Young's double slit — fringe width derivation: Visual walkthrough

Step 0 — The words we will use (so no symbol is a stranger)

Before any maths, meet the cast. Every one of these is a length you could measure with a ruler.

Two ideas we borrow from other notes:

Light from the two slits is coherent — see Coherence and coherent sources — meaning the two waves keep a fixed rhythm relative to each other, so their meeting is always the same story.
Waves add up or cancel depending on how they line up — see Interference of light and Path difference and phase difference.

Step 1 — The stage: two slits, one screen

WHAT. We place two slits $S_{1}$ (lower) and $S_{2}$ (upper), a gap $d$ apart. Straight across, a distance $D$ away, stands the screen. The point directly opposite the midpoint of the slits we call $O$ — the "centre".

WHY. Interference is a geometry problem: two waves race from two starting points to the same finish point. To measure who travels farther, we must first pin down where everything sits.

PICTURE. Below, the two slits are on the left, the screen on the right. $O$ is dead centre. The point $P$ is where we choose to look, a height $y$ above $O$ .

Step 2 — Two right triangles give the two distances

WHAT. Draw the two light-paths $S_{1} P$ and $S_{2} P$ . Each path is the slanted side (hypotenuse) of a right triangle whose flat side is $D$ and whose vertical side is "how far $P$ is above that slit".

WHY. We need the length of each path. A right triangle is the only tool where knowing two sides gives the third exactly — that is what Pythagoras does: $(hypotenuse)^{2} = (base)^{2} + (height)^{2}$ . We reach for it because both paths are naturally right triangles (horizontal $D$ , vertical offset).

PICTURE. Look at the two shaded triangles. Their bases are both $D$ . Their heights differ: $P$ is a bit farther above $S_{2}$ ? No — $P$ is farther above $S_{1}$ (the lower slit) than above $S_{2}$ . Watch the vertical legs.

Step 3 — The clever subtraction: square, then take the difference

WHAT. Instead of subtracting two ugly square roots, we subtract their squares first.

WHY. Square roots are hard to subtract directly; squares are easy. The messy $D^{2}$ appears in both, so it will cancel when we subtract — leaving something clean.

PICTURE. The animation-style panel shows $S_{2} P^{2}$ and $S_{1} P^{2}$ as areas; the shared $D^{2}$ block cancels, and only the height-difference survives.

Step 4 — Factor to reveal the path difference $Δ$

WHAT. We use the algebra identity $a^{2} - b^{2} = (a - b) (a + b)$ on the left side.

WHY. Because $(a - b)$ is exactly $Δ$ — the thing we want. Factoring surgically extracts it.

PICTURE. The bar shows $S_{2} P^{2} - S_{1} P^{2}$ split into two factors: the difference $Δ$ (what we chase) times the sum $S_{2} P + S_{1} P$ (which we will approximate next).

Step 5 — The far-screen approximation

WHAT. We say: since the screen is far ( $D$ huge) and the slit-gap and look-height are tiny, both paths are almost exactly $D$ long. So their sum $\approx 2 D$ .

WHY. A metre-long path barely changes when you nudge its endpoint by a millimetre. This is the small-angle regime — see Small angle approximation. We use it only in the sum (where a tiny error is harmless) and never in the delicate difference $Δ$ (where every scrap matters).

PICTURE. Both slanted paths flatten onto the horizontal $D$ ; the panel shows how, when $D ≫ d, y$ , the two lengths visually collapse onto the same $D$ .

Step 6 — Turn path difference into bright fringe positions

WHAT. A bright fringe happens where the two waves arrive in step — i.e. where the extra path $Δ$ is a whole number of wavelengths.

WHY. If one wave is exactly one (or two, or three...) full ripples behind, its crest still lands on the other's crest — they reinforce. That is the bright condition $Δ = nλ$ .

PICTURE. The screen shows the bright bands; the $n$ -th band sits where $Δ = nλ$ . Below each band, the two arriving waves are drawn crest-on-crest.

Step 7 — Fringe width: subtract neighbours, watch $n$ vanish

WHAT. The fringe width $β$ is the gap between band $n$ and band $n + 1$ .

WHY. Because the bands sit at $y_{n} = \frac{nλ D}{d}$ , evenly spaced, the distance between neighbours is what we call the fringe width. Subtracting kills the $n$ .

PICTURE. A ruler laid across the screen: consecutive bright bands are equally spaced; the arrow marks $β$ , the same length everywhere.

Step 8 — The degenerate & edge cases (the reader must never be surprised)

WHAT. We check the corners the pretty formula might hide.

WHY. A derivation you can trust must survive its extremes: dead centre, below-centre, huge slit-gap, and small-angle failure.

PICTURE. Four mini-panels, one per corner case.

The one-picture summary

Here is the whole journey on a single canvas: two right triangles → subtract squares → factor out $Δ$ → far-screen approximation → bright condition → subtract neighbours → $β = \frac{λ D}{d}$ .

Recall Feynman retelling — the walkthrough in plain words

Two little lamps sit a hair apart and blink in perfect time. On a far wall you look at a spot a little off-centre. Light from the top lamp has to walk a touch farther than light from the bottom lamp — that extra walk is $Δ$ . To find it, I draw two right triangles (each lamp to the spot) and, instead of wrestling two square roots, I subtract their squares — the big shared side $D^{2}$ politely cancels, leaving just $2 y d$ . Factoring turns that into " $Δ$ times (sum of the two long paths)". Since the wall is far, both paths are basically $D$ , so the sum is $2 D$ and $Δ = \frac{y d}{D}$ . Now, a bright stripe happens when the extra walk is a whole number of ripples, $nλ$ , which pins each stripe at $y_{n} = \frac{nλ D}{d}$ . The stripes march in equal steps, so the gap between two of them — the fringe width — is just $\frac{λ D}{d}$ : ripple-size times wall-distance, divided by lamp-gap. And the very centre, where both walks are equal, is bright no matter what.

Active Recall

Recall Why do we subtract the

squares of the paths instead of the paths themselves? Because squares kill the square roots and the shared $D^{2}$ cancels, leaving the clean $2 y d$ — direct subtraction of two roots is intractable.

Recall Where exactly is the small-angle approximation used, and where is it NOT?

Used only in the sum $S_{1} P + S_{2} P \approx 2 D$ ; never in the difference $Δ$ , where full precision is kept.

Recall Why do the fringes come out equally spaced?

Because $y_{n} = \frac{nλ D}{d}$ is linear in $n$ , so $y_{n + 1} - y_{n}$ is a constant $\frac{λ D}{d}$ — the $n$ cancels.

What is the surviving path-difference after cancelling

D^{2}

and

\frac{d ^{2}}{4}

2 y d

Master path difference formula?

Δ = \frac{y d}{D}

Final fringe width?

β = \frac{λ D}{d}

Why is the centre always bright?

O

both paths are equal, so

Δ = 0 = 0 \cdot λ

— constructive for every wavelength.

Connections

Young's double slit — fringe width derivation — the parent this page illustrates.
Interference of light — why crest-on-crest brightens.
Coherence and coherent sources — why two slits from one source work.
Path difference and phase difference — the meaning of $Δ$ .
Small angle approximation — licences $S_{1} P + S_{2} P \approx 2 D$ .
Refractive index — how immersing the setup shrinks $λ$ and $β$ .
Diffraction grating — the many-slit cousin of this two-slit story.