Young's double slit — evidence for wave nature of light — Physics

Before this

Superposition
Wave properties
Sine, cosine and tangent · coming soon

This leads to

Light plus light can make darkness

Shine two torches at a wall. Where the two patches of light overlap, the wall is brighter. That is what everyone expects: more light, more brightness.

Now send light from one laser through two narrow slits, very close together, and let it fall on a screen a few metres away. You do not see two bright patches. You see a row of bright and dark stripes, evenly spaced. At the dark stripes, light is arriving from both slits, and the screen is dark. Cover one slit, and those places become lit.

So adding light from a second slit can make a place darker. No stream of particles can do that. Two streams of particles only ever add up to more. But waves can cancel, as the last lecture showed with water and sound. The last lecture promised to use the same rule on light passing through two narrow slits. This lecture does it, and it turns the stripes into a way of measuring the wavelength of light.

Particles or waves? The question in 1800

By 1800 there were two ideas about light. Isaac Newton's followers thought light was a stream of tiny particles. Christiaan Huygens and others thought it was a wave. Newton's great authority made the particle idea the more popular one.

Thomas Young, an English doctor and physicist, described interference of light in lectures to the Royal Society in 1801 and 1803. In his 1803 lecture he let sunlight through a hole made with a fine needle in a window shutter. He held a slip of card about one-thirtieth of an inch wide (about 0.85 mm) in the narrow beam, so that light passed on both sides of it. Fringes appeared in the card's shadow. When he blocked the light on one side of the card, the fringes inside the shadow disappeared. Light from the two sides was interfering. He later described the same idea with two slits. Historians are not sure exactly which of his experiments he carried out and which he only described.

The set-up, and why one source

Young's double slit. Light from one laser reaches two slits, S₁ and S₂, a distance

a

apart. A screen at distance

D

shows bright fringes, a distance

x

apart. The central bright fringe is at O. Light from both slits reaches every point, such as P.

The superposition lecture explained why two torches never show bands of light and dark. Each torch gives out light in short, random bursts, so the two are not coherent. Their phase difference changes too fast for any pattern to stay still.

The double slit solves this problem. The two slits are lit by the same wavefronts from one source. Whatever the source does, both slits do the same, at the same moment. So S₁ and S₂ act as two coherent sources, in phase with each other. A laser makes this easy. With an ordinary lamp, the light first goes through a single narrow slit, which acts as one small source, and that slit lights the double slit.

Each slit is very narrow, so the light spreads out as it passes through. This spreading is called diffraction, and the next lecture is about it. Here you only need to know that it happens, so the light from each slit reaches a wide area of the screen, and the two spread-out waves overlap.

Bright fringes and dark fringes

Every point on the screen gets a wave from S₁ and a wave from S₂. They left the slits in phase. They travelled different distances. So the rule from the last lecture decides what each point gets.

Fringes, for slits in phase

Bright fringe: path difference $= nλ$ , with $n = 0, 1, 2, \dots$ (constructive interference).

Dark fringe: path difference $= (n + \frac{1}{2}) λ$ (destructive interference).

The point O on the centre line is the same distance from both slits. The path difference is zero, so O is the central bright fringe, with $n = 0$ . Moving away from O, the path difference grows steadily: $\frac{1}{2} λ$ at the first dark fringe, $1 λ$ at the next bright one, $1 \frac{1}{2} λ$ , $2 λ$ , and so on. This is why the fringes alternate.

Stop and think

The fringes are showing on the screen. You cover slit S₂. What happens at a place that was a dark fringe?

Show the answer

It becomes lit. With only S₁ open, there is no second wave to cancel the first, so the light from S₁ arrives and nothing removes it. The screen becomes one broad patch of light. Many students expect a dark fringe to stay dark, "because no light goes there". But light from each slit does go there. A dark fringe is a place where two waves arrive in antiphase and cancel. This test is the strongest evidence that the fringes come from interference.

From path difference to fringe spacing

The rule tells you which points are bright. To use the fringes, you also need to know where those points are. This is the hard part of the lecture. Spend time here.

The geometry, drawn with

a

hugely enlarged. P is a distance

y

from the centre O. N is on S₂P, with NP = S₁P, so the path difference is S₂N. In a real experiment the angle θ is far smaller than drawn.

In a real experiment, $a$ is about half a millimetre and $D$ is about two metres. So $D$ is thousands of times larger than $a$ . This one fact makes the geometry simple. Step through it below.

Step through — the fringe spacing formula

Press Next to add one step. Read the reason before you go on. Use the figure above.

Each step gives one line and the reason it is allowed.

P is a distance y from the centre O. The line from the middle of the slits to P makes an angle θ with the centre line. (This sets up the geometry.)
Mark N on S₂P so that NP = S₁P. Then the path difference is S₂N. (From N and S₁ onwards the two paths are equal.)
Because D is thousands of times larger than a, the two paths are almost parallel, so the triangle S₁S₂N has a right angle at N. (This is the only approximation.)
In that small triangle the angle at S₁ is θ, so S₂N = a sin θ. (S₂N is opposite θ, and a is the hypotenuse.)
In the big triangle, tan θ = y/D. For small angles sin θ ≈ tan θ, so the path difference ≈ ay/D. (In a real experiment θ is well under 1°.)
A bright fringe needs ay/D = nλ, so yₙ = nλD/a. (The in-phase rule.)
Neighbouring bright fringes are at nλD/a and (n + 1)λD/a, so they are x = λD/a apart. (The same for every pair, so the fringes are evenly spaced.)

That is it. That is the whole idea. Every fringe sits where the path difference $a y / D$ is a whole number of wavelengths, and those places are equally spaced.

Fringe spacing in Young's double slit

$x = \frac{λ D}{a}$

$x$ is the fringe spacing, from one bright fringe to the next. $λ$ is the wavelength, $D$ the distance from the slits to the screen, and $a$ the slit separation, all in metres. It holds when $D$ is much larger than $a$ . Some books write $w = λ D / s$ , or $s = λ D / d$ , for the same formula.

How good is the approximation? Take $λ = 633$ nm, $a = 0.50$ mm and $D = 2.0$ m. The third bright fringe is predicted at $y = 3 \times 2.53 \approx 7.6$ mm, where θ is only about 0.22°. Working out the two paths exactly gives a path difference of $2.99998 λ$ , not $3 λ$ . The formula is very good indeed.

The formula also explains the pattern in words. A longer wavelength gives wider fringes. Moving the screen further away gives wider fringes. Slits closer together give wider fringes, because the path difference grows more slowly as you move across the screen. Try each one below, one slider at a time.

Explore — the fringes on the screen

The strip shows the middle 30 mm of the screen. Change one quantity at a time and watch the spacing. Then switch to white light, as Young used.

Wavelength λ: 630 nm

Slit separation a: 0.50 mm

Slits to screen D: 2.0 m

White light (every colour from 400 nm to 700 nm at once)

Fringe spacings from $x = λ D / a$ , with one quantity changed at a time from 630 nm, 0.50 mm and 2.0 m.

630 nm, 0.50 mm, 2.0 m: $x \approx 2.52$ mm.
Wavelength 400 nm (violet): $x = 1.60$ mm. Wavelength 700 nm (red): $x = 2.80$ mm.
Slit separation 0.25 mm: $x \approx 5.04$ mm. Slit separation 1.00 mm: $x \approx 1.26$ mm.
Screen at 3.0 m: $x \approx 3.78$ mm.
White light: the central fringe is white, because every colour has a bright fringe at the centre. Further out the colours separate, violet nearest the centre, and the fringes soon blur.

What should you have seen? Each change did what the formula says, and nothing else changed: the fringes stayed evenly spaced, with a bright one at the centre. With white light, only the middle few fringes were clear. The central one was white, and the others had violet inner edges and red outer edges. Young's sunlight fringes looked like that.

The real pattern also fades towards the edges, which the strip does not show. That fading comes from diffraction at each slit, and the next lecture explains it.

Measuring a wavelength in practice

The formula turns a ruler into an instrument for measuring wavelengths of a few hundred nanometres. Rearranged, it gives $λ = \frac{x a}{D}$ . Measure $D$ with a metre rule. The slit separation is printed on the slide, or measured with a travelling microscope. The hardest part is $x$ , because one fringe spacing is only a few millimetres.

So measure across many fringes and divide. A ruler can be read to about 1 mm. Across one spacing of 2.5 mm, that is an uncertainty of about 40%. Across ten spacings, 25 mm, it is about 4%. Counting carefully matters: ten spacings span eleven bright fringes, from the first to the eleventh.

Never look into a laser beam, or at its reflection from a shiny surface. Even a low-power laser can damage the eye.

Worked examples

Worked Example 1 — Fringe spacing for a red laser

A helium–neon laser gives red light of wavelength 633 nm. The slits are 0.50 mm apart and the screen is 2.0 m away. Find the fringe spacing.

Convert to metres: $λ = 633 \times 1 0^{- 9}$ m, $a = 0.50 \times 1 0^{- 3}$ m, $D = 2.0$ m.

x = \frac{λ D}{a} = \frac{633 \times 1 0 ^{- 9} \times 2.0}{0.50 \times 1 0 ^{- 3}} \approx 2.53 \times 1 0^{- 3} m

So the fringes are about 2.53 mm apart. Check the size: a few millimetres, easy to see on a screen and to measure with a ruler. ✓

Worked Example 2 — Finding a wavelength

With slits 0.30 mm apart and a screen 1.5 m away, the fringe spacing is 3.1 mm. Find the wavelength.

λ = \frac{x a}{D} = \frac{3.1 \times 1 0 ^{- 3} \times 0.30 \times 1 0 ^{- 3}}{1.5} = 6.2 \times 1 0^{- 7} m

That is 620 nm, orange-red light. Check: it lies between 400 nm and 700 nm, inside the visible range. ✓

Worked Example 3 — Measuring across ten spacings

Kofi measures the distance across ten fringe spacings as 28 mm. The slits are 0.40 mm apart and the screen is 1.8 m away. Find the wavelength.

First the spacing: $x = 28 \div 10 = 2.8$ mm $= 2.8 \times 1 0^{- 3}$ m.

λ = \frac{x a}{D} = \frac{2.8 \times 1 0 ^{- 3} \times 0.40 \times 1 0 ^{- 3}}{1.8} \approx 6.2 \times 1 0^{- 7} m

So $λ \approx 620$ nm. Check: using 28 mm without dividing would give about 6200 nm, which is not visible light at all. The size of the answer catches the mistake. ✓

Worked Example 4 — Changing one thing at a time

Start from Worked Example 1, where $x \approx 2.53$ mm. Find the new spacing if (a) a green laser of wavelength 532 nm is used, (b) the slit separation is doubled, (c) the screen is moved to 4.0 m.

(a) $x$ is proportional to $λ$ : $2.53 \times \frac{532}{633} \approx 2.13$ mm. The fringes get closer, but not half as close: green light is not half the wavelength of red.

(b) $x$ is inversely proportional to $a$ : doubling $a$ halves $x$ , to about 1.27 mm.

Check (a) directly: $532 \times 1 0^{- 9} \times 2.0 \div (0.50 \times 1 0^{- 3}) \approx 2.13 \times 1 0^{- 3}$ m. ✓

Three classic mistakes

Trap 1 — Forgetting to divide

A measurement across several fringes is not the fringe spacing. Divide by the number of spacings. If your wavelength comes out in thousands of nanometres, you have probably forgotten.

Trap 2 — Counting fringes instead of spacings

Ten bright fringes, from the first to the tenth, span nine spacings, not ten. Count the gaps between fringes, or count from the first fringe to the eleventh for ten spacings.

Trap 3 — Mixing millimetres, nanometres and metres

Put every quantity in metres before you substitute: nm × 10⁻⁹, mm × 10⁻³. A slit separation left in millimetres makes the answer 1000 times too big or too small.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Blue light of wavelength 450 nm passes through two slits 0.25 mm apart. The screen is 1.2 m away. Find the fringe spacing.

Hint 1
Use $x = λ D / a$ , with everything in metres.

Hint 2
$λ = 450 \times 1 0^{- 9}$ m and $a = 0.25 \times 1 0^{- 3}$ m.

Full answer
$x = \frac{450 \times 1 0 ^{- 9} \times 1.2}{0.25 \times 1 0 ^{- 3}} = 2.16 \times 1 0^{- 3}$ m, so about 2.16 mm. Check: $a$ is half that of Worked Example 1, which widens the fringes, while the shorter wavelength and nearer screen narrow them. A few millimetres is sensible. ✓
In the set-up of question 1, how far from the central bright fringe is the third dark fringe? What is the path difference there?

Hint 1
Dark fringes lie halfway between bright ones.

Hint 2
The first dark fringe is at $0.5 x$ , the second at $1.5 x$ .

Full answer
The third dark fringe is at $2.5 x = 2.5 \times 2.16 \approx 5.4$ mm from the centre. The path difference there is $2.5 λ = 2.5 \times 450 = 1125$ nm, a whole number plus a half of wavelengths, so the waves arrive in antiphase. Check with $a y / D$ : $0.25 \times 1 0^{- 3} \times 5.4 \times 1 0^{- 3} \div 1.2 = 1.125 \times 1 0^{- 6}$ m. ✓
White light passes through slits 0.50 mm apart onto a screen 2.0 m away. Find where the first bright fringe beside the centre lies for violet light (400 nm) and for red light (700 nm). Explain why the central fringe is white and the next fringes have coloured edges.

Hint 1
Each colour makes its own pattern, with its own spacing.

Hint 2
At the centre, the path difference is zero for every wavelength.

Full answer
Violet: $x = 400 \times 1 0^{- 9} \times 2.0 \div (0.50 \times 1 0^{- 3}) = 1.6$ mm. Red: $700 \times 1 0^{- 9} \times 2.0 \div (0.50 \times 1 0^{- 3}) = 2.8$ mm. At the centre the path difference is zero, so every colour is bright there, and they add to white. The first bright fringe of each colour is at a different place, from 1.6 mm (violet) to 2.8 mm (red), so the fringe is spread into colours, violet on the inside. By the second fringe (violet at 3.2 mm) the colours of neighbouring fringes start to overlap, so the pattern blurs. Check: $2.8 \div 1.6 = 1.75 = 700 \div 400$ . ✓
The whole experiment of Worked Example 1 is repeated under water, with refractive index 1.33. Use the refraction lecture: the frequency of the light stays the same, but its speed falls to $c /1.33$ . Find the new wavelength and the new fringe spacing.

Hint 1
With $v = f λ$ and $f$ fixed, the wavelength falls by the same factor as the speed.

Hint 2
The wavelength in water is $633 \div 1.33$ nm. Then use $x = λ D / a$ .

Full answer
The wavelength in water is $633 \div 1.33 \approx 476$ nm. So $x = 476 \times 1 0^{- 9} \times 2.0 \div (0.50 \times 1 0^{- 3}) \approx 1.90 \times 1 0^{- 3}$ m, about 1.90 mm. The fringes move closer together by the factor 1.33, but the light still looks red: colour is set by frequency, which has not changed. Check: $2.53 \div 1.33 \approx 1.90$ . ✓

The bigger idea

Young's fringes settled one question and later opened a stranger one. In 1909 G. I. Taylor showed that fringes still form with light so faint that, by later understanding, only a tiny amount of light is in the apparatus at a time. In 1989 Akira Tonomura and his team sent electrons through a similar apparatus one at a time. Each electron arrived at a single point on the screen, but after many thousands the dots built up into fringes. Light and matter travel like waves and arrive like particles. That puzzle sits at the centre of quantum physics, which you will meet later.

For now, the message of this lecture is enough. A pattern of dark places where light from two slits overlaps can only come from waves. And with two slits, a ruler and the formula $x = λ D / a$ , you can measure a length a thousand times smaller than a millimetre.

Key concepts in this lesson

Two narrow slits lit by one source produce evenly spaced bright and dark fringes, and the dark fringes show that light can cancel light, which only waves can do.
The two slits act as coherent sources because they are lit by the same wavefronts, which is why two separate lamps never show fringes.
A bright fringe forms where the path difference is a whole number of wavelengths, and a dark fringe where it is a whole number plus a half, for slits in phase.
When D is much larger than a, the path difference at a distance y from the centre is about ay/D, which gives the fringe spacing x = λD/a.
Measuring across many fringe spacings and dividing makes the wavelength measurement far more precise, as long as you count spacings, not fringes.