Single-slit diffraction or two-slit interference? Read the pattern

1. Read the aperture diagram

For a single slit, light from different parts of the same opening combines. For two slits, the contributions from the two openings combine too. Both patterns arise from superposition; the useful distinction is which parts of the aperture you are comparing.

A broad central bright region bounded by zeros is a single-slit feature. Closely spaced bright and dark fringes are a two-slit interference feature. Real two-slit patterns can show both.

Before calculating: a drawing labels a clear gap of 0.060 mm between two openings, each 0.020 mm wide. What is d? It is 0.080 mm: half a width, the clear gap, then half a width. The gap itself is not d.

2. State the model and quantities

Assume monochromatic light in a uniform medium, incident normally and in phase across the aperture. The long, parallel slits are uniformly illuminated; we study spreading across their widths. For two slits, assume identical widths and equal illumination with a stable relative phase of zero.

Use a far-field (Fraunhofer) pattern. For free propagation, a useful sufficient check is $L\gg D^2/\lambda$, where D is the full transverse aperture size: a for one slit, or d+a for this two-slit arrangement. A suitable lens can instead produce a far-field pattern in its focal plane; that configuration needs its own screen geometry. Being far from the aperture does not automatically make every observation angle small.

• $\lambda$: wavelength in the propagation medium, in metres (m).
• $a>0$: width of each slit, in m.
• $d>a$: centre-to-centre slit separation, in m, for two non-overlapping openings with a gap.
• $L>0$: perpendicular distance from aperture plane to screen, in m.
• $y$: signed screen position from the central axis, in m; positive above the centre in our side-view geometry.
• $\theta$: signed angle from the central axis, positive toward positive y; use radians in small-angle approximations.

On a flat screen, $y=L\tan\theta$. Near the centre, $\sin\theta\approx\tan\theta\approx\theta$, so $y\approx L\sin\theta$. Check $|y|/L\ll1$ before using this shortcut. If it is not small, retain the trigonometry. Remember $1\,\mathrm{nm}=10^{-9}\,\mathrm m$ and $1\,\mathrm{mm}=10^{-3}\,\mathrm m$.

If wavelength is unfamiliar, review wavelength and period on wave graphs before continuing.

3. One slit: locate dark minima

For a uniformly illuminated slit, the far-field dark minima satisfy

Here n labels a diffraction minimum. Zero is excluded: straight ahead, the contributions add to form the central maximum. Only orders with $|n\lambda/a|<1$ reach a finite point on our screen.

At the first positive minimum, the edge-to-edge path difference is one wavelength. Pair each point in one half of the slit with the corresponding point in the other half: the paired paths differ by half a wavelength and cancel. Comparing only the two edges would miss this cancellation across the full opening.

For small angles,

W is the full width of the central maximum between its first dark boundaries, from $y_{-1}$ to $y_{+1}$. It is not the centre-to-first-minimum distance and not the full width at half maximum. The side maxima are weaker; do not treat them as equally bright copies of the centre.

4. Two slits: locate interference fringes

In the ideal narrow-slit model, constructive interference occurs when the path difference between the two slit centres is a whole number of wavelengths:

Here m labels an interference maximum, including the central bright fringe at m=0. The ideal dark interference fringes satisfy $d\sin\theta=(m+\tfrac12)\lambda$, with m any integer. These are different conditions from the single-slit minima.

Near the centre, successive ideal bright fringes have spacing

$\Delta y>0$ is the distance from one bright fringe to the next on the screen. It is not a signed position. Equal spacing is a small-angle approximation, not an exact statement across an arbitrarily wide flat screen.

5. Worked comparison: keep wavelength and screen distance fixed

Use $\lambda=480\,\mathrm{nm}$ and $L=3.00\,\mathrm m$. Experiment A uses one slit of width $a=0.0240\,\mathrm{mm}$. Experiment B uses two identical slits of that width with centre separation $d=0.120\,\mathrm{mm}$. Find A’s first dark positions and central width, then B’s ideal bright-fringe spacing.

Step 1: convert before substituting

Step 2: use a for the single slit

By symmetry, $y_{-1}\approx-60.0\,\mathrm{mm}$. The full central width is $W\approx120\,\mathrm{mm}$. The negative position marks the opposite side of the centre; the width is positive.

Step 3: use d for the two slits

The ideal m=+2 bright position is $y_2\approx+24.0\,\mathrm{mm}$. The result is a position measured from the centre, twice the adjacent-fringe spacing.

Step 4: check the approximation and the units

At A’s first minimum, $\sin\theta=0.0200$. Using $y=L\tan(\arcsin0.0200)$ gives 60.012 mm, about 0.020% above the approximation. The full two-slit aperture is $D=d+a=0.144\,\mathrm{mm}$, giving $D^2/\lambda=0.0432\,\mathrm m$, much smaller than 3.00 m. The stated far-field model is appropriate here. In both screen formulas, the units reduce to $\mathrm m\times\mathrm m/\mathrm m=\mathrm m$.

6. Predict first, then calculate

7. Real two-slit patterns have an envelope

Each slit has a finite width. Its diffraction changes the brightness of the two-slit fringes: fine interference structure sits inside a broad diffraction envelope. Increasing d at fixed a packs the fringes more closely; increasing a at fixed d narrows the envelope.

For our worked numbers, d/a=5. The nominal m=±5 interference maxima coincide with n=±1 diffraction minima. Those orders are missing: the envelope is zero there. The central envelope contains nominal orders −4 through +4.

The simple spacing formula locates ideal interference orders. Multiplying by a varying envelope can shift the actual local intensity peaks slightly, so do not claim exact measured peak spacing for finite-width slits.

8. Independent retry

Cover the answers. Monochromatic light has wavelength 540 nm and the screen is 2.40 m away. A single slit has width 0.0360 mm. A separate two-slit aperture has this same slit width and centre separation 0.180 mm. Assume the same far-field conditions.

1. Find the single slit’s first dark positions and full central width.
2. Find the ideal two-slit fringe spacing and the m=−2 bright position.
3. Only the slit width doubles. What happens to the central diffraction width, the two-slit envelope width and the ideal fringe spacing?
4. Why does setting n=0 in the single-slit minimum condition give the wrong interpretation?

If your answer differs, first check whether you used a or d, whether the question asks for a half-width or full width, and whether all lengths use the same unit.

Bring the diagram to tutoring

For optics tutoring or CEGEP waves and optics tutoring, bring your aperture sketch, your first equation and the step that is unclear. A tutor can help distinguish geometry, model choice and calculation. Share your current chapter so Nablio can confirm fit and availability. Sessions are offered in English.

Physics references

For further reading, see OpenStax University Physics Volume 3 on single-slit diffraction and finite-width double-slit patterns. The explanations, numerical exercises and figure specifications here were written for this lesson; no textbook figure or exercise is reproduced.