \text{negligible diffraction, ray optics valid.}

where: $a$ is the width of the slit or obstacle (metres) and $\lambda$ is the wavelength of the wave (metres). This single ratio is why sound, with wavelengths of order a metre, diffracts noticeably around a metre-wide doorway, while visible light, with $\lambda \sim 5\times10^{-7}\ \text{m}$ , does not visibly bend around that same doorway — the doorway is millions of times larger than a light wavelength. When the ratio is small enough to matter, a single narrow slit illuminated by monochromatic light throws a characteristic pattern onto a distant screen: a bright central band, much wider and far more intense than anything else in the pattern, then a dark band (first minimum), then a secondary maximum much dimmer than the centre, then another dark band, and so on symmetrically on both sides. Three qualitative facts describe this pattern completely at this stage:

the central maximum is roughly twice as wide (in angle) as each secondary maximum,
the secondary maxima fall off rapidly in intensity moving away from the centre — the pattern is dominated by its central band, unlike YDSE fringes which are all comparably bright,
the whole pattern narrows as the slit is widened and broadens as the slit is narrowed — the pattern's width scales inversely with $a$ .

Figure

A single narrow slit illuminated from the left by plane wavefronts that pass through and spread out, forming a screen pattern with a wide bright central band, a dark band, a dimmer secondary bright band, another dark band, and the mirror image on the other side

Interference and diffraction, the same wavelet idea

It helps to see diffraction and Young's double-slit interference as two versions of one calculation, differing only in how many coherent sub-sources are combined. In YDSE, light passes through two narrow slits, each acting as essentially one coherent point source, and the two overlapping wavelet trains produce the familiar equally spaced, equally bright fringes of width $\beta = \lambda D/d$ . In single-slit diffraction, the slit itself is wide enough to contain a continuum of coherent secondary sources across its width $a$ ; every one of those wavelets superposes with every other, and the resulting pattern — a dominant central band with weak, rapidly fading side bands — looks qualitatively nothing like YDSE fringes, precisely because it is built from many sources spread across one aperture rather than two isolated point sources.

Central maximum width from wavelet cancellation

Derivation AAngular half-width of the central maximum in single-slit diffraction

The goal is to find the angle $\theta_1$ at which the very first dark band appears on either side of the bright centre, using nothing beyond geometry and the idea of pairing up wavelets that cancel — no calculus and no intensity formula, matching the syllabus's qualitative-treatment ceiling while still giving the one number ("how wide is the central band") that the prescribed scope explicitly names. The trick is to split the slit's width into two equal halves and show that, at the right angle, every wavelet in the top half is cancelled by a matching wavelet exactly half a slit-width below it.

Step 1. Set up the path difference. Consider a slit of width $a$ illuminated by a plane wave arriving straight on, and look at light leaving the slit at angle $\theta$ to the straight-ahead direction. The wavelet leaving the very top edge of the slit travels a distance $a\sin\theta$ farther to a distant screen point than the wavelet leaving the very bottom edge, because the two edges are separated by $a$ measured along the wavefront and the extra path is the projection of that separation onto the direction of travel.

Step 2. Split the slit into two equal halves, each of width $a/2$ . Pair every point in the lower half with the point exactly $a/2$ above it in the upper half. Each such pair is separated by $a/2$ along the slit, so the path difference between the two wavelets in a pair, at angle $\theta$ , is $(a/2)\sin\theta$ .

Step 3. Demand that every pair cancels. Two wavelets cancel (destructive interference) when their path difference is exactly half a wavelength, $\lambda/2$ . Setting the pair's path difference equal to this, $(a/2)\sin\theta_1 = \lambda/2$ , and cancelling the common factor of $2$ gives the condition for the first minimum:

a\sin\theta_1 = \lambda. \qquad (1)

Step 4. Read off the central maximum's width. Because every point in the lower half of the slit has exactly one cancelling partner in the upper half at this angle, the entire slit's contribution cancels — a true dark band, the first minimum. For small angles (screen far from the slit, or narrow slit) $\sin\theta_1 \approx \theta_1$ in radians, so $\theta_1 \approx \lambda/a$ . This is the half-angle from the centre to the first dark band on either side, so the central maximum's full angular width is $2\lambda/a$ — the qualitative "width of the central maximum" the syllabus asks you to describe, now pinned to one clean formula. Because $\theta_1$ grows as $a$ shrinks, a narrower slit spreads the pattern wider, exactly the qualitative trend stated in the physical picture above. Consistency check: as $a \to \infty$ (a very wide opening), $\theta_1 = \lambda/a \to 0$ — the pattern collapses onto the straight-ahead direction and the shadow becomes sharp, matching the ray-optics limit for large apertures.

Worked examples

Board level - sound versus light through the same doorway

A doorway of width $a = 1.0\ \text{m}$ is crossed by (i) sound waves of wavelength $\lambda = 1.0\ \text{m}$ , typical of audible sound, and (ii) visible light of wavelength $\lambda = 5\times10^{-7}\ \text{m}$ . Using the diffraction condition (aperture size compared with wavelength), explain why sound is heard clearly around the corner outside the doorway while light does not visibly bend around its edges.

Step 1. Compute the ratio $a/\lambda$ for sound. $\dfrac{a}{\lambda} = \dfrac{1.0\ \text{m}}{1.0\ \text{m}} = 1.$ The doorway is comparable in size to the sound wavelength.

Step 2. Compute the same ratio for light. $\dfrac{a}{\lambda} = \dfrac{1.0\ \text{m}}{5\times10^{-7}\ \text{m}} = 2\times10^{6}.$ The doorway is about two million times larger than the light wavelength.

Step 3. Interpret using the diffraction criterion. Because $a/\lambda \approx 1$ for sound, diffraction is strong: sound wavelets spread well beyond the geometric shadow of the doorway, so sound is heard clearly around corners. Because $a/\lambda \approx 2\times10^6 \gg 1$ for light, diffraction is negligible at this scale, so light travels through essentially as straight rays and the doorway casts a sharp-edged shadow.

Step 4. State the final answer. Sound diffracts strongly around the doorway ( $a/\lambda = 1$ ) while light's diffraction is utterly negligible there ( $a/\lambda = 2\times10^{6}$ ), which is exactly why one is heard around corners and the other is not seen to bend.

Competitive level - narrowing a slit and tracking the pattern quantitatively

Monochromatic light of wavelength $\lambda = 6000\ \text{Å} = 6\times10^{-7}\ \text{m}$ falls on a single slit of width $a = 0.30\ \text{mm}$ , and the pattern is viewed on a screen far away (small-angle regime). (a) Find the angular half-width $\theta_1$ of the central maximum in this setup. (b) If the slit width is then reduced to half its original value, find the new $\theta_1$ and describe what happens to (i) the width and (ii) the brightness of the central maximum.

Step 1. Apply $\theta_1 \approx \lambda/a$ for the original slit. $\theta_1 = \dfrac{6\times10^{-7}\ \text{m}}{0.30\times10^{-3}\ \text{m}} = 2.0\times10^{-3}\ \text{rad}.$ The full angular width of the central maximum is $2\theta_1 = 4.0\times10^{-3}\ \text{rad}$ , a little over $0.2^\circ$ — narrow, but not zero, confirming light still diffracts measurably through a fraction-of-a-millimetre slit.

Step 2. Halve the slit width, $a' = 0.15\ \text{mm} = 1.5\times10^{-4}\ \text{m}$ , and recompute. $\theta_1' = \dfrac{6\times10^{-7}}{1.5\times10^{-4}} = 4.0\times10^{-3}\ \text{rad}$ , exactly double the original $\theta_1$ — consistent with $\theta_1 \propto 1/a$ from Step 4 of the derivation above.

Step 3. Interpret (i), the width. Halving $a$ doubles $\theta_1$ , so the central maximum becomes twice as wide — the pattern spreads out further on the screen, matching the qualitative rule that narrower slits diffract more strongly.

Step 4. Interpret (ii), the brightness. The same total light energy passing through a slit of half the width is now spread over a central band of about twice the angular width, so the energy per unit angle (the brightness) roughly halves — the central maximum becomes wider and correspondingly dimmer. This qualitative brightness trend is as far as the syllabus requires; the underlying intensity formula is not examined.

Step 5. State the final answers. (a) $\theta_1 = 2.0\times10^{-3}\ \text{rad}$ (full width $4.0\times10^{-3}\ \text{rad}$ ). (b) Halving the slit width doubles $\theta_1$ to $4.0\times10^{-3}\ \text{rad}$ : the central maximum becomes twice as wide and roughly half as bright. Timing note: this problem is two short computations chained together — budget under two minutes total; compute $\theta_1 = \lambda/a$ once, note the inverse proportionality, and simply halve $a$ mentally to double $\theta_1$ without recomputing from scratch on the exam.

Common traps

Thinking a narrower slit gives a narrower, sharper central maximum — it is the reverse: $\theta_1 \approx \lambda/a$ grows as $a$ shrinks, so a narrower slit spreads the pattern wider.
Treating diffraction and interference as unrelated phenomena — both are superposition of Huygens' secondary wavelets; the only difference is the number and layout of coherent sub-sources (two isolated slits for YDSE, a continuum across one slit for diffraction).
Applying the YDSE fringe-width formula $\beta = \lambda D/d$ to a single-slit diffraction problem, or vice versa applying $\theta_1 = \lambda/a$ to a double-slit problem — the two setups and formulas belong to different experiments and are not interchangeable.
Assuming diffraction needs a microscopically tiny gap — it only needs $a$ comparable to $\lambda$ , which is why sound diffracts obviously at door-sized openings while light needs sub-millimetre slits.
Forgetting that the small-angle approximation $\sin\theta_1 \approx \theta_1$ (radians) is what makes $\theta_1 \approx \lambda/a$ valid — it fails if the slit is so narrow or the wavelength so long that $\theta_1$ is not small.

You should be able to…

State the general condition for diffraction to be noticeable and explain it using Huygens' principle.
Explain why light casts sharp shadows around everyday objects while sound bends audibly around corners, using the ratio $a/\lambda$ .
Describe the qualitative shape of a single-slit diffraction pattern: wide bright centre, dark minima, dim secondary maxima.
Derive, using only the two-half wavelet-cancellation argument, the first-minimum condition $a\sin\theta_1 = \lambda$ and the small-angle result $\theta_1 \approx \lambda/a$ .
Predict how the central maximum's width and brightness change when the slit width is increased or decreased.
Distinguish single-slit diffraction from Young double-slit interference by the number of coherent sub-sources involved and by the resulting pattern's shape.

Ask the tutor: Want to see how this angular half-width connects directly to the full single-slit intensity pattern, or work through more slit-width numericals with the AI tutor?

Is diffraction only something light does, or do all waves do it?

All waves diffract — sound, water ripples, radio waves, even electrons and other matter waves at the quantum level. The reason light's diffraction is so much harder to notice in daily life is purely a matter of scale: visible light's wavelength is about half a micrometre, so it only bends noticeably around obstacles or through openings of that same tiny size. Sound, with wavelengths of the order of a metre, diffracts around door-sized openings that you walk through every day, which is why this chapter often uses the sound-versus-light doorway comparison to make the effect concrete.

Why does the central maximum in single-slit diffraction look nothing like the fringes in Young double-slit experiment?

Because the two experiments combine wavelets from very different numbers of sources. YDSE combines light from just two narrow slits, each acting like a single point source, giving equally spaced, roughly equally bright fringes described by $\beta = \lambda D/d$ . Single-slit diffraction instead combines wavelets from every point across one continuous slit of width $a$ ; the huge number of interfering sub-sources produces one dominant, wide central band and only weak, rapidly fading secondary bands, not a row of equally bright stripes. Same wave principle, very different source geometry, very different pattern.

Do I need to memorise or derive the full single-slit intensity formula for the board exam?

No. The CBSE 2025-26 prescription explicitly marks single-slit diffraction and the width of the central maximum as a qualitative treatment, so you are not expected to derive or apply the detailed Fraunhofer intensity distribution across the whole pattern. The one quantitative result worth knowing, $\theta_1 \approx \lambda/a$ for the first minimum (and hence $2\lambda/a$ for the central maximum's width), comes from the simple wavelet-pair cancellation argument in this article and sits comfortably within the qualitative-treatment expectation, since it needs only geometry and no calculus.

Why does a wider slit make everyday light behave like it is not diffracting at all?

Because $\theta_1 \approx \lambda/a$ shrinks toward zero as $a$ grows large compared with $\lambda$ . For a doorway or window of ordinary size, $a$ is millions of times larger than light's wavelength, so the predicted spreading angle is immeasurably small — the central "maximum" collapses onto the dead-straight direction a ray picture predicts, and the shadow looks perfectly sharp. Diffraction is always happening in principle; it is only the ratio $a/\lambda$ that decides whether it is large enough to see.