Double-Slit Interference Simulation

About this experiment

In 1801 Thomas Young shone light through two narrow slits and saw a row of bright and dark bands on a screen. Particles of light flying in straight lines could not explain it, but waves could: light from the two slits adds up where crests meet crests and cancels where crests meet troughs. Change the wavelength, the slit separation and the distance to the screen, measure the fringe spacing, and see how the width of each slit adds a second, wider pattern on top.

Learning goals

Explain the bright and dark fringes using path difference and superposition
Use Δy = λL/d to relate fringe spacing to wavelength, screen distance and slit separation
Measure a wavelength from a double-slit pattern
Describe the single-slit diffraction pattern and how it limits the double-slit fringes
Explain why white light gives a white central fringe with coloured fringes beside it

How to use it

1With the default settings, use the ruler to measure the distance across 5 fringe spacings and divide by 5. Compare with the readout.
2Change the wavelength from red to blue. Do the fringes move closer together or further apart?
3Double the slit separation d. What happens to the fringe spacing? Then double the screen distance L.
4Switch on Show average brightness. Find the lines along which the light from the two slits cancels.
5Switch to one slit and make it narrower. Why does the central band get wider?
6Back with two slits, set d to exactly 5 times a. Count the bright fringes inside the central band. Which order is missing?
7Choose White light. Which colour is on the inside of each coloured fringe, and why?

The model

d sin θ = mλ

Bright fringes: the path difference is a whole number of wavelengths

d sin θ = (m + \frac{1}{2}) λ

Dark fringes: the path difference is an odd number of half wavelengths

Δ y = \frac{λ L}{d}

Fringe spacing on a distant screen (small angles)

a sin θ = nλ

Dark bands of a single slit of width a (n = 1, 2, 3, ...)

I = I_{0} cos^{2} (\frac{π d sin θ}{λ}) (\frac{sin β}{β})^{2}, β = \frac{π a sin θ}{λ}

Brightness of the double-slit pattern: interference fringes inside a diffraction envelope

Assumptions

The screen is far away compared with the slit separation (Fraunhofer diffraction), and the light is coherent and arrives as a plane wave.
Both slits have the same width and are lit equally. The brightness of each fringe is shown relative to the central maximum.
White light is modelled as 31 equally bright wavelengths from 400 to 700 nm, each making its own pattern; the patterns add.
The wave picture on the left is a sketch: the real wavelength is far too small to see, so it is drawn tens of thousands of times larger.

Try this in class

Q1Light of wavelength 600 nm passes through slits 0.30 mm apart onto a screen 2.0 m away. Find the fringe spacing.
Q2Why must the light from the two slits be coherent for fringes to appear?
Q3A student measures 10 fringe spacings as 21 mm with L = 1.5 m and d = 0.45 mm. Calculate the wavelength.
Q4What would you see if one slit were covered?
Q5Why do fringes get dimmer away from the centre even though the double-slit formula alone says they are all equally bright?

Frequently asked questions

Why does a smaller slit separation give wider fringes?

The path difference between the two slits is d sinθ. If d is small, you have to go to a larger angle before the path difference reaches one whole wavelength, so the next bright fringe is further from the centre.

What is the difference between interference and diffraction?

Both are waves adding up. Interference usually means waves from a few separate sources, like the two slits. Diffraction means waves from all the points across one opening, which spread out and make the wide envelope.

Does the experiment work with electrons?

Yes. Electrons, neutrons and even large molecules sent one at a time through two slits build up the same fringe pattern, with a wavelength given by the de Broglie relation λ = h / p.