Young's Double-Slit Experiment

Spec mapping (OCR H556 Module 4.5): Apply the two-source interference formula $w = \lambda D / s$w=λD/s to measure the wavelength of visible light. Describe the apparatus, justify the small-angle approximation, and account for the role of coherence. This lesson is a PAG 3 anchor — measurement of wavelength using a laser and a double slit is one of the required practical activities.

In 1801, the English natural philosopher Thomas Young performed an experiment that helped settle a century-long debate about the nature of light. He passed a single beam of light through two closely-spaced slits and observed, on a screen placed some distance beyond, a pattern of alternating bright and dark fringes. The pattern could only be explained if light were a wave — particles would have produced two isolated bright patches, one behind each slit, not an extended fringe pattern. Young's experiment remains the canonical demonstration of wave interference and is one of the most influential experiments in the history of physics.

At A-Level you are expected to describe the experiment, explain the fringe pattern qualitatively and quantitatively, derive the small-angle formula, and apply it numerically. The OCR formula sheet uses the symbols $w$w (fringe spacing), $\lambda$λ (wavelength), $D$D (slit-screen distance) and $s$s (slit separation):

$w = \frac{\lambda D}{s}$w=sλD​

Some textbooks use $x$x for the fringe spacing and $a$a for the slit separation. The physics is the same; only the lettering varies. OCR question papers settle the notation in the question stem; you should match it exactly when responding.

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The Set-Up

A typical modern Young's experiment uses the following apparatus.

1. A monochromatic, coherent light source — most commonly a laser, because it is naturally monochromatic and spatially coherent. Historically, Young used sunlight passed through a narrow primary slit to create a quasi-coherent source.
2. Two parallel slits, narrow and close together, separated by a small distance $s$s (typically $0.1$0.1–$0.5\text{ mm}$0.5 mm).
3. A screen placed a distance $D$D beyond the slits (typically $1$1–$3\text{ m}$3 m).
4. The resulting fringe pattern — alternating bright and dark fringes — observed on the screen. The bright-to-bright spacing is the fringe spacing $w$w.

flowchart LR
    L[Laser: coherent and monochromatic] --> S1[Slit 1]
    L --> S2[Slit 2]
    S1 --> P[Point P on screen]
    S2 --> P
    P --> V[Resultant depends on path difference]
    V --> BR[Bright fringes: path difference n lambda]
    V --> DK[Dark fringes: path difference n plus half lambda]

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How the Pattern Forms

The two slits act as sources of secondary waves (Huygens's principle). Because both slits are illuminated by the same primary wave, they emit light that is automatically coherent — same frequency, constant phase relationship — regardless of any random fluctuations in the laser, because both slits feel any wobble of the source simultaneously.

At any point on the screen, waves from the two slits arrive having travelled slightly different distances. The path difference between the two waves determines whether they interfere constructively or destructively at that point:

• Bright fringe (constructive): path difference $\Delta = n\lambda$Δ=nλ, where $n = 0, \pm 1, \pm 2, \ldots$n=0,±1,±2,…
• Dark fringe (destructive): path difference $\Delta = (n + \tfrac{1}{2})\lambda$Δ=(n+21​)λ, where $n = 0, \pm 1, \pm 2, \ldots$n=0,±1,±2,…

Where they are in phase, amplitudes double (intensity quadruples) — a bright fringe. Where they are in antiphase, the amplitudes cancel — a dark fringe.

The result is a regular series of equally-spaced bright and dark fringes on the screen, symmetric about the central maximum (the point directly between the two slits).

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The Double-Slit Formula: Derivation

Let the two slits be separated by a distance $s$s, and the screen be at a distance $D$D from the slits. Consider a point $P$P on the screen at a perpendicular distance $y$y from the central maximum.

The line from each slit to $P$P makes a small angle $\theta$θ with the central axis. The path difference between the two waves is approximately the small extra distance one wave travels, which from the geometry of two parallel rays a distance $s$s apart, both pointing at $P$P, is:

$\Delta = s \sin\theta$Δ=ssinθ

For small angles ($D \gg s$D≫s and $D \gg y$D≫y, always satisfied in a laboratory Young's experiment with $s \sim 0.1\text{ mm}$s∼0.1 mm and $D \sim 1\text{ m}$D∼1 m):

$\sin\theta \approx \tan\theta = \frac{y}{D}$sinθ≈tanθ=Dy​

so

$\Delta \approx \frac{s\,y}{D}$Δ≈Dsy​

The condition for the $n$n-th bright fringe is $\Delta = n\lambda$Δ=nλ:

$\frac{s\,y_n}{D} = n\lambda \quad \Rightarrow \quad y_n = \frac{n\lambda D}{s}$Dsyn​​=nλ⇒yn​=snλD​

Adjacent bright fringes therefore differ in $y$y by:

$w = y_{n+1} - y_n = \frac{\lambda D}{s}$w=yn+1​−yn​=sλD​

Rearranging for wavelength:

$\lambda = \frac{s\,w}{D}$λ=Dsw​

This is the formula OCR expects you to know.

Variables in the Formula

Symbol	Meaning	Typical value
$\lambda$λ	Wavelength of the light	$400$400–$700\text{ nm}$700 nm for visible
$s$s	Slit separation	$0.1$0.1–$1.0\text{ mm}$1.0 mm
$w$w	Fringe spacing (bright-to-bright or dark-to-dark)	$1$1–$10\text{ mm}$10 mm
$D$D	Slit-to-screen distance	$1$1–$3\text{ m}$3 m

The formula is valid when $s \ll D$s≪D and $y \ll D$y≪D, which together guarantee $\sin\theta \approx \tan\theta \approx \theta$sinθ≈tanθ≈θ. At A-Level this approximation is always assumed for the double-slit geometry (it is not used for diffraction gratings — see Lesson 9).

Exam tip. OCR question papers consistently use $w$w for fringe spacing and $s$s for slit separation. Some older textbooks use $x$x and $a$a. Read the question, then stick to the lettering in the stem.

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Worked Example 1 — Finding the Wavelength

Q. In a Young's double-slit experiment, the slit separation is $s = 0.40\text{ mm}$s=0.40 mm and the distance to the screen is $D = 1.50\text{ m}$D=1.50 m. The fringe spacing is measured to be $w = 2.25\text{ mm}$w=2.25 mm. Calculate the wavelength of the light.

A. Using $\lambda = sw/D$λ=sw/D with $s = 4.0\times 10^{-4}\text{ m}$s=4.0×10−4 m, $w = 2.25\times 10^{-3}\text{ m}$w=2.25×10−3 m, $D = 1.50\text{ m}$D=1.50 m:

$\lambda = \frac{(4.0\times 10^{-4})(2.25\times 10^{-3})}{1.50} = 6.00\times 10^{-7}\text{ m} = 600\text{ nm}$λ=1.50(4.0×10−4)(2.25×10−3)​=6.00×10−7 m=600 nm

This lies in the orange part of the visible spectrum.

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Worked Example 2 — Predicting Fringe Spacing

Q. Light of wavelength $\lambda = 450\text{ nm}$λ=450 nm illuminates two slits $0.25\text{ mm}$0.25 mm apart. A screen is placed $1.20\text{ m}$1.20 m away. Calculate the fringe spacing.

A.

$w = \frac{\lambda D}{s} = \frac{(4.50\times 10^{-7})(1.20)}{2.50\times 10^{-4}} = 2.16\times 10^{-3}\text{ m} = 2.16\text{ mm}$w=sλD​=2.50×10−4(4.50×10−7)(1.20)​=2.16×10−3 m=2.16 mm

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Worked Example 3 — Ratio of Fringe Spacings

Q. A double-slit apparatus is illuminated first with red light ($\lambda = 700\text{ nm}$λ=700 nm) and then with blue light ($\lambda = 450\text{ nm}$λ=450 nm). The slit separation and screen distance are unchanged. Calculate the ratio of the fringe spacings.

A. Since $w \propto \lambda$w∝λ:

$\frac{w_{\text{red}}}{w_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} = \frac{700}{450} = 1.56$wblue​wred​​=λblue​λred​​=450700​=1.56

The red fringes are $1.56$1.56 times wider than the blue fringes.

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Worked Example 4 — Counting Fringes Correctly

Q. In a Young's experiment, a student measures $10$10 adjacent bright fringes spanning a distance of $18.0\text{ mm}$18.0 mm. If the slit separation is $s = 0.20\text{ mm}$s=0.20 mm and $D = 1.00\text{ m}$D=1.00 m, calculate the wavelength.

A. Watch the off-by-one. Ten bright fringes span $10 - 1 = 9$10−1=9 fringe spacings, so

$w = \frac{18.0}{9} = 2.00\text{ mm} = 2.00\times 10^{-3}\text{ m}$w=918.0​=2.00 mm=2.00×10−3 m

$\lambda = \frac{sw}{D} = \frac{(2.00\times 10^{-4})(2.00\times 10^{-3})}{1.00} = 4.00\times 10^{-7}\text{ m} = 400\text{ nm}$λ=Dsw​=1.00(2.00×10−4)(2.00×10−3)​=4.00×10−7 m=400 nm

Violet, at the short-wavelength end of the visible spectrum.

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Worked Example 5 — Effect of Changing the Slits

Q. A Young's apparatus initially uses slits $s = 0.50\text{ mm}$s=0.50 mm apart and produces fringes $w = 1.50\text{ mm}$w=1.50 mm apart. The slits are replaced by ones $0.20\text{ mm}$0.20 mm apart, with all other parameters unchanged. Calculate the new fringe spacing.

A. $w \propto 1/s$w∝1/s, so

$w_{\text{new}} = w_{\text{old}} \times \frac{s_{\text{old}}}{s_{\text{new}}} = 1.50 \times \frac{0.50}{0.20} = 3.75\text{ mm}$wnew​=wold​×snew​sold​​=1.50×0.200.50​=3.75 mm

Narrower slit separation gives wider fringes — easier to measure.

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Improving the Accuracy (PAG 3 Method Notes)

To measure wavelength accurately using Young's fringes, good practice (and the standard expectation for OCR PAG 3) is:

1. Use a monochromatic laser — narrow linewidth and high spatial coherence.
2. Make the fringes as widely spaced as practicable by using a small $s$s and a large $D$D. This reduces the percentage uncertainty in $w$w.
3. Measure across many fringes (typically $10$10 or $20$20) and divide. If you measure across $n$n fringes you must divide by $n - 1$n−1 spacings — that off-by-one is the single most common arithmetic slip.
4. Measure from the centre of a bright fringe to the centre of a bright fringe — not bright to dark.
5. Repeat the measurement and take a mean; estimate the uncertainty as half the range or as the standard error on the mean.
6. Use a vernier calliper, travelling microscope or calibrated camera for precision better than $0.1\text{ mm}$0.1 mm on the fringe-position measurement.
7. Estimate the uncertainty in $s$s (typically the manufacturer's tolerance, $\pm 5\%$±5% for a cheap card slit, $\pm 0.5\%$±0.5% for a precision pair) and in $D$D (typically $\pm 1$±1–$2\text{ mm}$2 mm).
8. Combine fractional uncertainties: $\frac{\delta\lambda}{\lambda} = \frac{\delta s}{s} + \frac{\delta w}{w} + \frac{\delta D}{D}$λδλ​=sδs​+wδw​+DδD​.

A well-executed Young's experiment can recover the laser's quoted wavelength to better than $\pm 2\%$±2%.

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Worked Example 6 — Uncertainty Calculation

Q. A student measures $w = 2.0 \pm 0.1\text{ mm}$w=2.0±0.1 mm, $s = 0.30 \pm 0.02\text{ mm}$s=0.30±0.02 mm, $D = 1.50 \pm 0.01\text{ m}$D=1.50±0.01 m. Calculate the wavelength and its percentage uncertainty.

A.

$\lambda = \frac{sw}{D} = \frac{(3.0\times 10^{-4})(2.0\times 10^{-3})}{1.50} = 4.0\times 10^{-7}\text{ m} = 400\text{ nm}$λ=Dsw​=1.50(3.0×10−4)(2.0×10−3)​=4.0×10−7 m=400 nm

Fractional uncertainties:

$\frac{\delta\lambda}{\lambda} = \frac{0.02}{0.30} + \frac{0.1}{2.0} + \frac{0.01}{1.50} \approx 0.067 + 0.050 + 0.007 = 0.124$λδλ​=0.300.02​+2.00.1​+1.500.01​≈0.067+0.050+0.007=0.124

So $\lambda = (4.0 \pm 0.5) \times 10^{-7}\text{ m}$λ=(4.0±0.5)×10−7 m, i.e. $\pm 12\%$±12%. The slit-separation tolerance dominates the error budget; that is the term to attack to improve accuracy.

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Young's Experiment as Evidence for the Wave Nature of Light

Before Young, Newton had argued (on the basis of refraction and of sharp shadows) that light was made of particles (corpuscles). Huygens had argued that it was a wave. The question was difficult to resolve because the wavelengths of visible light were so small that diffraction effects are rarely obvious. Young's experiment provided definitive evidence for the wave nature of light:

• The fringe pattern is inexplicable on a particle theory. Particles passing through two slits should form two bright regions directly behind the slits, not an extended pattern of bright and dark fringes.
• The fringe spacing depends on wavelength in exactly the way wave theory predicts.
• Young could measure the wavelength of light directly from the fringe spacing — the first reliable determination of $\lambda$λ for visible light.

Later quantum theory showed that the full story is subtler: light has both wave and particle aspects (wave-particle duality), and the same fringe pattern appears even when photons are sent through the apparatus one at a time (the single-photon Young's experiment performed in many laboratories from the 1970s onwards). But at A-Level you can treat Young's experiment firmly as evidence of the wave nature of light.

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Variations on the Young Experiment