Single Slit Diffraction Pattern

When light passes through a single narrow slit, it doesn't just project a sharp shadow of the slit on a distant screen. Instead, it spreads out and forms a pattern of bright and dark bands — a diffraction pattern. The central bright band is much wider than the slit, and it's flanked by progressively weaker secondary maxima separated by dark minima.

This is different from the double-slit experiment (Problem 154), which produces evenly-spaced fringes of nearly equal brightness. Single-slit diffraction produces a dominant central maximum that contains most of the light, with much weaker side fringes. Understanding the difference — and understanding that both effects coexist when you have two slits — is one of the keys to mastering wave optics.

Why does a slit cause diffraction?

Huygens' principle says that every point on a wave front acts as a source of secondary wavelets. When a plane wave hits a slit, every point across the slit width becomes a tiny new source of circular waves. These wavelets all have the same frequency and wavelength, but they travel different distances to reach any given point on the screen. At some angles, they add up constructively; at others, they cancel destructively.

The crucial insight: the slit has a finite width $a$. If $a$ were infinitely narrow, the light would spread out uniformly in all directions (a point source). If $a$ were infinitely wide, there'd be no diffraction — just a geometric shadow. It's the interplay between $a$ and $\lambda$ that creates the pattern.

Finding the minima

The positions of destructive interference (dark fringes / minima) are easiest to find. Consider dividing the slit into two equal halves. For each point in the top half, there's a matching point in the bottom half that is $a/2$ below it. If the path difference between these paired points equals $\lambda/2$, their contributions cancel perfectly.

This happens when:

$a \sin\theta = m\lambda, \quad m = \pm 1, \pm 2, \pm 3, \ldots$

Note: $m = 0$ is not a minimum — it's the central maximum (zero path difference → everything adds up).

For small angles ($\sin\theta \approx \theta \approx y/L$):

$y_{m} = \dfrac{m\lambda L}{a}$

For our problem: $\lambda = 500$ nm, $a = 0.05$ mm $= 5 \times 10^{-5}$ m, $L = 2$ m.

First minimum ($m = 1$): $y_{1} = \dfrac{(1)(500 \times 10^{-9})(2)}{5 \times 10^{-5}} = \dfrac{10^{-6}}{5 \times 10^{-5}} = 0.02 \text{ m} = 20 \text{ mm}$

The width of the central maximum = $2y_{1} = 40$ mm = 4 cm. That's huge compared to the slit width (0.05 mm) — the light has spread out by a factor of 800!

Second minimum ($m = 2$): $y_{2} = 40$ mm. Third minimum ($m = 3$): $y_{3} = 60$ mm.

The intensity pattern (method 2)

The complete intensity formula from the Fraunhofer diffraction integral is:

$I(\theta) = I_{0} \left[\dfrac{\sin\beta}{\beta}\right]^{2}$

where $\beta = \dfrac{\pi a \sin\theta}{\lambda}$.

This is the sinc-squared function ($\text{sinc}^{2}$). At $\theta = 0$, $\beta = 0$ and $\sin\beta/\beta \to 1$, giving maximum intensity $I_{0}$. The zeros occur at $\beta = m\pi$, i.e., $a\sin\theta = m\lambda$ — exactly the minima we found above.

The secondary maxima (bright spots between minima) are much weaker than the central peak:

• Central maximum: $I_{0}$ (100%)
• First secondary maximum: approximately $0.045\,I_{0}$ (4.5%)
• Second secondary maximum: approximately $0.016\,I_{0}$ (1.6%)
• Third: approximately $0.008\,I_{0}$ (0.8%)

So the central band contains about 84% of the total light energy. The remaining 16% is split among all the side fringes.

The Rayleigh criterion and resolution

Diffraction sets a fundamental limit on how small a detail any optical instrument can resolve. The Rayleigh criterion says that two point sources are just barely resolvable when the central maximum of one falls on the first minimum of the other. This gives a minimum resolvable angle:

$\theta_{\min} = \dfrac{1.22\,\lambda}{D}$

where $D$ is the diameter of the circular aperture (for a circular opening, the factor is 1.22 instead of 1.00 for a slit).

This is why telescopes need large mirrors — a bigger $D$ means a smaller $\theta_{\min}$ and sharper images. It's also why electron microscopes use electrons instead of light — electrons have much shorter wavelengths, so the resolution limit is much finer.

Single slit vs double slit

In the double-slit experiment, two effects happen simultaneously:

1. Each slit individually produces a single-slit diffraction pattern (the broad envelope)
2. The two slits interfere with each other (the fine fringes)

The actual observed pattern is the product: the fine double-slit fringes, modulated by the single-slit diffraction envelope. Where the envelope reaches a minimum, several double-slit fringes vanish ("missing orders"). This is why, in real double-slit experiments, the fringes far from center are dimmer than the ones near center.

Common mistakes

• Confusing the minima condition with double-slit maxima. For a single slit, $a\sin\theta = m\lambda$ gives minima. For double slits, $d\sin\theta = m\lambda$ gives maxima. Same equation form, opposite meanings. Pay attention to which problem you're solving.
• Forgetting the central maximum is twice as wide. The distance from the first minimum on one side to the first minimum on the other is $2\lambda L/a$, not $\lambda L/a$. The central peak is double-width because it spans from $m = -1$ to $m = +1$.
• Using $d$ instead of $a$. In single-slit problems, the relevant quantity is the slit width $a$, not the slit separation $d$ (which doesn't exist — there's only one slit).

Try it in the visualization

Widen the slit ($a$) and watch the central peak narrow — more slit width means less spreading. Shorten the wavelength and the pattern also tightens. Toggle the intensity graph to see the sinc² shape with its rapidly decaying side lobes. Compare with the double-slit pattern from Problem 154 to see how the single-slit envelope modulates the interference fringes.