Worked examples — Young's double slit — fringe width derivation

This page is the exhaustive problem-drill child of the fringe-width derivation. The parent gave you the two master tools:

$\Delta =\frac{yd}{D}\text{and}\beta =\frac{\lambda D}{d}.$

Here we hit every kind of number these formulas can throw at you: the point above centre, the point below centre, the exact centre, the tiny-input and huge-input limits, a change of medium, a real-world word problem, and an exam twist. Each example says which cell of the matrix it fills, and — crucially — makes you forecast the answer before I show it.

---

The apparatus — every symbol, in words and a picture

Before any formula, meet the physical setup. The figure below is the schematic every example silently assumes; look at it once and the letters $d$, $D$, $S_1$, $S_2$, $P$, $O$, $y$ stop being mysterious.

---

Notation we will lean on — defined once, in words

Two more symbols do the counting work, so let us pin them down before we ever use them in a sum.

We will also need the dark-fringe positions. A dark band sits exactly halfway between two bright ones, so its order is a half-integer:

---

The scenario matrix

Before working examples, let us list every case-class a YDSE problem can be. If a cell is not covered you would meet a scenario you have never seen — that is exactly what we forbid.

#	Case class	What is weird / what to watch	Example that covers it
A	Point above centre, $y>0$	Straight $\Delta =+yd/D$	Ex 1
B	Point below centre, $y<0$	$\Delta$ is negative → same fringe by symmetry	Ex 2
C	Point at centre, $y=0$	$\Delta =0$ → the zeroth (central) bright fringe	Ex 2
D	Find $\lambda$ from a known fringe	Rearrange $y_n$, not $\beta$	Ex 3
E	Bright or dark? classify a given $y$	Compute $\Delta /\lambda$, integer vs half-integer	Ex 4
F	Medium change (water/oil)	$\lambda \to \lambda /n$ shrinks $\beta$	Ex 5
G	Real-world word problem	Reading a lab ruler, count $N$ fringes	Ex 6
H	Exam twist — shift screen distance	Careful re-reading of which variable changed	Ex 7
I	Degenerate / limiting input ($d\to 0$, $D\to \infty$, $\lambda \to 0$)	Does the formula stay sane?	Ex 8

We now fill every cell.

---

Example 1 — Cell A (point above centre)

Step 1 — Convert to SI. $\lambda =500\times 10^{-9}\text{m}$, $d=0.20\times 10^{-3}\text{m}$, $D=1.0\text{m}$. Why this step? The formula $\beta =\lambda D/d$ is only true if all lengths use the same ruler; mixing nm and mm gives nonsense.

Step 2 — Fringe width first. $\beta =\frac{\lambda D}{d}=\frac{(500\times 10^{-9})(1.0)}{0.20\times 10^{-3}}=2.5\times 10^{-3}\text{m}=2.5\text{mm}.$ Why this step? $\beta$ is the "unit gap"; once we have it, the $n$-th fringe is just $n$ of them.

Step 3 — Position of 3rd bright fringe. $y_3=3\beta =3(2.5\text{mm})=7.5\text{mm}.$ Why this step? $y_n=n\lambda D/d=n\beta$, so counting fringes is literally counting $\beta$'s.

The figure below turns this arithmetic into a picture: the blue bars are the bright bands at $y_0=0,y_1=2.5,y_2=5.0,\dots$ mm; notice each is exactly one green arrow ($\beta$) above the last, and the orange bar is our answer $y_3=7.5$ mm — the third rung up the ladder.

Verify: Directly, $y_3=3\lambda D/d=3(500\times 10^{-9})(1)/(0.2\times 10^{-3})=7.5\times 10^{-3}\text{m}$ ✓. Millimetres, as forecast — sensible for a lab bench.

---

Example 2 — Cells B & C (below centre, and the centre itself)

Step 1 — Below centre means $y<0$. For the point $2\beta$ below $O$, $y_{-2}=-2\beta =-5.0\text{mm}$. Why this step? $y$ is a signed height measured from $O$; below centre is negative. The path difference becomes $\Delta =yd/D<0$, meaning now $S_{1}P$ (the distance from slit $S_1$ to the point) is the longer path instead of $S_{2}P$.

Step 2 — Negative $\Delta$ is still a valid bright condition. $\Delta =-2\lambda$ satisfies $\Delta =n\lambda$ with $n=-2$. A whole number of wavelengths — positive or negative — is still constructive. Why this step? The pattern is symmetric about $O$; the fringe $2\beta$ below looks identical to the one $2\beta$ above. That is why we allow $n=0,\pm 1,\pm 2,\dots$

The figure makes the mirror symmetry explicit: the blue slits $S_1,S_2$ on the left send dashed rays to three screen points. The green dot is $O$ (the $n=0$ central bright fringe at $y=0$); the orange dot is a point above ($y>0$, $n=+2$) and the red dot its mirror image below ($y<0$, $n=-2$). The two double-headed arrows are equal in length — that equality is the symmetry.

Step 3 — The centre. At $O$, $y=0\Rightarrow \Delta =0=0\cdot \lambda$. That is $n=0$: a bright fringe. Why this step? Zero path difference means the two waves arrive perfectly in step — maximum brightness. The centre is always bright in a symmetric YDSE.

Verify: $|y_{-2}|=2\beta =5.0\text{mm}$ matches $y_{+2}=+5.0\text{mm}$ by symmetry ✓. Central fringe $\Delta =0$ → constructive → bright ✓ (a common trap: people say the centre is dark).

---

Example 3 — Cell D (find the wavelength)

Step 1 — Choose the right formula. We know a specific fringe's position, so use $y_n=n\lambda D/d$, not $\beta$ alone. Why this step? $\beta$ tells you spacing; here we are handed the absolute position of fringe number $n=5$, so the $y_n$ form is the direct fit.

Step 2 — Rearrange for $\lambda$. $\lambda =\frac{y_{n}d}{nD}=\frac{(6.0\times 10^{-3})(0.40\times 10^{-3})}{(5)(0.80)}.$ Why this step? Solve algebraically for the unknown before plugging numbers — fewer arithmetic slips.

Step 3 — Compute. $\lambda =\frac{2.4\times 10^{-6}}{4.0}=6.0\times 10^{-7}\text{m}=600\text{nm}.$

Verify: $600\text{nm}$ is orange-red visible light ✓ — inside 400–700 nm as forecast. Back-check: $y_5=5(600\text{e-}9)(0.8)/(0.4\text{e-}3)=6.0\text{mm}$ ✓.

This uses the small-angle result $y_n=n\lambda D/d$ derived above.

---

Example 4 — Cell E (bright or dark? classify a point)

Step 1 — Compute the path difference. $\Delta =\frac{yd}{D}=\frac{(4.0\times 10^{-3})(0.30\times 10^{-3})}{1.2}=1.0\times 10^{-6}\text{m}.$ Why this step? $\Delta$ is the master quantity — everything (bright/dark) is decided by comparing $\Delta$ to $\lambda$. See Path difference and phase difference.

Step 2 — Divide by $\lambda$ to count wavelengths. $\frac{\Delta }{\lambda }=\frac{1.0\times 10^{-6}}{600\times 10^{-9}}=\frac{5}{3}\approx \mathrm{1.667.}$ Why this step? Compare against the two exact conditions written as equations: $\text{bright}⟺\frac{\Delta }{\lambda }=n\in \{0,\pm 1,\pm 2,\dots \},\text{dark}⟺\frac{\Delta }{\lambda }=m+\frac{1}{2}\in \{\pm \frac{1}{2},\pm \frac{3}{2},\dots \}.$

Step 3 — Classify. $\frac{5}{3}=1.667$ is neither an integer nor a half-integer. It lies between $\frac{3}{2}$ (dark) and $2$ (bright), so the point is partially lit — brighter than a minimum, dimmer than a maximum. Why this step? Real detectors read intensity smoothly; only the exact integer/half-integer points are pure max/min.

Verify: $\Delta /\lambda =5/3\approx 1.667$ ✓; not in $\{0,1,2,\dots \}$ nor $\{0.5,1.5,\dots \}$, so "neither" is correct ✓.

---

Example 5 — Cell F (immerse in a medium)

Step 1 — What changes in water? Frequency is fixed (set by the source); light slows by factor $n_w$, so ${\lambda }^{\prime }=\lambda /n_w$. Why this step? $\beta$ depends on the in-medium wavelength, and only $\lambda$ changes — $d$ and $D$ are physical distances that do not care about the medium.

Step 2 — Scale the fringe width. ${\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d}=\frac{\lambda D}{n_{w}d}=\frac{\beta }{n_w}=\frac{1.8\text{mm}}{1.33}.$ Why this step? $\beta \propto \lambda$, so dividing $\lambda$ by $n_w$ divides $\beta$ by the same $n_w$.

Step 3 — Compute. ${\beta }^{\prime }=1.35\text{mm}(\text{to 3 s.f.}).$

Verify: $1.8/1.33=1.353\text{mm}$ ✓ — narrower, as forecast (higher index squeezes the pattern).

---

Example 6 — Cell G (real-world word problem)

Step 1 — Count the gaps, not the fringes. From fringe number 1 to fringe number 10 there are $10-1=9$ gaps (each gap is one $\beta$). Picture nine fence-panels between ten fence-posts: ten posts, but only nine panels. So $9\beta =16.2\text{mm}⟹\beta =\frac{16.2\text{mm}}{9}=1.80\text{mm}.$ Why this step? This is the single most common lab error. The distance between fringe $a$ and fringe $b$ is $(b-a)\beta$, because $y_b-y_a=(b-a)\beta$ from $y_n=n\beta$. Counting fringes (ten) instead of gaps (nine) inflates the spacing.

Step 2 — Solve $\beta =\lambda D/d$ for $\lambda$. $\lambda =\frac{\beta d}{D}=\frac{(1.80\times 10^{-3})(0.25\times 10^{-3})}{1.5}.$ Why this step? Everything else is measured; only $\lambda$ is unknown, so isolate it.

Step 3 — Compute. $\lambda =\frac{4.5\times 10^{-7}}{1.5}=3.0\times 10^{-7}\text{m}=300\text{nm}.$

Verify: $\lambda =300\text{nm}$ ✓ arithmetic — but this is ultraviolet, invisible to the eye. In a real lab that flags something: either the light truly is UV (detected on film, not by eye), or the student mis-counted and used 10 gaps instead of 9. Had they wrongly used $10\beta =16.2$ mm they would get $\beta =1.62$ mm and $\lambda =270\text{nm}$ — still UV, and further from any visible value. Lesson: count gaps = (last − first), then sanity-check the colour band.

---

Example 7 — Cell H (exam twist)

Step 1 — Compute ${\beta }_1$. ${\beta }_1=\frac{\lambda D_1}{d}=\frac{(650\times 10^{-9})(1.0)}{0.50\times 10^{-3}}=1.30\times 10^{-3}\text{m}=1.30\text{mm}.$ Why this step? Anchor the "before" value.

Step 2 — Use proportionality, don't recompute from scratch. $\beta \propto D$, so ${\beta }_2={\beta }_1\cdot (D_2/D_1)=1.30\text{mm}\times 2=2.60\text{mm}$. Why this step? Recognising $\beta \propto D$ is faster and exam-safe: only $D$ changed (the slits $S_1,S_2$ and their spacing $d$ are untouched), so scale.

Step 3 — The shift. $\Delta \beta ={\beta }_2-{\beta }_1=2.60-1.30=1.30\text{mm}.$ Why this step? Exams often ask for the change, testing whether you know the linear dependence.

Verify: ${\beta }_2=650\text{e-}9\cdot 2.0/0.5\text{e-}3=2.6\text{mm}$ ✓ ; doubling $D$ doubled $\beta$ ✓ as forecast.

---

Example 8 — Cell I (degenerate & limiting inputs)

Step 1 — $d\to 0$ (slits merge). $\beta =\lambda D/d\to \infty$. Why this step? As the two slits $S_1,S_2$ merge into one, fringes spread infinitely far apart — meaning the whole screen becomes a single bright blur (no distinguishable pattern). This is the two-source pattern collapsing back to one source. Physically: you cannot have interference from a single opening (that is single-slit diffraction instead).

Step 2 — $D\to \infty$ (screen very far). $\beta \to \infty$ too, but the angular width $\theta =\lambda /d$ stays fixed and finite. Why this step? The linear spacing grows without bound, yet the angle between fringes is independent of $D$ — that is why the parent note prized the angular form $\theta =\lambda /d$.

Step 3 — $\lambda \to 0$ (imagine light with no wavelength). $\beta \to 0$: fringes crowd infinitely close, washing into uniform brightness — no visible pattern. Why this step? Interference is a wave effect; a zero-wavelength wave behaves like a straight ray (geometric optics), and rays do not interfere. This is why we never see interference with, say, a cricket ball.

The figure plots $\beta$ against $d$ for fixed $\lambda ,D$: watch the blue curve rocket toward the sky as $d$ shrinks to the left (the $d\to 0$ blow-up, red dashed line), and flatten toward small $\beta$ as $d$ grows to the right (fringes crowding, green arrow). It is the shape of $\beta \propto 1/d$ made visible.

Verify: All three limits are consistent with $\beta =\lambda D/d$: numerator $\to 0$ gives $\beta \to 0$; denominator $\to 0$ gives $\beta \to \infty$; $D\to \infty$ gives $\beta \to \infty$ while $\beta /D=\lambda /d$ stays finite ✓.

---

Recap of the matrix

Recall Which example filled each cell?

A→Ex1, B→Ex2, C→Ex2, D→Ex3, E→Ex4, F→Ex5, G→Ex6, H→Ex7, I→Ex8. Every cell covered.

Recall The single question behind every problem

Compute $\Delta =yd/D$, divide by $\lambda$: integer = bright, half-integer = dark, else in-between. ::: Yes — that one comparison drives all nine cells.

Recall How do you convert "distance across N fringes" into

$\beta$? Count gaps = (last fringe number − first fringe number), then divide the measured distance by that number of gaps. ::: From fringe 1 to fringe 10 there are 9 gaps, so $\beta =\text{distance}/9$.

---

Connections

• Young's double slit — fringe width derivation — the parent derivation these drills exercise.
• Path difference and phase difference — the $\Delta$ we compute in every example.
• Small angle approximation — why $\sin \theta \approx \tan \theta =y/D$ throughout.
• Refractive index — Example 5's medium change.
• Coherence and coherent sources — needed for fringes to exist at all (Ex 8a).
• Diffraction grating — the $d\to 0$ single-opening limit.
• Interference of light — the umbrella phenomenon.