Double-Slit Interference Calculator
Calculate the fringe spacing and the position of bright fringes in Young's double-slit experiment. Enter the wavelength, slit separation, screen distance and fringe order.
Find the Spacing and Position of Bright Fringes
Double-Slit Interference
Where:
- \(\Delta y\) — fringe spacing, the distance between two neighbouring bright fringes on the screen (m)
- \(y_m\) — position of the bright fringe of order \(m\), measured from the centre of the pattern (m)
- \(\lambda\) — wavelength of the light (m)
- \(L\) — distance from the slits to the screen (m)
- \(d\) — separation between the two slits (m)
- \(m\) — fringe order: \(m = 0\) is the central maximum, \(m = 1, 2, 3, \ldots\) the side fringes
The first formula gives the spacing of the interference pattern, and the second gives how far the \(m\)-th bright fringe lies from the centre. Both assume the small-angle regime (\(\lambda \ll d\)), which holds in virtually every laboratory and classroom setup.
Double-Slit Interference – Calculation Example
Example: a red laser (\(\lambda = 650\) nm) shines on two slits separated by \(d = 0.25\) mm, and the pattern is observed on a screen \(L = 1\) m away. Where does the third bright fringe (\(m = 3\)) fall?
\[ \Delta y = \frac{\lambda L}{d} = \frac{650 \times 10^{-9}\ \text{m} \times 1\ \text{m}}{0.25 \times 10^{-3}\ \text{m}} = 2.6 \times 10^{-3}\ \text{m} = 2.6\ \text{mm} \]
\[ y_3 = m \, \Delta y = 3 \times 2.6\ \text{mm} = 7.8\ \text{mm} \]
Result: the bright fringes are 2.6 mm apart, and the third-order bright fringe lies 7.8 mm from the centre of the pattern on either side. These are exactly the calculator's default values, so you can reproduce the result above instantly.
How to use the calculator
- Enter the Wavelength (λ) of the light and pick its unit (nm, µm or mm). A red laser pointer is about 650 nm, a green one 532 nm.
- Enter the Slit separation (d) — the distance between the centres of the two slits — and choose µm, mm or cm.
- Enter the Screen distance (L) from the slits to where the pattern is observed, in cm, m or ft.
- Optionally enter a Fringe order (m) (0, 1, 2, 3 …). If you leave it blank, the calculator only reports the fringe spacing.
- Read the results: the fringe spacing \(\Delta y\) in several units, and — if you entered an order — the position \(y_m\) of that bright fringe measured from the centre of the pattern.
What is double-slit interference?
In 1801 Thomas Young sent light through two narrow, closely spaced slits and saw something a stream of particles could never produce: alternating bright and dark bands on a screen. Waves reaching the screen from the two slits travel slightly different distances. Wherever the path difference is a whole number of wavelengths — \(d \sin\theta = m\lambda\) — the waves arrive in step and reinforce each other, producing a bright fringe. Where the path difference is an odd number of half-wavelengths, they cancel and produce a dark fringe.
The pattern is a direct fingerprint of the light's wavelength, which is what makes the experiment so useful. Measuring the fringe spacing on a screen with a ruler gives \(\lambda\) through \(\lambda = \Delta y \, d / L\); this is still one of the most common ways to determine the wavelength of a laser in teaching labs. Conversely, with a known wavelength the same geometry is used to measure tiny distances — such as the spacing of tracks or thin structures — far smaller than a ruler can resolve.
The fringe spacing \(\Delta y\) tells you how "stretched" the pattern is: it grows linearly with the wavelength and with the screen distance, and shrinks as the slits are moved apart. The position result \(y_m\) is symmetric — a fringe of order \(m\) appears at \(+y_m\) on one side of the centre and at \(-y_m\) on the other, because both orders \(+m\) and \(-m\) satisfy the bright-fringe condition. The central maximum (\(m = 0\)) sits on the axis and is twice as wide as the side fringes.
Interference is not limited to light. The same principle governs noise-cancelling headphones, the colours on a soap film, and the diffraction of water and sound waves — you can explore the wave side of sound further with our Sound Intensity Level calculator. For a broader collection, see also the Simple Pendulum calculator and the Faraday's Law calculator.
Assumptions, limitations and practical tips
- Small-angle approximation: the formulas use \(\sin\theta \approx \tan\theta \approx \theta\), which is excellent when \(\lambda \ll d\) and fringes stay close to the axis. For high orders near the maximum allowed order, the calculator compares the approximate position with the exact geometry and notes the difference when it exceeds 1%.
- Maximum order: since \(m\lambda \le d\), only orders up to \(m = \lfloor d/\lambda \rfloor\) exist. If you request a higher order, the calculator tells you the highest possible one.
- Coherent light: the two slits must be illuminated by the same coherent source (a laser or a single slit in front of a lamp). Two independent light sources will not produce a stable pattern.
- Point slits, no diffraction envelope: the model treats each slit as infinitely narrow. Real slits of finite width add a diffraction envelope that modulates the fringe brightness — very high orders may be too dim to see even though the math allows them.
- Accuracy tip: because \(\Delta y\) is proportional to \(L\), measuring the screen distance carefully matters. Pushing the screen further away spreads the pattern out and makes the fringe spacing easier to measure precisely.
- Wavelength must be smaller than the slit separation: if \(\lambda \ge d\) the first bright fringe would need an angle of 90° or more, and no distinct pattern forms; the calculator flags this input as invalid.
Frequently Asked Questions
How do I calculate the position of dark fringes?
Dark fringes lie exactly halfway between neighbouring bright fringes, so their positions are \(y = \left(m + \tfrac{1}{2}\right) \frac{\lambda L}{d}\) for \(m = 0, 1, 2, \ldots\). For example, the first dark fringe sits at \(\tfrac{1}{2} \Delta y\) from the centre, and the spacing between consecutive dark fringes is the same \(\Delta y = \lambda L / d\) as for bright ones.
Can I use this calculator to find the wavelength of an unknown light source?
Yes — that is the classic lab method. Measure the fringe spacing \(\Delta y\) on a screen with a ruler, then rearrange the formula to \(\lambda = \Delta y \, d / L\). For instance, fringes 2.0 mm apart with \(d = 0.25\) mm slits and \(L = 1\) m give \(\lambda = 500\) nm. Accuracy improves with larger \(L\) and averaging over many fringes.
What are the most common unit mistakes in double-slit calculations?
Mixing units is the number-one error: wavelength is usually given in nanometres while slit separation is in millimetres, and they differ by a factor of a million. Always convert everything to metres first: 1 nm = \(10^{-9}\) m, 1 µm = \(10^{-6}\) m, 1 mm = \(10^{-3}\) m. Another frequent slip is entering the slit width instead of the slit separation.
What is the difference between slit separation and slit width?
Slit separation \(d\) is the distance between the centres of the two slits and it controls the fringe spacing. Slit width \(a\) is the size of each individual opening and it controls the diffraction envelope — how bright the fringes are at different angles. This calculator only needs \(d\); if you also want the intensity envelope, you need \(a\) and the single-slit diffraction formula \(a \sin\theta = n\lambda\).
Why can't I see a fringe of high order, or why does the calculator reject my order?
A bright fringe of order \(m\) requires \(m\lambda \le d\), because the path difference cannot exceed the slit separation. The highest possible order is \(m = \lfloor d/\lambda \rfloor\). With 650 nm light and \(d = 0.25\) mm, that is order 384 — but in practice real slits of finite width make very high orders extremely dim, so only the first dozen or so are usually visible.
Does the double-slit experiment work with white light instead of a laser?
Yes, but each wavelength forms its own fringe pattern with its own spacing, so the colours spread out: you see a white central fringe flanked by rainbows of colour that blur together after a few orders. A single bright colour (monochromatic light, e.g. a laser or a sodium lamp with a filter) gives the crisp, evenly spaced dark and bright bands this calculator models.
Is the small-angle approximation accurate enough for my setup?
For typical classroom and lab setups, yes — it is usually accurate to a fraction of a percent. It breaks down only for high fringe orders at large angles. The calculator checks this automatically: it compares the approximate position \(y_m = m\lambda L/d\) with the exact value \(L\tan(\arcsin(m\lambda/d))\) and warns you when the difference exceeds 1%.