The two-slit formula treats each slit as a point. Real slits have a width $a$, and each one, by itself, makes a single-slit pattern. The real double-slit pattern is the product of the two: fine two-slit fringes, inside the broad bright band of one slit.
Key idea
For two slits of width $a$ whose centers are $d$ apart, $$I = I_{\max}\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right)\left(\frac{\sin\beta}{\beta}\right)^2,\quad \beta = \frac{\pi a\sin\theta}{\lambda}.$$ The single-slit factor is the envelope. A bright fringe of order $m$ that lands on a zero of the envelope is missing: this happens when $\frac{m}{d/a}$ is a whole number, so if $d = 3a$, orders $\pm 3, \pm 6, \ldots$ vanish.
Why it works
Divide each slit into many thin strips. All the strips in one slit add up to the single-slit amplitude $E_0\frac{\sin\beta}{\beta}$, which depends on $\theta$. The second slit is the same, shifted by $d$, so its light is the same amplitude with the extra phase $\frac{2\pi d\sin\theta}{\lambda}$. Adding the two gives the two-slit factor times the one-slit amplitude; squaring gives the product of intensities. The envelope is zero where $a\sin\theta = p\lambda$ ($p = \pm1, \pm2, \ldots$). A two-slit bright fringe sits where $d\sin\theta = m\lambda$. Both happen at the same angle when $\frac{m}{d} = \frac{p}{a}$, that is $m = p\,\frac{d}{a}$.
Worked examples
Worked example: Missing orders and the central band
Two slits each 0.100 mm wide have centers 0.300 mm apart. Which orders are missing, and how many bright fringes fit inside the central bright band of the envelope?
- $d/a = 0.300/0.100 = 3$.
- Missing: $m = 3p$, so $m = \pm 3, \pm 6, \ldots$
- The central band of the envelope runs to the first zero, $a\sin\theta = \lambda$, which is where $m = 3$ would be.
- Inside it: $m = 0, \pm1, \pm2$.
Answer: Orders $\pm3, \pm6, \ldots$ are missing; 5 bright fringes fit in the central band
Worked example: How bright is the first fringe?
For the same slits, how bright is the $m = 1$ fringe compared with the center?
- At the $m = 1$ fringe, $d\sin\theta = \lambda$, so the two-slit factor is $\cos^2\pi = 1$.
- There, $\beta = \frac{\pi a\sin\theta}{\lambda} = \pi\frac{a}{d} = \frac{\pi}{3}$.
- Envelope: $\left(\frac{\sin(\pi/3)}{\pi/3}\right)^2 = \left(\frac{0.866}{1.047}\right)^2 = 0.684$.
Answer: $0.684$ of the central intensity
Worked example: Read the slit width from a pattern
A double slit with $d = 0.250$ mm shows its first missing order at $m = 5$. How wide are the slits?
- The first missing order is $m = d/a$ (the first zero of the envelope).
- $a = d/5 = 0.250/5 = 0.0500$ mm.
Answer: $a = 50.0$ μm
Common mistake
Treating each slit as a point, so that every order $d\sin\theta = m\lambda$ appears with full brightness. Real slits have width, so their own diffraction dims the outer fringes and wipes out the orders that land on the envelope's zeros. A quick check: the fringe spacing depends on $d$, while the size of the envelope depends on $a$. The narrower the slits, the wider the envelope and the more fringes fit in it.