Exercises — Diffraction — single slit intensity pattern derivation

First dark fringe → smallest non-zero $m$, so $m=1$. Use $a\sin \theta =m\lambda$: $\sin \theta =\frac{m\lambda }{a}=\frac{1\cdot 500\times 10^{-9}}{0.20\times 10^{-3}}=2.5\times 10^{-3}.$ Since this is tiny, $\theta \approx \sin \theta$ in radians: $\theta \approx 2.5\times 10^{-3}\text{rad}=0.1432^{\circ }.$ What it looks like: a very small opening angle — the first dark line is almost straight ahead, because $a$ is far bigger than $\lambda$.

$a\sin \theta =2\lambda$ matches $a\sin \theta =m\lambda$ with $m=2$, a whole number $\ne 0$. $\Rightarrow \text{DARK (second minimum).}$ Check with $\beta$: $\beta =\frac{\pi a\sin \theta }{\lambda }=\frac{\pi \cdot 2\lambda }{\lambda }=2\pi$, and $\sin (2\pi )=0$, so $I=0$. Confirmed dark.

A point on the screen at angle $\theta$ sits at height $y=L\tan \theta$ (exact geometry). In the reference figure this is where the red ray from the slit strikes the screen. Because the diffraction angles here are tiny ($\theta \sim 10^{-3}$ rad), we use the small-angle approximation $\tan \theta \approx \sin \theta \approx \theta (\theta \text{ in radians, }\theta \ll 1),$ so $y\approx L\theta$. This is the single approximation that lets us combine the screen geometry ($y=L\tan \theta$) with the diffraction formula ($a\sin {\theta }_1=\lambda$) consistently. Half-width $y_1$ reaches the first minimum where ${\theta }_1\approx \lambda /a$: $y_1=L{\theta }_1=L\frac{\lambda }{a}=1.5\cdot \frac{500\times 10^{-9}}{0.20\times 10^{-3}}=3.75\times 10^{-3}\text{m}.$ Full width = both sides: $W=2y_1=7.5\times 10^{-3}\text{m}=7.5\text{mm}.$

We reuse the same small-angle approximation as in L2.1: for tiny angles $\tan \theta \approx \sin \theta \approx \theta$, so the screen height and the diffraction angle share the same $\theta$. Then ${\theta }_1\approx \lambda /a$ and $y_1=L{\theta }_1$ combine to $y_1=\frac{L\lambda }{a}\Rightarrow a=\frac{L\lambda }{y_1}=\frac{2.0\cdot 650\times 10^{-9}}{6.5\times 10^{-3}}.$ $a=2.0\times 10^{-4}\text{m}=0.20\text{mm}.$ Check the approximation is valid: ${\theta }_1\approx \lambda /a=650\times 10^{-9}/0.20\times 10^{-3}=3.25\times 10^{-3}$ rad — indeed $\ll 1$, so $\tan \approx \sin \approx \theta$ holds to better than $0.001\%$.

Convert the path difference to $\beta$: $\beta =\frac{\pi a\sin \theta }{\lambda }=\frac{\pi \cdot \frac{2}{3}\lambda }{\lambda }=\frac{2\pi }{3}.$ Now plug into $I/I_0=(\sin \beta /\beta {)}^2$: $\frac{I}{I_0}={\left(\frac{\sin (2\pi /3)}{2\pi /3}\right)}^2={\left(\frac{\sqrt{3}/2}{2\pi /3}\right)}^2={\left(\frac{3\sqrt{3}}{4\pi }\right)}^2\approx \mathrm{0.1710.}$ So about $17\%$ of the central peak — we are on the falling shoulder, heading toward the first dark fringe at $a\sin \theta =\lambda$ ($\beta =\pi$).

Take $\beta =1.4303\pi \approx 4.4934$ rad. $\frac{I}{I_0}={\left(\frac{\sin (4.4934)}{4.4934}\right)}^2\approx {\left(\frac{-0.9775}{4.4934}\right)}^2\approx \mathrm{0.0472.}$ That is about $4.7\%$ — the side bands are dramatically fainter than the centre. The envelope $1/{\beta }^2$ is what crushes them. What it looks like: on the figure, the tallest side bump barely clears the axis compared to the towering central peak.

A double-slit bright fringe disappears when it lands exactly on a single-slit minimum.

• Double-slit bright: $d\sin \theta =p\lambda$.
• Single-slit dark: $a\sin \theta =m\lambda$.

Divide the two conditions (same $\theta$): $\frac{d}{a}=\frac{p}{m}\Rightarrow \frac{p}{m}=\frac{0.50}{0.10}=5.$ So $p=5m$: orders $p=\pm 5,\pm 10,\dots$ are missing (the first missing order is $p=5$, killed by $m=1$). Why: at that angle the two slits are in phase (double-slit says bright), but each individual slit has zero output (single-slit envelope is dark), so bright $\times$ zero $=$ zero.

A photon of wavelength $\lambda$ carries total momentum $p=h/\lambda$. Emerging within the central cone of half-angle ${\theta }_1$, its transverse (sideways) momentum spans roughly $\Delta p_y\approx p{\theta }_1=\frac{h}{\lambda }\cdot \frac{\lambda }{a}=\frac{h}{a}.$ Multiply by the position spread $\Delta y\sim a$: $\Delta y\Delta p_y\sim a\cdot \frac{h}{a}=h.$ What this does and does not say. The $a$ cancels — narrower slit → wider angular spread — and the product comes out at the scale of Planck's constant, i.e. $\Delta y\Delta p_y\sim h\sim 2\pi \hslash$. This is only an order-of-magnitude argument: our $\Delta y\sim a$ and $\Delta p_y\sim h/a$ are rough "sizes," not the precise statistical standard deviations. The exact, rigorous statement of the uncertainty principle is $\Delta y\Delta p_y\ge \frac{\hslash }{2},$ and our estimate ($\sim h=2\pi \hslash$) sits comfortably above that floor — as it must. The takeaway is the scaling ($\Delta p_y\propto 1/a$), not a claim that the product equals $h$ exactly.

What we do: Taylor-expand $\sin \beta$ for small $\beta$: $\sin \beta =\beta -\frac{{\beta }^3}{6}+\frac{{\beta }^5}{120}-\cdots$ Divide by $\beta$: $\frac{\sin \beta }{\beta }=1-\frac{{\beta }^2}{6}+\cdots \stackrel{\beta \to 0}{\to }1.$ Squaring keeps the limit $1$, so $I(0)=I_0\cdot 1=I_0$. Why / what it looks like (phasor picture). Recall from the parent derivation that the slit is imagined as a large number $N$ of thin strips, each radiating a tiny wavelet of amplitude $A_0/N$, where $A_0$ is the total amplitude you would get if all $N$ wavelets were perfectly in phase. As $\theta \to 0$ the path difference across the slit vanishes, so every wavelet is in step. Adding these $N$ equal-length arrows tip-to-tail with no phase turn between them gives a straight line of full length $A_0$ — nothing is lost to cancellation. That maximal, undiminished sum ($I_0=A_0^2$) is the towering central peak.

First minimum needs $\sin {\theta }_1=\lambda /a$.

• $a=\lambda$: $\sin {\theta }_1=1\Rightarrow {\theta }_1=90^{\circ }$. The first dark fringe is pushed all the way to the edge of the visible half-plane. The central bright band now fills the entire forward screen.
• $a<\lambda$: $\sin {\theta }_1=\lambda /a>1$, which has no real solution. There is no dark fringe at all — the intensity just falls off smoothly to the sides with a single broad hump. The slit is so narrow it behaves almost like one point source (a lone Huygens wavelet), radiating in all forward directions with no cancellation.

The lesson: the "fringe" picture requires $a>\lambda$. Once $a\le \lambda$ the pattern degenerates to a single spread-out glow — the extreme of "narrow slit ⇒ wide pattern."

L1.1 first minimum ::: $\theta \approx 0.1432^{\circ }$ L1.2 $a\sin \theta =2\lambda$ ::: dark (2nd minimum) L2.1 full central width ::: $7.5\text{mm}$ L2.2 slit width ::: $0.20\text{mm}$ L3.1 $I/I_0$ at $a\sin \theta =\frac{2}{3}\lambda$ ::: $\approx 0.171$ L3.2 first secondary peak ::: $\approx 0.047$ (4.7%) L4.1 first missing double-slit order ::: $p=5$ (family $5m$) L4.2 $\Delta y\Delta p_y$ ::: $\sim \hslash$ (order of magnitude; exact bound $\ge \hslash /2$) L5.1 $\lim (\sin \beta /\beta {)}^2$ ::: $1$, so $I(0)=I_0$ L5.2 $a\le \lambda$ ::: no dark fringes (single broad hump)