Young’s double-slit experiment

Point a laser at two narrow, closely spaced slits and let the light fall on a distant screen, and instead of two bright lines you see a whole ladder of evenly spaced bright and dark bands. That pattern is the single most convincing piece of evidence that light is a wave. Thomas Young first performed this experiment in 1801, splitting one wavefront into two coherent sources at the slits so their phase relationship never drifts, then let the two beams overlap and interfere. Where crest meets crest the screen lights up; where crest meets trough it goes dark — energy is not simply added, it is redistributed. This node builds the geometry of the two-slit path difference, the bright and dark fringe conditions, the fringe-width formula, and the full two-beam intensity law $I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$ . Multiple-beam (grating) interference and formal coherence theory are out of scope — the treatment stays with elementary two-beam interference and the small-angle path-difference geometry, exactly as in the PH11001 syllabus.

The core insight

Splitting one wavefront at two slits gives two coherent sources whose path difference $\Delta \approx dy/D$ sets the phase difference $\delta = 2\pi\Delta/\lambda$ , producing an intensity pattern $I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$ with bright fringes at $\Delta=m\lambda$ spaced by $\beta=\lambda D/d$ .

The core idea

Why coherence is essential. A stationary interference pattern needs a constant phase difference between the two waves at every point on the screen, at every instant. Two independent light sources — say two separate bulbs — have phases that drift randomly relative to each other, so any interference pattern they might momentarily form washes out within nanoseconds, far faster than any detector or eye can register it. Young's trick is to derive both sources from a single wavefront: the two slits $S_1,S_2$ are illuminated by the same incoming wave, so their outputs are locked together in phase forever, and the pattern becomes stable. The slit separation $d$ also needs to be small enough that the arriving wavefront is essentially flat across both slits, which is the spatial-coherence half of the requirement.

Geometry of path difference. Let the slits be separated by $d$ and the screen sit at distance $D$ , with $D \gg d$ . Consider a point on the screen at height $y$ from the central axis. The two rays converging on that point start from $S_1$ and $S_2$ and therefore travel slightly different distances. Using the small-angle approximation valid when $D$ is much larger than $d$ and $y$ , this path difference is

\Delta = d\sin\theta \approx \frac{dy}{D}

where: $\Delta$ is the extra distance travelled by one ray relative to the other, $\theta$ is the angle the point subtends at the midpoint of the slits, $d$ is the slit separation, $y$ is the distance from the screen's central axis, and $D$ is the slit-to-screen distance.

Figure

Two slits separated by distance d send rays to a point on a distant screen where they interfere, with bright and dark fringes marked along the screen

Fringe conditions. Constructive interference (a bright fringe) occurs when the path difference is an integer number of wavelengths, and destructive interference (a dark fringe) when it is a half-integer number:

\Delta = m\lambda \quad (\text{bright}), \qquad \Delta = \left(m+\tfrac12\right)\lambda \quad (\text{dark})

where: $m=0,\pm1,\pm2,\dots$ is the fringe order and $\lambda$ is the wavelength of the light.

Two-beam intensity. Superposing the two waves (adding field amplitudes, then squaring) rather than simply adding intensities gives

I = I_1+I_2+2\sqrt{I_1I_2}\cos\delta, \qquad \delta=\frac{2\pi}{\lambda}\Delta

where: $I_1,I_2$ are the individual beam intensities at the screen and $\delta$ is the phase difference corresponding to path difference $\Delta$ . For equal-width slits, $I_1=I_2=I_0$ , so $I=4I_0\cos^2(\delta/2)$ , swinging between $4I_0$ at bright fringes and $0$ at dark fringes — the energy lost from dark regions reappears in bright ones, which is the real signature of interference, not just $I_1+I_2$ . Note that the average intensity over many fringes is $2I_0$ , exactly what incoherent addition would give — interference redistributes energy across the screen without creating or destroying it overall.

Derivation AFringe width from the path-difference condition

The goal is to combine the path-difference geometry $\Delta \approx dy/D$ with the bright-fringe condition $\Delta = m\lambda$ to obtain the spacing between consecutive bright fringes, called the fringe width, and to show that this spacing is the same everywhere on the screen regardless of the fringe order $m$ .

Step 1. Locate the $m$ -th bright fringe. Setting the path difference equal to the bright-fringe condition, $\Delta = dy_m/D = m\lambda$ , and solving for the screen position gives $y_m = \dfrac{m\lambda D}{d}$ .

Step 2. Locate the next bright fringe. The $(m+1)$ -th bright fringe sits at $y_{m+1} = \dfrac{(m+1)\lambda D}{d}$ , obtained the same way with $m$ replaced by $m+1$ .

Step 3. Subtract to get the spacing. The fringe width $\beta$ is the distance between these two consecutive bright fringes: $\beta = y_{m+1}-y_m = \dfrac{(m+1)\lambda D}{d}-\dfrac{m\lambda D}{d} = \dfrac{\lambda D}{d}$ .

Step 4. Notice $m$ has cancelled out entirely, so $\beta=\lambda D/d$ is the same for every pair of neighbouring bright fringes — and, by an identical calculation starting from the dark-fringe condition, for every pair of neighbouring dark fringes too. The fringes are therefore equally spaced across the whole pattern, which is what makes $\beta$ measurable directly with a ruler on the screen.

Worked example

Light of wavelength $\lambda=600\ \text{nm}$ illuminates two slits separated by $d=0.5\ \text{mm}$ , with a screen $D=1.5\ \text{m}$ away.

Step 1. Find the fringe width using $\beta=\lambda D/d$ : $\beta = \dfrac{(600\times10^{-9})(1.5)}{0.5\times10^{-3}} = 1.8\times10^{-3}\ \text{m} = 1.8\ \text{mm}$ .

Step 2. Locate the 4th dark fringe. Dark fringes occur at $y_m=(m+\tfrac12)\beta$ , and counting the first dark fringe as $m=0$ , the 4th dark fringe corresponds to $m=3$ : $y_3=(3.5)(1.8\ \text{mm})=6.3\ \text{mm}$ from the centre.

Step 3. Extend to an intensity check. Suppose instead a point sits at $y=1.0\ \text{mm}$ from the centre with $d=1.0\ \text{mm}$ , $D=1.0\ \text{m}$ and $\lambda=650\ \text{nm}$ (the values from a companion measurement). The phase difference is $\delta = \dfrac{2\pi dy}{\lambda D} \approx 9.66\ \text{rad} \equiv 3.377\ \text{rad}\pmod{2\pi}$ , giving $I/I_{\max}=\cos^2(\delta/2)=\cos^2(1.689\ \text{rad})\approx 0.013$ .

The fringe width is 1.8 mm, the 4th dark fringe lies 6.3 mm from the centre, and a point 1.0 mm off-centre under different slit conditions sits near a dark fringe at only about 1.3 percent of maximum intensity.

Common traps

Using $\sin\theta \approx y/D$ far from the axis, where $D \gg d,y$ no longer holds and the exact $\Delta=d\sin\theta$ must be used instead.
Mixing up dark-fringe counting conventions — some problems index the first dark fringe as $m=0$ , others start at $m=1$ ; always check the wording.
Adding intensities as $I=I_1+I_2$ , which is only the incoherent result; coherent superposition adds amplitudes first, producing the cross-term $2\sqrt{I_1I_2}\cos\delta$ .
Assuming any two light sources will interfere; two independent lamps or lasers have uncorrelated phases and give no stable pattern — coherence requires splitting one wavefront.

Ask the tutor: if you want to work through a variant where the slits are illuminated at an angle or immersed in a medium of refractive index n, ask the AI tutor to walk through it with you.

Why does the two-slit pattern have equally spaced fringes instead of getting closer together farther from the centre?

The bright-fringe positions are $y_m = m\lambda D/d$ , which is linear in $m$ , so consecutive fringes are always separated by the same $\beta=\lambda D/d$ no matter how large $m$ gets — as long as the small-angle approximation $D \gg d,y$ still holds. Very far from the centre, where $y$ becomes comparable to $D$ , the exact $\sin\theta$ geometry takes over and the spacing does eventually change, but within the syllabus range the pattern is treated as uniformly spaced.

If I use white light instead of monochromatic light, what happens to the fringes?

Each wavelength produces its own fringe pattern with its own fringe width $\beta=\lambda D/d$ , and since $\beta$ depends on $\lambda$ , the patterns for different colours line up only at the central fringe (where $\Delta=0$ for every wavelength) and progressively separate away from the centre. The result is a white central fringe flanked by coloured fringes that blur into a continuous wash of colour after only a few orders — this is why monochromatic sources are used for clean, sharp fringe counting in this syllabus.

Does the double-slit pattern prove light cannot also behave as a particle?

No — this experiment demonstrates the wave nature of light through interference, but it does not rule out particle behaviour; that resolution belongs to quantum mechanics, which is out of scope here. Within this course, Young's experiment is used purely to establish that light obeys the superposition principle like other waves, producing the path-difference and intensity relations derived above.

Young’s double-slit experiment

The core idea

Worked example

Common traps

Before this

Where this leads