Young's Double-Slit Experiment

Young's two-beam interferometer is the ancestor of every precision optical instrument that counts wavelengths. Split one beam into two, recombine them, and the fringes report the path difference between the two arms to a fraction of a wavelength — the working principle of the Michelson interferometer, of the Fabry–Pérot etalons that lock laser frequencies, and, at the grandest scale, of LIGO, whose kilometre-long arms detect gravitational waves by watching interference fringes shift by less than a thousandth of a proton's width. The same fringe geometry calibrates the flatness of optical surfaces, measures the thickness of transparent films, and reads out the tiny displacements in fibre-optic sensors. And the experiment's deepest legacy is conceptual: fire the double slit one photon at a time, or one electron at a time, and the interference pattern still builds up dot by dot — the founding demonstration of quantum superposition, and the experiment Richard Feynman called the one that holds "the only mystery" of quantum mechanics. Young's barcode of light turned out to be a barcode of reality itself.

Start from what the previous article established: two coherent sources — same frequency, a fixed phase relationship — produce a sustained interference pattern, bright where the waves arrive in step and dark where they arrive exactly out of step. Young's genius was in how he made two coherent sources out of ordinary light. Two separate lamps will never do it: each lamp is a crowd of independent atoms whose phases jitter randomly billions of times a second, so their pattern washes out instantly. Young's trick was to take one source and split it. The light passing through slit $S_1$ and the light passing through slit $S_2$ began life at the same wavefront, so whatever random phase wobble the source has, both slits inherit it identically — the phase difference between them stays locked. Two slits, one parent beam: guaranteed coherence. Now follow the two beams to the screen. Consider a point $P$ a height $y$ above the central axis. Light from $S_2$ (the farther slit) travels a slightly longer road to reach $P$ than light from $S_1$. That extra distance is the path difference $\Delta$. If $\Delta$ happens to be a whole number of wavelengths, the two waves arrive crest-aligned and reinforce — a bright fringe. If $\Delta$ is a half-integer number of wavelengths, they arrive crest-against-trough and cancel — a dark fringe. Straight ahead ($y = 0$) the two paths are exactly equal, $\Delta = 0$, so the centre of the pattern is always bright: the central maximum. As you move up the screen, $\Delta$ grows steadily, sweeping through $\lambda, 2\lambda, 3\lambda, \dots$ and passing a dark fringe halfway between each — a regular ladder of light and shadow.

Two length scales make this work, and their ratio is the whole magic. The slit separation $d$ and the wavelength $\lambda$ are microscopic — $d$ is a fraction of a millimetre, $\lambda$ is under a thousandth of that. But the screen distance $D$ is huge by comparison — a metre or more. The geometry amplifies the tiny wavelength up to the visible fringe spacing by exactly the factor $D/d$, which in a typical bench setup is around a thousand. That is how a length of $600\ \mathrm{nm}$ — utterly invisible — becomes a $0.6\ \mathrm{mm}$ fringe you can measure with a travelling microscope.

Everything rests on one small-angle approximation. Because the screen is far away ($D \gg d$), the two rays from $S_1$ and $S_2$ to the point $P$ are very nearly parallel. Drop a perpendicular from $S_1$ onto the ray from $S_2$: the little right triangle it forms has the slit separation $d$ as its hypotenuse and the path difference $\Delta$ as the short side opposite the angle $\theta$ that the rays make with the axis. So the extra path is

where: $d$ — the separation between the two slits; $y$ — the height of $P$ above the central axis; $D$ — the slit-to-screen distance; $\theta$ — the small angle the rays make with the axis, so small that $\sin\theta \approx \tan\theta = y/D$. The physical reading: the farther up the screen you look (larger $y$), the longer the detour the far slit's light must make, and the path difference grows in exact proportion to $y$. This linearity is why the fringes come out evenly spaced.

A bright fringe appears wherever the two waves arrive in step — that is, wherever the path difference is a whole number of wavelengths. Setting $\Delta = n\lambda$ in equation (1) and solving for the position:

where: $n$ is the order of the fringe — $n = 0$ is the central bright fringe on the axis, $n = 1$ the first bright fringe on either side, and so on. A dark fringe appears where the waves arrive exactly out of step, which is where the path difference is an odd number of half-wavelengths, $\Delta = (n - \tfrac12)\lambda$:

where: the first dark fringe ($n = 1$) sits at $\tfrac12$ a fringe-spacing above the centre, the second at $1\tfrac12$, and so on — each dark fringe nestles exactly halfway between two bright ones. Notice the bright and dark ladders are interleaved and both are evenly spaced. That regularity is the experiment's signature and the reason a photograph of the fringes looks like a barcode.

The single most useful quantity is the fringe spacing $\beta$ — the centre-to-centre distance between two neighbouring bright fringes (equivalently, between two neighbouring dark ones). Subtract the position of the $n$-th bright fringe from the $(n{+}1)$-th using equation (2):

where: $\beta$ (fringe spacing, sometimes called the fringe width) is independent of $n$ — every gap in the pattern is the same size, which is exactly why the fringes are uniform. This one line is the engine of every double-slit numerical: measure $\beta$, $D$ and $d$ with a ruler and you have solved for $\lambda$; know $\lambda$, $D$ and $d$ and you have predicted the whole pattern. Take the standard bench values $\lambda = 600\ \mathrm{nm}$ (red-orange light), $d = 1\ \mathrm{mm}$ and $D = 1\ \mathrm{m}$:

So the bright fringes sit at $y_1 = 0.6\ \mathrm{mm}$, $y_2 = 1.2\ \mathrm{mm}$, $y_3 = 1.8\ \mathrm{mm}$, and the dark fringes at $0.3\ \mathrm{mm}$, $0.9\ \mathrm{mm}$, and so on. The wavelength was $600\ \mathrm{nm} = 0.0006\ \mathrm{mm}$; the fringe is $0.6\ \mathrm{mm}$ — a magnification of exactly $D/d = 1000$. There is also a clean angular version: the angular spacing between fringes is

which is handy when the screen distance is not given: the fringes subtend a fixed angle $\lambda/d$ regardless of how far away you put the screen.

Reading equation (4) as a set of dials tells you how to control the fringes. Longer wavelength $\lambda$ (redder light) spreads the fringes; shorter (bluer) crowds them — which is why white light gives a coloured pattern, with each colour's fringes at its own spacing and only the central fringe pure white. Larger screen distance $D$ spreads them linearly. Larger slit separation $d$ shrinks them — counter-intuitively, pushing the slits farther apart packs the fringes closer, because $d$ is in the denominator. And the most-tested variation: immerse the whole apparatus in a transparent liquid of refractive index $n$. The frequency is fixed by the source, but the wavelength inside the medium shrinks to $\lambda_{\text{medium}} = \lambda/n$, so every fringe contracts:

where: $\beta'$ is the fringe spacing in the medium and $\beta$ the value in air. For the $0.6\ \mathrm{mm}$ pattern dunked in water ($n = 1.33$), $\beta' = 0.6/1.33 = 0.451\ \mathrm{mm}$ — the fringes crowd together by a third. This is the same $\lambda_{\text{medium}} = \lambda/n$ rule that governed refraction; here it shows up as a shrinking barcode.

Board level — the standard fringe pattern (CBSE and ICSE pattern)

In a double-slit experiment the slits are $d = 1\ \mathrm{mm}$ apart and the screen is $D = 1\ \mathrm{m}$ away. Monochromatic light of wavelength $\lambda = 600\ \mathrm{nm}$ is used. Find the fringe spacing and the positions of the first three bright fringes and the first two dark fringes.

Step 1. Fringe spacing from equation (4): $\beta = \lambda D/d = (600\times10^{-9})(1)/(1\times10^{-3}) = 6\times10^{-4}\ \mathrm{m} = \boxed{0.6\ \mathrm{mm}}$.

Step 2. Bright fringes at $y_n = n\beta$: $y_1 = \boxed{0.6\ \mathrm{mm}}$, $y_2 = \boxed{1.2\ \mathrm{mm}}$, $y_3 = \boxed{1.8\ \mathrm{mm}}$.

Step 3. Dark fringes at $y_n = (n-\tfrac12)\beta$: the first at $0.5\beta = \boxed{0.3\ \mathrm{mm}}$ and the second at $1.5\beta = \boxed{0.9\ \mathrm{mm}}$ — exactly halfway between the bright fringes, as they must be.

Step 4. Sense-check: the fringes are equally spaced ($0.6\ \mathrm{mm}$ apart) and the wavelength ($0.0006\ \mathrm{mm}$) has been magnified up to a visible $0.6\ \mathrm{mm}$ by the factor $D/d = 1000$. If your $\beta$ had come out in micrometres you divided by $D$ instead of multiplying.

Board level — a second apparatus and the angular fringe (JEE-Main and NEET pattern)

A double slit with $d = 0.5\ \mathrm{mm}$ is lit by green light $\lambda = 500\ \mathrm{nm}$ and the pattern is caught on a screen $D = 2\ \mathrm{m}$ away. Find the fringe spacing. What angle does one fringe subtend at the slits?

Step 1. $\beta = \lambda D/d = (500\times10^{-9})(2)/(0.5\times10^{-3}) = 2\times10^{-3}\ \mathrm{m} = \boxed{2.0\ \mathrm{mm}}$ — more than three times wider than the previous bench, because the light is bluer but the slits are closer and the screen farther.

Step 2. The angular fringe spacing does not need $D$: $\theta_{\text{fringe}} = \lambda/d = (500\times10^{-9})/(0.5\times10^{-3}) = \boxed{1\times10^{-3}\ \mathrm{rad}}$.

Step 3. Cross-check with $\beta$: $\theta = \beta/D = (2\times10^{-3})/2 = 1\times10^{-3}\ \mathrm{rad}$ — the two roads agree.

Competitive level — the whole pattern dunked in water (JEE-Advanced and NEET pattern)

The first apparatus ($\beta = 0.6\ \mathrm{mm}$ in air) is submerged in a tank of water, $n = 1.33$, with the source, slits and screen all under water. What is the new fringe spacing?

Step 1. Immersion does not touch $D$ or $d$; it shrinks the wavelength. Inside water $\lambda_{\text{water}} = \lambda/n = 600/1.33 = 451\ \mathrm{nm}$, the frequency unchanged.

Step 2. Feed the shrunk wavelength into equation (4), or use the shortcut $\beta' = \beta/n$ directly: $\beta' = 0.6/1.33 = \boxed{0.451\ \mathrm{mm}}$.

Step 3. Physical reading: the fringes crowd inward by a third, because the light "measures out" its bright bands in the shorter in-water wavelength. If a distractor offers $0.6\times1.33 = 0.80\ \mathrm{mm}$ it has multiplied by $n$ instead of dividing — immersion always shrinks the pattern. Timing note: under 45 seconds once you recognise the immersion shortcut $\beta' = \beta/n$. Do not re-derive from $\lambda_{\text{water}}$ under exam pressure — spot the "in a medium" cue and divide the air fringe spacing by $n$ in one step.

$d$ is the separation, $a$ is the width. Young's fringe spacing uses the slit separation $d$ in $\beta = \lambda D/d$. The slit width $a$ belongs to single-slit diffraction. Writing $\beta = \lambda D/a$ is the single most common double-slit error — check which length the question gives before you plug.

Dark fringes are at $(n-\tfrac12)\lambda$, not $n\lambda$. Bright fringes need a whole number of wavelengths of path difference; dark fringes need a half-integer. Mixing these up shifts every dark position by half a fringe. Memory hook: whole $\lambda$ = crest on crest = bright.

Immersion divides $\beta$ by $n$; it does not multiply. Sinking the rig in a medium shrinks the wavelength to $\lambda/n$, so $\beta' = \beta/n$. The fringes get closer, never wider. A slab of glass in one path is a different effect (a fringe shift, not a rescaling) — that is the archetype article, not this one.

Fringe spacing is uniform — do not "average" it. Because $\beta = \lambda D/d$ is independent of the order $n$, every gap is identical. A question giving "ten fringes span $6.0\ \mathrm{mm}$" means $\beta = 6.0/10 = 0.6\ \mathrm{mm}$ straight away; you do not need to find individual positions.

Board scope. Young's double slit is examinable on CBSE, ICSE, JEE, NEET, AP Physics 2, both IB levels, and A-Level 9702 (which writes it as $\lambda = ax/D$, with $a$ their symbol for the slit separation and $x$ for the fringe spacing — a genuine notation clash with the width-$a$ convention, so read each board's symbol table). IGCSE 0625 has no wave optics at all — do not revise this for an IGCSE paper. The explicit intensity formula $I = 4I_0\cos^2(\phi/2)$ is CBSE/JEE/NEET depth; AP, IB and A-Level stay at the path-difference conditions and a qualitative intensity picture.

Texts. NCERT Physics Class 12, Ch. 10 ("Wave Optics"), the section on Young's experiment and fringe width · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves", double-slit interference · Halliday, Resnick and Walker, "Interference", Young's experiment · Cambridge A-Level 9702 topic 15 (two-source interference, $\lambda = ax/D$).

Provenance. Standard results. Recompute: $\beta = \lambda D/d = (600\times10^{-9})(1)/(1\times10^{-3}) = 0.6\ \mathrm{mm}$; bright $y_1, y_2, y_3 = 0.6, 1.2, 1.8\ \mathrm{mm}$; dark $y_1, y_2 = 0.3, 0.9\ \mathrm{mm}$; angular fringe $\lambda/d = 6\times10^{-4}\ \mathrm{rad} = 0.034^\circ$; magnification $D/d = 1000$; second bench $\beta = (500\times10^{-9})(2)/(0.5\times10^{-3}) = 2.0\ \mathrm{mm}$ with $\theta = \lambda/d = 1\times10^{-3}\ \mathrm{rad}$; immersion $\beta' = 0.6/1.33 = 0.451\ \mathrm{mm}$ (with $\lambda_{\text{water}} = 600/1.33 = 451\ \mathrm{nm}$); "ten fringes span $6.0\ \mathrm{mm}$" gives $\beta = 0.6\ \mathrm{mm}$; distractor $0.6\times1.33 = 0.80\ \mathrm{mm}$.

Prerequisites. Coherence & Interference: Path Difference, Phase & Intensity — where coherent sources, path↔phase, and the bright/dark conditions come from. Leads to. ARCHETYPE: YDSE & Fringe-Width Numericals — the slab-shift, immersion and missing-order problem gym · Single-Slit Diffraction — what the finite slit width $a$ does to the envelope.

Go deeper (UG). The full field treatment of the double slit — the exact intensity as the coherent sum of two spherical waves, the finite-slit envelope, and the coherence conditions that make the pattern visible — is developed in Young's Double-Slit Experiment in the university optics track. Enrichment only; nothing above depends on it.