Preview

Single-Slit Diffraction

Q: The slit width is halved. How does the width of the central bright fringe change?

Central fringe width ∝ λL/a. If a halves, what happens to the width?

Q: Red light (700 nm) replaces green light (550 nm) through the same slit. How does the diffraction pattern change?

Longer wavelength → larger sinθ for minima → wider pattern.

Q: Why is the central maximum much wider and brighter than the secondary maxima?

You can work it out this way: consider how many wavelets add constructively at θ = 0 versus at a secondary maximum. Also compare the areas under each lobe of the sinc² curve.

Explore how slit width and wavelength shape the diffraction intensity pattern

Hold two fingers almost together in front of a bright light and squint through the gap: the light does not stay in a neat finger-width stripe. It fans out into a broad central blaze flanked by dimmer, narrower echoes — and here is the part your intuition will fight: make the gap NARROWER and the blaze gets WIDER. Squeeze the light and it spreads. This is single-slit diffraction, and the mechanism is one of the deepest ideas in wave physics: Huygens' principle says every point inside the slit re-radiates its own wavelet, and what lands on the screen is the interference of all of them at once. Straight ahead, every wavelet arrives in step — that is the central maximum. Off to the side, at exactly the angle where a·sinθ = λ, the slit can be split in two halves whose wavelets cancel pair by pair, and the screen goes dark: the first minimum. Everything about the pattern flows from that one condition. Narrower slit means a smaller a, which needs a LARGER angle to reach one wavelength of path difference — so the dark fringe moves outward and the central maximum swells. The width of the whole pattern is set by the ratio λ/a, and the intensity follows the sinc² curve: the central lobe takes almost everything, twice as wide as any side lobe, while the first side lobe manages only about 4.5% of the peak. Red light, with its longer wavelength, spreads more than blue. And when the slit is thousands of wavelengths wide? The wavelets never stop interfering — the dark fringes just crowd so close to the forward direction that the pattern collapses into what looks like a sharp geometric shadow. Diffraction never turns off; it only gets small. In this lab you get the slit, the wavelength, and the screen distance as live controls, with the exact sinc² engine underneath. Start by testing the claim your intuition is already arguing with.

What you'll be able to do

Apply the minima condition a·sinθ = mλ quantitatively on the panel's own rows: at defaults (a = 0.5 mm, λ = 550 nm, L = 1.0 m) the first minimum sits at θ₁ = asin(λ/a) = 0.063° and the central maximum spans W = 2L·tanθ₁ = 2.20 mm; halve the slit to 0.25 mm and both rows answer the inverse law — θ₁ = 0.126°, W = 4.40 mm, the pattern doubling
Read the sinc² intensity distribution as a budget, not a shape: the central maximum is twice as wide as any side maximum ('the second maximum is only about half as wide as the central maximum', OpenStax §4.1 verbatim) and the first two side lobes peak at just I₁ ≈ 0.045I₀ and I₂ ≈ 0.016I₀ (§4.2 verbatim) — almost all the light lands in the center lobe, which is why narrowing the slit costs you brightness as it buys you spread
Explain the pattern from Huygens' principle: every point in the slit is a wavelet source, the first minimum is the angle where wavelets cancel pair-wise ('each ray from the slit interferes destructively with another ray', §4.1 verbatim), and diffraction never switches off — when a ≫ λ the minima crowd so close to the axis that the pattern compresses into a geometric shadow, but the wavelets still interfere (the HTML quiz Q2 scenario)

Formulas

a \sin\theta = m\lambda \quad (m = \pm1, \pm2, \ldots)

Dark fringe condition for single-slit diffraction

\Delta y \approx \frac{\lambda L}{a}

Approximate fringe spacing on a distant screen

I(\theta) \propto \left[\frac{\sin\alpha}{\alpha}\right]^2 \quad \alpha = \frac{\pi a \sin\theta}{\lambda}

Intensity distribution (sinc² envelope)

Make a prediction

Default state: slit a = 0.5 mm, λ = 550 nm, screen at L = 1.0 m — the panel reads first minimum θ₁ = 0.063° and central maximum width W = 2.20 mm. Now halve the slit width to 0.25 mm (λ and L untouched). What happens to the central maximum?

No grading here — commit to a guess, then scroll down and test it yourself.

This is a Pro experiment

Upgrade to Pro to access this experiment — or keep learning with one of our free labs.

View Pro plans Browse free experiments

Explore

Run the gate live: from defaults (a = 0.5 mm, θ₁ = 0.063°, W = 2.20 mm), drag Slit Width down to 0.25 mm and watch both rows answer the inverse law — θ₁ = 0.126°, W = 4.40 mm, the whole texture fanning wider as the barrier visibly closes. Then push the other way, 0.5 → 2 mm: the pattern squeezes to a needle. Squeeze light, it spreads.
Sweep Wavelength 400 → 700 nm at fixed slit: the W row climbs 1.60 → 2.80 mm while the wavelet rods and the pattern migrate through the spectrum's colors. Red diffracts most, violet least — λ sits in the numerator of every angle in this lab.
Separate angle from distance: drag Screen Distance 0.5 → 3 m and watch the θ₁ row freeze at 0.063° while W stretches linearly, 1.10 → 6.60 mm. L moves fringes across the screen; it cannot move them across the sky — angular positions belong to λ/a alone.
Audit the intensity budget on the chart: the central lobe spans ±θ₁ (twice as wide as any side lobe), the first side lobe peaks at ≈0.045 I₀, the second at ≈0.016 I₀. Then tour the three presets — Narrow, Wide (a = 3636λ: a needle that is still a sinc²), and Red — and read each pattern against its numbers.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It doubles — θ₁ = 0.126°, W = 4.40 mm: a·sinθ = λ is an inverse law, so a smaller slit needs a LARGER angle to reach one wavelength of path difference

The first minimum sits where the path difference across the slit reaches exactly one wavelength: a·sinθ₁ = λ (OpenStax Eq. 4.1). Halve a and sinθ₁ must double to keep the product — θ₁ = asin(550 nm / 0.25 mm) = 0.126°, and the panel's central width W = 2L·tanθ₁ follows to 4.40 mm (exactly double at these small angles; this lab's machine gate brackets the ratio at 2.000 on the engine's own arithmetic). The mechanism makes it inevitable: the slit is a row of Huygens wavelet sources, and with half as many of them, cancellation off-axis becomes harder — you must look at a steeper angle before the wavelets can pair up and annihilate. Option B is the window-clipping instinct — the sim's own misconception card names it: 'Narrower slit → narrower pattern. Wrong! Smaller a → larger α → wider central max' — and OpenStax states the law flat: 'an increase in the slit width results in a decrease in the width of the central peak.' A slit does not clip a pre-existing beam; it defines how many sources get to vote. Option C misassigns the variables: λ and L scale the pattern, but its SHAPE is priced by the ratio λ/a — the slit width is exactly the knob that changed.

Quiz (0/3)

The slit width is halved. How does the width of the central bright fringe change?

Red light (700 nm) replaces green light (550 nm) through the same slit. How does the diffraction pattern change?

Why is the central maximum much wider and brighter than the secondary maxima?

You can now

Apply the minima condition a·sinθ = mλ quantitatively on the panel's own rows: at defaults (a = 0.5 mm, λ = 550 nm, L = 1.0 m) the first minimum sits at θ₁ = asin(λ/a) = 0.063° and the central maximum spans W = 2L·tanθ₁ = 2.20 mm; halve the slit to 0.25 mm and both rows answer the inverse law — θ₁ = 0.126°, W = 4.40 mm, the pattern doubling
Read the sinc² intensity distribution as a budget, not a shape: the central maximum is twice as wide as any side maximum ('the second maximum is only about half as wide as the central maximum', OpenStax §4.1 verbatim) and the first two side lobes peak at just I₁ ≈ 0.045I₀ and I₂ ≈ 0.016I₀ (§4.2 verbatim) — almost all the light lands in the center lobe, which is why narrowing the slit costs you brightness as it buys you spread
Explain the pattern from Huygens' principle: every point in the slit is a wavelet source, the first minimum is the angle where wavelets cancel pair-wise ('each ray from the slit interferes destructively with another ray', §4.1 verbatim), and diffraction never switches off — when a ≫ λ the minima crowd so close to the axis that the pattern compresses into a geometric shadow, but the wavelets still interfere (the HTML quiz Q2 scenario)