Young's double slit — fringe width derivation

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WHAT are we deriving?

We want a formula for $\beta$ in terms of:

• $\lambda$ = wavelength of light,
• $d$ = separation between the two slits,
• $D$ = distance from the slits to the screen.

WHY these three? Because the geometry of "two sources sending waves to a far screen" depends only on how far apart the sources are ($d$), how far the screen is ($D$), and the wave's natural ruler ($\lambda$).

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HOW the geometry works

Set up: two slits $S_1$ and $S_2$, separated by $d$. A screen sits at distance $D$. We measure a point $P$ on the screen a distance $y$ from the central point $O$ (the point directly opposite the midpoint of the slits).

Step 1 — Find the path difference. The wave from $S_2$ has to travel farther than from $S_1$ to reach $P$ (when $P$ is above centre).

Step 2 — Subtract to get path difference $\Delta$. $S_2P^2 - S_1P^2 = \left(y+\tfrac d2\right)^2 - \left(y-\tfrac d2\right)^2 = 2yd$

Also $S_2P^2 - S_1P^2 = (S_2P - S_1P)(S_2P + S_1P)$.

Why? This factoring lets us isolate the difference $\Delta = S_2P - S_1P$, which is exactly the path difference we need.

Since $D \gg d$ and $D \gg y$, both distances $\approx D$, so $S_2P + S_1P \approx 2D$: $\Delta = S_2P - S_1P = \frac{2yd}{S_2P + S_1P} \approx \frac{2yd}{2D} = \frac{yd}{D}$

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HOW we get bright and dark fringes

Position of the $n$-th bright fringe: set $\dfrac{y_n d}{D} = n\lambda$: $y_n = \frac{n\lambda D}{d}$

Why? We just plugged the bright condition into the master path-difference formula and solved for $y$.

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Deriving fringe width $\beta$

The fringe width is the gap between consecutive bright fringes, $n$ and $n+1$: $\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d}$

The $n$'s cancel — this is the deep reason fringes are equally spaced!

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Angular fringe width

Since $y = D\tan\theta \approx D\theta$ for small angles, the angular spacing is: $\theta_{\text{fringe}} = \frac{\beta}{D} = \frac{\lambda}{d}$

Why useful? It removes $D$ — the angular pattern depends only on the source geometry $\lambda/d$.

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Worked examples

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Forecast-then-Verify

Recall Forecast: if you double

$d$, what happens to $\beta$? Forecast: Halves. Verify: $\beta\propto 1/d$, so $d\to 2d$ gives $\beta\to\beta/2$. ✓ Fringes crowd together.

Recall Forecast: switch red (700 nm) to violet (400 nm). Wider or narrower?

Forecast: Narrower. Verify: $\beta\propto\lambda$, smaller $\lambda$ → smaller $\beta$. ✓

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Common mistakes (Steel-manned)

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Active Recall

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Recall Feynman: explain to a 12-year-old

Imagine two kids tapping a pond at the exact same rhythm, making ripples. Where two ripple-crests meet, the water bounces up high (bright). Where a crest meets a dip, they cancel and the water stays flat (dark). On a wall far away you'd see bright-dark-bright-dark stripes. How far apart the stripes are depends on three things: how big the ripples are ($\lambda$), how far the wall is ($D$), and how far apart the two tappers are ($d$). Put the tappers closer and tap gentle big ripples on a far wall → fat stripes. The recipe is "ripple-size times wall-distance, divided by tapper-gap."

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Connections

• Interference of light — YDSE is the textbook case.
• Coherence and coherent sources — why a single source + two slits is needed.
• Path difference and phase difference — $\Delta\phi = \frac{2\pi}{\lambda}\Delta$.
• Diffraction grating — many slits → sharper version of the same idea.
• Refractive index — explains the water-immersion shrink.
• Small angle approximation — the key simplifying tool here.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Young's double slit ka idea bahut simple hai. Do choti slits se light nikalti hai aur wo do "coherent sources" ki tarah behave karti hai — matlab dono ki light hamesha same rhythm me chalti hai. Screen pe kisi point par dono slits se aane wali light ka raasta thoda alag hota hai. Is raaste ke farak ko hum path difference $\Delta$ kehte hain. Jahan $\Delta$ poore wavelength ka multiple hota hai ($n\lambda$), wahan crest pe crest milta hai → bright fringe. Jahan $\Delta = (n+\tfrac12)\lambda$ hota hai, wahan cancel ho jaata hai → dark fringe.

Geometry se nikalta hai ki $\Delta = \dfrac{yd}{D}$, jahan $y$ screen par centre se distance hai, $d$ slit gap, aur $D$ screen ki doori. Bright fringe ke liye $\dfrac{yd}{D}=n\lambda$ rakho to position milti hai $y_n = \dfrac{n\lambda D}{d}$. Do consecutive bright fringes ka gap nikalo to $n$ cancel ho jaata hai aur milta hai fringe width $\beta = \dfrac{\lambda D}{d}$.

Yeh formula bahut kuch bata deta hai: bada $\lambda$ ya bada $D$ → fringes door door (mote), aur bada $d$ → fringes paas paas (patle). Agar pura setup paani me daal do to wavelength $n$ se chhota ho jaata hai, isliye fringes bhi $n$ guna paas aa jaate hain. Exam me yaad rakhna: dark fringe ki condition $n\lambda/2$ nahi, balki $(2n+1)\lambda/2$ hoti hai — yeh common galti hai. Bas "lambda-D-over-d" ka mantra yaad rakho aur units ko SI me convert karna mat bhoolna.