Worked examples — Diffraction — single slit intensity pattern derivation

This page is the problem gym for the single-slit intensity derivation. The parent note built the master formula from scratch:

Before any numbers, one habit: every quantity here is a length or an angle, so units must survive every line. If you ever divide a length by a length and get "metres", you made an error.

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The scenario matrix

Diffraction problems look varied but they are all cells of one grid. Every worked example below is tagged with the cell it fills, so by the end you have hit every corner.

• Cell A — find a minimum angle. Given $\lambda ,a$, solve $a\sin \theta =m\lambda$. → Ex 1
• Cell B — find a minimum position on a screen. Add screen distance $L$, use $y\approx L\theta$. → Ex 2
• Cell C — intensity at a given angle. Plug $\beta$ into $(\sin \beta /\beta {)}^2$. → Ex 3
• Cell D — secondary maximum. Not halfway: solve $\tan \beta =\beta$. → Ex 4
• Cell E — degenerate / limiting inputs. $\theta =0$; $a=\lambda$; $a<\lambda$ (no dark fringe); $a\to \infty$ (diffraction vanishes). → Ex 5
• Cell F — inverse problem. Given the pattern, find $a$ or $\lambda$. → Ex 6
• Cell G — real-world word problem. Laser through a hair, wall projection. → Ex 7
• Cell H — exam twist: two colours / combined constraint. When do minima of two wavelengths coincide? → Ex 8

Two edge cases deserve a flag before we start, because they trap people:

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Cell A — Angle of a minimum

1. Write the dark-fringe condition. Why this step? Minima — not maxima — are the sharp landmarks; the parent note showed pair-cancellation gives $a\sin \theta =m\lambda$. $\sin {\theta }_m=\frac{m\lambda }{a}=\frac{m(650\times 10^{-9})}{0.12\times 10^{-3}}=m\times 5.417\times 10^{-3}.$

2. $m=1$. Why: this is the inner edge of the central blob. $\sin {\theta }_1=5.417\times 10^{-3}\Rightarrow {\theta }_1=\arcsin (0.005417)=0.3103^{\circ }.$

3. $m=2$. Why: the next dark ring out. $\sin {\theta }_2=1.0833\times 10^{-2}\Rightarrow {\theta }_2=0.6207^{\circ }.$

Forecast answer: yes, almost exactly double here — because $\theta$ is tiny, $\sin \theta \approx \theta$, so doubling $\sin \theta$ doubles $\theta$. This coincidence breaks at large angles.

Recall Verify

Sanity: $a\gg \lambda$ (120 μm vs 0.65 μm), so angles must be tiny — they are (sub-degree). ✓ Ratio check ::: ${\theta }_2/{\theta }_1=0.6207/0.3103=2.00$, matches small-angle doubling.

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Cell B — Position on a screen

1. Convert angle to screen position, and justify the small-angle chain. Why this step? The screen is flat and far, so a ray at angle $\theta$ hits height $y=L\tan \theta$ exactly. Here ${\theta }_1=0.31^{\circ }=5.42\times 10^{-3}$ rad — a tiny number. For such tiny angles the Taylor expansions in radians give $\tan \theta =\theta +\frac{1}{3}{\theta }^3+\cdots \approx \theta$ and $\sin \theta =\theta -\frac{1}{6}{\theta }^3+\cdots \approx \theta$, both accurate to better than one part in $10^5$. So $\tan \theta \approx \sin \theta \approx \theta$ and we may use whichever is handy — here $\sin {\theta }_1$ is already known from Ex 1. See the figure: the red ray leaves the slit at ${\theta }_1$ and strikes the screen at $y_1$. $y_1\approx L{\theta }_1\approx L\sin {\theta }_1=1.8\times 5.417\times 10^{-3}=9.75\times 10^{-3}\text{m}=9.75\text{mm}.$

2. Full central width. Why: the central bright band runs from the $m=-1$ dark fringe to the $m=+1$ dark fringe — symmetric, so double $y_1$. $W=2y_1=19.5\text{mm}.$

3. Compare to spacing of outer fringes. Why: outer dark fringes are spaced by $\Delta y=L\lambda /a=9.75\text{mm}$ each. So the central band ($19.5$ mm) is twice as wide as the gap between any two neighbouring outer fringes.

Forecast answer: wider — the central maximum is the signature "fat middle" of every diffraction pattern.

Recall Verify

Units ::: $[\text{m}]\times [\text{dimensionless}]=[\text{m}]$ ✓ — $\sin \theta$ (and $\theta$ in radians) carries no units. $W=2\times 9.75=19.5$ mm, and outer spacing $=L\lambda /a=9.75$ mm, ratio exactly $2$. ✓

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Cell C — Intensity at an arbitrary angle

1. Find $\beta$. Why this step? Intensity depends only on $\beta$, not on $\theta$ directly. $\beta =\frac{\pi a\sin \theta }{\lambda }=\pi \cdot \frac{3}{4}=\frac{3\pi }{4}.$

2. Plug into sinc-squared. Why: that IS the master formula. $\frac{I}{I_0}={\left(\frac{\sin (3\pi /4)}{3\pi /4}\right)}^2={\left(\frac{0.7071}{2.356}\right)}^2=(0.3001{)}^2=\mathrm{0.0901.}$

3. Interpret. Why: $\approx 9\%$ — already very dim, heading fast to zero at $\beta =\pi$.

Forecast answer: below half. The sinc-squared curve falls steeply, so "three-quarters of the way in angle" is already down to ~9%, not ~25%.

Recall Verify

$\sin (3\pi /4)=\frac{\sqrt{2}}{2}=0.7071$, divide by $2.356$, square → $0.0901$. ✓ Cross-check against parent's Ex 3: at $a\sin \theta =\lambda /2$ ($\beta =\pi /2$), $I/I_0=4/{\pi }^2=0.405$. Our point is further out, so must be dimmer — $0.09<0.405$ ✓.

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Cell D — Secondary maximum (the "not halfway" case)

1. Differentiate — see where the slope is zero. Why this step? A peak of $f(\beta )={\left(\frac{\sin \beta }{\beta }\right)}^2$ is where $f^{\prime }(\beta )=0$. Write $g(\beta )=\frac{\sin \beta }{\beta }$; then $f=g^2$ and $f^{\prime }=2g{g}^{\prime }$, so we need $g=0$ (that's the dark fringes, not what we want) OR $g^{\prime }=0$. Using the quotient rule on $g=\sin \beta /\beta$: $g^{\prime }(\beta )=\frac{\beta \cos \beta -\sin \beta }{{\beta }^2}.$ Set the numerator to zero: $\beta \cos \beta -\sin \beta =0$. Divide by $\cos \beta$: $\beta -\tan \beta =0\Rightarrow \boxed{\tan \beta =\beta }.$ Geometrically (see the figure) this is where the straight line $y=\beta$ crosses the branch of $y=\tan \beta$. Because $\tan \beta$ shoots up steeply from $\pi$ toward its asymptote at $1.5\pi$, it meets the gently-rising line $y=\beta$ before reaching $1.5\pi$ — so the peak is pulled toward the centre.

2. Solve numerically. Why: $\tan \beta =\beta$ has no clean closed form. ${\beta }_1\approx 1.4303\pi =4.4934\text{rad}.$

3. Intensity there. Why: plug ${\beta }_1$ into the envelope. $\frac{I}{I_0}={\left(\frac{\sin (4.4934)}{4.4934}\right)}^2=0.04718\approx 4.7\%.$

Forecast answer: NOT at $1.5\pi$ — it is at $1.430\pi$, shifted inward. The side lobe carries under 5% of the central brightness, which is why diffraction blobs have one bright core and faint ears.

Recall Verify

${\beta }_1=4.4934$: check $\tan (4.4934)=4.4934$ (self-consistent root). Intensity $(\sin (4.4934)/4.4934{)}^2=0.0472$. ✓ This matches the parent's stated $\approx 4.5\%$ (parent rounded $\beta$ to $1.5\pi$; exact root gives $4.7\%$).

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Cell E — Degenerate & limiting inputs

(a) Dead centre, $\theta =0$. Why: everyone assumes brightest but should prove it. $\beta =0\Rightarrow \frac{\sin \beta }{\beta }\to 1\Rightarrow I=I_0.$ The ratio $\frac{\sin \beta }{\beta }\to 1$ as $\beta \to 0$ is the small-angle limit ($\sin x\approx x$). All wavelets in phase → peak. Not a dark fringe (that's the $m=0$ trap from the parent note).

(b) $a=\lambda$. Why: the first minimum needs $\sin \theta =m\lambda /a=1$. $\sin {\theta }_1=\frac{\lambda }{a}=1\Rightarrow {\theta }_1=90^{\circ }.$ The first dark fringe is pushed to the very edge of "forward". Only one minimum ($m=1$) is physically reachable; $m=2$ would need $\sin \theta =2>1$ → impossible.

(c) $a=0.5\lambda$. Why: now check existence. $\sin {\theta }_1=\frac{\lambda }{a}=\frac{\lambda }{0.5\lambda }=2>1.$ No solution. There are zero dark fringes — the central maximum spreads across the entire forward hemisphere. The slit is so tight it barely diffracts into a beam at all; it behaves almost like a single point source radiating everywhere.

(d) $a\to \infty$ (very wide slit). Why: the opposite extreme — what happens as the slit becomes huge compared with $\lambda$? The first minimum sits at $\sin {\theta }_1=\frac{\lambda }{a}\underset{a\to \infty }{\to }0\Rightarrow {\theta }_1\to 0.$ The central bright band collapses to zero angular width — all the dark fringes crowd in toward $\theta =0$, so the pattern shrinks to a single infinitely-narrow spike straight ahead. Physically: diffraction vanishes and the light just travels straight through as a plain beam (the ray-optics / geometric-optics limit). A hole much bigger than the wavelength — a window, a doorway — casts a sharp shadow with no visible fringes. This is why we never see diffraction from everyday openings: for them $a\gg \lambda$ and $\lambda /a$ is unmeasurably small.

Forecast answer: as $a\to \infty$ the light marches straight through — diffraction disappears and geometry (sharp shadows) takes over.

Recall Verify

(a) ${\lim }_{\beta \to 0}\sin \beta /\beta =1$ ✓. (b) $\lambda /a=1\Rightarrow \theta =90^{\circ }$ ✓. (c) $\lambda /a=2>1$, no real $\arcsin$ ✓. (d) ${\lim }_{a\to \infty }\lambda /a=0\Rightarrow {\theta }_1\to 0$ ✓.

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Cell F — Inverse problem

1. Half-width first. Why this step? Our formulas use the distance to one fringe; $W$ is the full width. $y_1=\frac{W}{2}=13\text{mm}=0.013\text{m}.$

2. Relate to the slit. Why: $y_1=L\lambda /a$ (small-angle first minimum). Rearrange for $a$. $a=\frac{L\lambda }{y_1}=\frac{2.5\times 546\times 10^{-9}}{0.013}.$

3. Compute. Why: just arithmetic now. $a=\frac{1.365\times 10^{-6}}{0.013}=1.05\times 10^{-4}\text{m}=0.105\text{mm}=105\mu \text{m}.$

Forecast answer: about a tenth of a millimetre — a hundred microns, a typical lab slit.

Recall Verify

Units ::: $[\text{m}][\text{m}]/[\text{m}]=[\text{m}]$ ✓. Back-substitute ::: $y_1=L\lambda /a=2.5\times 546\times 10^{-9}/1.05\times 10^{-4}=0.013$ m ✓.

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Cell G — Real-world word problem

1. Fringe spacing = distance between neighbouring minima. Why this step? Adjacent dark fringes ($m$ and $m+1$) are separated by $\Delta y=L\lambda /a$ in the small-angle regime. $\Delta y=\frac{L\lambda }{a}.$

2. Solve for $a$. Why: the hair width is what we want. $a=\frac{L\lambda }{\Delta y}=\frac{3.0\times 632.8\times 10^{-9}}{0.021}=9.04\times 10^{-5}\text{m}=90\mu \text{m}.$

Forecast answer + reality check: ~90 μm. Real hair is 50–100 μm — spot on. This is a genuine way to measure a hair with a laser and a ruler.

Recall Verify

$a=3.0\times 632.8\times 10^{-9}/0.021=9.04\times 10^{-5}$ m $=90.4\mu$m ✓, within the human-hair range.

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Cell H — Exam twist: two colours together

1. Coincidence condition. Why this step? A dark fringe of blue at order $m_1$ and orange at order $m_2$ share an angle when $a\sin \theta =m_1{\lambda }_1=m_2{\lambda }_2.$

2. Reduce the ratio. Why: the smallest integers satisfying $m_1{\lambda }_1=m_2{\lambda }_2$ come from $\frac{m_1}{m_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{640}{480}=\frac{4}{3}$. So the smallest pair is $m_1=4$ (blue), $m_2=3$ (orange).

3. Compute the angle. Why: plug either colour — they agree by construction. $\sin \theta =\frac{m_1{\lambda }_1}{a}=\frac{4\times 480\times 10^{-9}}{80\times 10^{-6}}=\mathrm{0.0240.}$ $\theta =\arcsin (0.0240)=1.3752^{\circ }.$

4. Cross-check with orange. Why: trust nothing until both colours match. $\frac{m_2{\lambda }_2}{a}=\frac{3\times 640\times 10^{-9}}{80\times 10^{-6}}=\mathrm{0.0240.}✓$

Forecast answer: the ratio is $4:3$, so blue's 4th dark fringe coincides with orange's 3rd, at $\theta \approx 1.38^{\circ }$.

Recall Verify

$4\times 480=1920=3\times 640$ ✓ (both $=1920$ nm·order). $\sin \theta =1920\times 10^{-9}/80\times 10^{-6}=0.024$, $\theta =1.3752^{\circ }$ ✓.

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Connections

• Contrast every "bright vs dark" rule with Young's Double Slit Experiment — opposite meaning for $a\sin \theta =m\lambda$.
• The pair-cancellation logic rests on Huygens Principle and Phasor Addition of Waves.
• The far-screen "$y=L\tan \theta$" geometry assumes the Fraunhofer limit.
• Many-slit generalisation → Diffraction Grating; resolving two blobs → Rayleigh Criterion and Resolution.
• The "narrow slit ⇒ wide spread" trend is the wave root of the Heisenberg Uncertainty Principle.