E. The Wave Nature of Light · The Wave Nature of LightYoung's Double-Slit Experiment
E2.1 E2.3 E2.4 E3.2 A1.8 A1.12 A1.13About 75 minutes
Learning goals- I can describe Young's double-slit experiment and explain why it showed that light is a wave.
- I can explain why the two slits must be lit by coherent light.
- I can use d sin θₘ = mλ (bright) and d sin θₙ = (n − ½)λ (dark) to find angles and wavelengths.
- I can use Δx = Lλ/d to find fringe spacing, wavelength, slit separation or screen distance.
Words to know
double slit
Two very narrow, closely spaced parallel openings that act as two in-phase sources of light.
fringe
One of the bright or dark bands in an interference or diffraction pattern.
coherent
Describes waves (or sources) with the same frequency and a constant phase relationship, so their interference pattern stays still.
order (m)
The number of a bright fringe counted out from the central bright fringe (m = 0).
fringe spacing (Δx)
The distance on the screen between the centres of neighbouring bright (or dark) fringes.
monochromatic
Of one wavelength (one colour).
Water waves, now with light
In the ripple tank, two in-phase sources made a fixed pattern of nodal and antinodal lines. If light is a wave, two in-phase sources of light should do the same thing: where the antinodal lines hit a screen, you should see bright bands, and where the nodal lines hit, dark bands.
The trouble is that light's wavelength is tiny, so the sources must be extremely close together, and they must stay perfectly in step. Around 1801 to 1803, the English scientist Thomas Young found a brilliant way to do both.
Young's set-up
- Light passes through a single narrow slit. Diffraction spreads it out, so the slit acts like one source.
- The spreading light reaches two narrow slits, very close together (a fraction of a millimetre). Both slits are lit by the same wavefront, so they act as two sources that are exactly in phase.
- Light from the two slits diffracts, overlaps and falls on a screen a metre or more away.
On the screen there is a row of evenly spaced bright and dark fringes. The dark fringes are where light from one slit cancels light from the other. Particles cannot do that, so the fringes were powerful evidence for the wave model.
Why not just use two light bulbs? Ordinary light sources send out short, random bursts of waves whose phases change millions of times per second. Two separate bulbs are not coherent: their pattern shifts so fast that it averages out to even lighting. Young's trick of lighting both slits from one source made them coherent. Today we use a laser, which produces coherent, monochromatic light.
The equations
The slits are a tiny distance d apart and the screen is far away, so this is exactly the far-field two-source pattern from the ripple tank. The path difference to a point at angle θ is d sin θ.
Bright fringes (constructive, like antinodal lines): path difference = whole number of wavelengths
d sin θₘ = mλ, m = 0, 1, 2, … (m = 0 is the central bright fringe)
Dark fringes (destructive, like nodal lines): path difference = (n − ½)λ
d sin θₙ = (n − ½)λ, n = 1, 2, 3, …
The angles are usually very small, so sin θ ≈ tan θ = x/L, where x is the distance on the screen from the centre of the central bright fringe. Then the bright fringes are at xₘ = mLλ/d, and neighbouring fringes are separated by the fringe spacing:
Δx = Lλ / d
The fringes are evenly spaced. They spread out more for longer λ, a larger L, or closer slits (smaller d).
Worked exampleExample 1: Fringe spacing
Red light (λ = 650 nm) passes through two slits 0.20 mm apart. The screen is 2.0 m away. How far apart are the bright fringes?
- Convert to metres: λ = 650 × 10⁻⁹ m, d = 0.20 × 10⁻³ m.
- Δx = Lλ/d = (2.0)(650 × 10⁻⁹) ÷ (0.20 × 10⁻³).
- Δx = 6.5 × 10⁻³ m.
The bright fringes are 6.5 mm apart.
Worked exampleExample 2: Measuring the wavelength of light
Slits 0.25 mm apart make a pattern on a screen 1.5 m away. A student measures 1.80 cm from the centre of the 1st bright fringe to the centre of the 6th bright fringe on the same side. Find the wavelength.
- From the 1st to the 6th bright fringe there are 5 spacings, so Δx = 1.80 cm ÷ 5 = 0.360 cm = 3.60 × 10⁻³ m. (Measuring across several fringes reduces the uncertainty.)
- Rearrange Δx = Lλ/d: λ = dΔx/L.
- λ = (0.25 × 10⁻³)(3.60 × 10⁻³) ÷ 1.5 = 6.0 × 10⁻⁷ m.
λ = 6.0 × 10⁻⁷ m = 600 nm (orange light).
Worked exampleExample 3: Angles of bright and dark fringes
Light of wavelength 500 nm falls on slits 0.050 mm apart. Find the angle to (a) the 3rd-order bright fringe and (b) the 2nd dark fringe.
- (a) sin θ₃ = 3λ/d = 3(5.00 × 10⁻⁷) ÷ (5.0 × 10⁻⁵) = 0.030, so θ₃ = 1.72°.
- (b) sin θ₂ = (2 − ½)λ/d = 1.5(5.00 × 10⁻⁷) ÷ (5.0 × 10⁻⁵) = 0.015, so θ₂ = 0.86°.
- Check: the 2nd dark fringe (0.86°) lies between the 1st and 2nd bright fringes (0.57° and 1.15°). ✓
White light
With white light, every colour makes its own fringes. All colours have a bright fringe at the centre (m = 0), so the central fringe is white. For other orders, Δx is larger for longer λ, so each bright fringe becomes a small spectrum: violet on the inside (closest to the centre) and red on the outside. After a few orders the colours overlap and wash out. This is one more way waves separate white light into colours.
Just for funThomas Young was also one of the first scholars to make progress on reading Egyptian hieroglyphs from the Rosetta Stone. Decoding fringes and decoding ancient writing: clearly a man who liked patterns.
Hands-on activityMeasure a wavelength with a simulation (and, optionally, a laser)
Main version (computer): open a free double-slit simulation, such as PhET Wave Interference ("Slits" screen, light source).
- Choose a colour and set the slit separation d and screen distance L. Record them.
- Turn on the screen and the intensity graph. Measure across as many bright fringes as you can and divide by the number of spacings to get Δx.
- Calculate λ = dΔx/L and compare with the wavelength the simulation states.
- Change one variable at a time (λ, d, then L) and record how Δx changes. Make a graph of Δx against 1/d and check that it is a straight line.
Optional version (laser, grown-up present): if you can borrow a double-slit slide from school, shine a low-power laser through it onto a wall 2 to 3 m away and measure Δx with a ruler.
Laser safety (grown-up needed): use only a low-power laser pointer labelled Class 2 (or Class II) with output under 1 mW, ideally a red one. Never look into the beam or point it at anyone, at a pet, out a window or toward aircraft or vehicles. Keep the beam below eye level, aim it only at a wall or a sheet of paper, and keep shiny objects (mirrors, glass, watches, jewellery) out of its path so it cannot reflect into someone's eye. Do not use a green or blue pointer of unknown power, and never look at the beam through binoculars or a lens. Switch it off as soon as you finish each measurement.
What to look for: your calculated λ should agree with the stated wavelength within a few percent. Doubling d should halve Δx; doubling L should double Δx; red light should give wider spacing than blue.
Your turn
1Why did Young pass light through a single slit before the double slit?
So both slits were lit by the same wavefront, making them coherent, in-phase sources.
2Light of λ = 550 nm passes through slits 0.15 mm apart onto a screen 3.0 m away. Find Δx.
Δx = Lλ/d = 3.0 × 550 × 10⁻⁹ ÷ 1.5 × 10⁻⁴ = 0.011 m = 11 mm.
3Slits 0.30 mm apart, screen 2.0 m away, Δx = 4.2 mm. Find λ.
λ = dΔx/L = 3.0 × 10⁻⁴ × 4.2 × 10⁻³ ÷ 2.0 = 6.3 × 10⁻⁷ m = 630 nm.
4Light of λ = 633 nm hits slits 0.10 mm apart. Find the angle to the 2nd-order bright fringe.
sin θ = 2 × 633 × 10⁻⁹ ÷ 1.0 × 10⁻⁴ = 0.01266, so θ = 0.725°.
5What is the path difference at the first dark fringe?
6How does the fringe pattern change if the slits are moved closer together?
Δx = Lλ/d gets larger: the fringes spread farther apart.
7What colour is the central fringe with white light, and why?
White: every wavelength has a bright fringe at m = 0, where the path difference is zero.
More practice: warm-up
1Write the equation for the angle of the m-th bright fringe.
2What is the order of the central bright fringe?
Same frequency and a constant phase relationship.
4Are double-slit fringes evenly spaced or unevenly spaced (for small angles)?
Evenly spaced (Δx = Lλ/d).
5Convert 0.25 mm to metres.
More practice: on your own
1A screen is 1.2 m from slits; Δx = 2.4 mm for light of λ = 600 nm. Find d.
d = Lλ/Δx = 1.2 × 600 × 10⁻⁹ ÷ 2.4 × 10⁻³ = 3.0 × 10⁻⁴ m = 0.30 mm.
2For λ = 633 nm and d = 0.10 mm, how far from the central fringe is the 1st dark fringe on a screen 3.0 m away?
x = (n − ½)Lλ/d = 0.5 × 3.0 × 633 × 10⁻⁹ ÷ 1.0 × 10⁻⁴ = 9.5 × 10⁻³ m (9.5 mm).
3Light of λ = 500 nm falls on slits 0.12 mm apart. The screen is 1.8 m away. How far from the centre is the 3rd dark fringe?
x = (3 − ½)Lλ/d = 2.5 × 1.8 × 5.0 × 10⁻⁷ ÷ 1.2 × 10⁻⁴ = 0.0188 m ≈ 1.9 cm.
4Slits are 0.020 mm apart and λ = 650 nm. What is the highest-order bright fringe that can exist, in theory?
sin θ ≤ 1 gives m ≤ d/λ = 2.0 × 10⁻⁵ ÷ 6.5 × 10⁻⁷ = 30.8, so m = 30.
5The whole double-slit apparatus is submerged in water (n = 1.33). The fringe spacing in air was 4.5 mm. What is it now? Explain.
The wavelength in water is λ/1.33, so Δx = Lλ/d also shrinks by 1.33: 4.5 ÷ 1.33 = 3.4 mm.
Challenge
1White light (400 nm to 700 nm) passes through slits 0.20 mm apart onto a screen 1.2 m away. How wide is the 1st-order spectrum on the screen (from violet edge to red edge)?
x(red) = 1.2 × 700 × 10⁻⁹ ÷ 2.0 × 10⁻⁴ = 4.2 mm; x(violet) = 1.2 × 400 × 10⁻⁹ ÷ 2.0 × 10⁻⁴ = 2.4 mm. Width = 1.8 mm.
2In a double-slit experiment, the 4th dark fringe for an unknown light falls exactly where the 3rd bright fringe for 600 nm light falls. Find the unknown wavelength.
Same angle: (4 − ½)λ = 3(600 nm), so λ = 1800 ÷ 3.5 = 514 nm.
3A thin piece of glass is placed over one slit so light through it is delayed by exactly half a cycle. Describe the new pattern and explain.
The two slits are now out of phase, so at the centre (zero path difference) crest meets trough: the centre becomes dark. Every bright fringe and dark fringe swaps places; the spacing Δx is unchanged.
Quiz
1. Young's experiment supported the wave model mainly because…
2. Why do two separate light bulbs not make a visible interference pattern?
3. What is Δx for λ = 600 nm, d = 0.30 mm and L = 1.5 m?
4. Which change makes the fringes spread farther apart?
5. The path difference to the 3rd-order bright fringe is…
6. With white light, the colour nearest the central fringe in each first-order spectrum is…
Big idea: Young's experiment is the centrepiece of this unit (E2.3 and E2.4; E3.2). Students should be able to set up the equations from the path-difference idea, not just memorize them.
How to help: the PhET simulation lets your teen collect real data without equipment. Ask them to predict what will happen before changing each variable.
Laser safety: if a laser is used at all, read the safety box with your teen first and supervise. Never allow anyone to look into the beam.
Common difficulties:
- Unit conversions: nm → ×10⁻⁹ m and mm → ×10⁻³ m.
- Mixing up m (bright, mλ) and n (dark, (n − ½)λ).
- Measuring from the edge of a fringe instead of its centre, and measuring a single spacing. Measuring across many spacings is more precise.
- Using Δx = Lλ/d at large angles, where sin θ ≈ x/L is no longer accurate.