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Depth
Derivation

Young's Double-Slit Interference

D-432 Home PU-206 Threads light · waves Depends on Optical Wave and Helmholtz Equations from Maxwell
Statement

Consider two identical, coherent, monochromatic point sources (slits) separated by distance \(d\), emitting waves of wavelength \(\lambda\) in phase. For a distant observation point at angle \(\theta\) from the central axis, where the screen distance \(L\) satisfies \(L \gg d\), the path difference is \(\Delta = d\sin\theta\). The resulting intensity distribution on the screen is given by \(I(\theta) = 4I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right)\), where \(I_0\) is the intensity from a single slit alone. Bright fringes occur at \(d\sin\theta = m\lambda\) and dark fringes at \(d\sin\theta = (m+\tfrac{1}{2})\lambda\), for integer \(m\). Under the small-angle approximation, the fringe position is \(y_m = m\frac{\lambda L}{d}\) and the fringe spacing is \(\Delta y = \frac{\lambda L}{d}\).

Why it matters

Young's double-slit experiment is the definitive demonstration of the wave nature of light. It provides a direct, quantitative link between the wavelength of light and observable spatial patterns, establishing the principle of superposition as a fundamental law of wave physics. The derivation introduces the concept of path difference and phase coherence, which are central to all of wave optics, from thin-film interference to diffraction gratings.

Furthermore, the result is foundational for understanding coherence, which is required for any interference phenomenon. The fact that the total intensity averages to \(2I_0\) rather than \(4I_0\) demonstrates that interference redistributes energy rather than creating or destroying it, a key insight into the conservation of energy in wave systems.

Assumptions
The sources are coherent.If the phase difference between the two sources fluctuates randomly in time (incoherent sources), the interference term averages to zero, and the intensity is simply the sum of the individual intensities \(I_1 + I_2\), with no fringes.
The sources are monochromatic.If the light contains a range of wavelengths, each wavelength produces its own fringe pattern with a different spacing \(\Delta y \propto \lambda\). The patterns overlap and wash out, especially at higher orders, leading to a blurred or colored fringe pattern.
The screen distance \(L\) is much greater than the slit separation \(d\) (\(L \gg d\)).This allows the use of the parallel-ray approximation, where the path difference is \(d\sin\theta\). If \(L\) is comparable to \(d\), the Fresnel diffraction regime applies, and the simple far-field formula is invalid.
The slits are narrow (point sources) or the single-slit diffraction envelope is accounted for.In reality, slits have a finite width \(a\). The double-slit interference pattern is modulated by a single-slit diffraction envelope \(\mathrm{sinc}^2(\frac{\pi a \sin\theta}{\lambda})\). The envelope narrows as \(a\) grows, so fewer bright fringes are seen, and orders that land on an envelope zero (\(d\sin\theta = m\lambda\) together with \(a\sin\theta = n\lambda\)) are missing.
Derivation
1
\[ E_1(t) = E_0 \cos(\omega t - k r_1) \]
The electric field at the observation point due to slit 1, where \(r_1\) is the distance from slit 1 to the point, \(k = 2\pi/\lambda\), and \(\omega = 2\pi f\). A
2
\[ E_2(t) = E_0 \cos(\omega t - k r_2) \]
The electric field due to slit 2, with distance \(r_2\). The amplitudes are equal because the slits are identical and the screen is far away. A
3
\[ E_{\text{tot}}(t) = E_1(t) + E_2(t) = E_0 \left[ \cos(\omega t - k r_1) + \cos(\omega t - k r_2) \right] \]
Principle of superposition: the total electric field is the vector sum of the fields from each source. A
4
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Trigonometric identity for the sum of cosines. Here \(A = \omega t - k r_1\) and \(B = \omega t - k r_2\). B
5
\[ \frac{A+B}{2} = \omega t - k\frac{r_1+r_2}{2}, \qquad \frac{A-B}{2} = \frac{k(r_2-r_1)}{2} \]
Substitute \(A\) and \(B\). The term \(\frac{r_1+r_2}{2}\) is the average distance, and \(r_2-r_1\) is the path difference. A
6
\[ E_{\text{tot}}(t) = 2 E_0 \cos\left(\omega t - k\frac{r_1+r_2}{2}\right) \cos\left(\frac{k(r_2-r_1)}{2}\right) \]
Apply the identity from Step 4. The first cosine oscillates rapidly in time, while the second is a spatially varying amplitude factor. B
7
\[ I \propto \langle E_{\text{tot}}^2 \rangle_t \]
The intensity is proportional to the time-averaged square of the electric field. For a sinusoidal wave, \(\langle \cos^2(\omega t + \phi) \rangle_t = \frac{1}{2}\). A
8
\[ I = \left\langle \left[ 2 E_0 \cos\left(\omega t - k\frac{r_1+r_2}{2}\right) \cos\left(\frac{k(r_2-r_1)}{2}\right) \right]^2 \right\rangle_t \]
Square the total field. The term \(\cos\left(\frac{k(r_2-r_1)}{2}\right)\) is constant in time for a fixed observation point, so it can be pulled out of the time average. C
9
\[ I = 4 E_0^2 \cos^2\left(\frac{k(r_2-r_1)}{2}\right) \left\langle \cos^2\left(\omega t - k\frac{r_1+r_2}{2}\right) \right\rangle_t \]
Expand the square and separate the time-dependent and time-independent parts. B
10
\[ \left\langle \cos^2\left(\omega t - k\frac{r_1+r_2}{2}\right) \right\rangle_t = \frac{1}{2} \]
The time average of \(\cos^2\) over a full cycle is \(1/2\). A
11
\[ I = 2 E_0^2 \cos^2\left(\frac{k(r_2-r_1)}{2}\right) \]
Substitute the time average. Define \(I_0 = \frac{1}{2} E_0^2\) as the intensity from a single slit. Then \(I = 4 I_0 \cos^2\left(\frac{k(r_2-r_1)}{2}\right)\). B
12
\[ r_2 - r_1 \approx d \sin\theta \]
In the far-field approximation (\(L \gg d\)), the path difference is the projection of the slit separation onto the direction of observation. This is derived from the geometry of the two rays to the distant point. C
13
\[ I(\theta) = 4 I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right) \]
Substitute \(k = 2\pi/\lambda\) and \(r_2-r_1 = d\sin\theta\) into the intensity formula. A
Result
\[ I(\theta) = 4 I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right) \]

Reading. The intensity varies as the square of the cosine of half the phase difference. Bright fringes (maxima) occur when the argument is \(m\pi\), i.e., \(d\sin\theta = m\lambda\). Dark fringes (minima) occur when the argument is \((m+\tfrac{1}{2})\pi\), i.e., \(d\sin\theta = (m+\tfrac{1}{2})\lambda\).

Units check. The argument of the cosine is \(\frac{d \sin\theta}{\lambda}\). Both \(d\) and \(\lambda\) are lengths, so the ratio is dimensionless. The intensity \(I\) has units of \(\mathrm{W/m^2}\), consistent with \(I_0\).

Limiting cases
  • Small angles (\(\theta \ll 1\)). \(\sin\theta \approx \theta \approx y/L\). The fringe position is \(y_m = m\frac{\lambda L}{d}\), and the fringe spacing is \(\Delta y = \frac{\lambda L}{d}\). This is the standard linear approximation used in most laboratory measurements.
  • Large slit separation (\(d \gg \lambda\)). The fringe spacing \(\Delta y = \frac{\lambda L}{d}\) becomes very small, and many fringes are visible across the screen. The pattern becomes fine and dense.
  • Small slit separation (\(d\) approaching \(\lambda\)). The fringes spread to large angles and only a few orders exist, since \(|\sin\theta| \le 1\) allows only \(|m| \le d/\lambda\). For \(d \lt \lambda\) there is no first-order maximum at all: only the broad central fringe remains.
  • Finite slit width \(a\). \(I(\theta) = 4I_0\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right)\left[\frac{\sin(\pi a\sin\theta/\lambda)}{\pi a\sin\theta/\lambda}\right]^2\): the two-slit fringes sit under a single-slit envelope, and order \(m\) is missing whenever it falls on an envelope zero, i.e. when \(m = (d/a)\,n\) for integer \(n\) (e.g. orders \(\pm3, \pm6, \dots\) vanish when \(d = 3a\)).
  • Incoherent sources. The interference term averages to zero, and the intensity is \(I = I_1 + I_2 = 2I_0\), a uniform illumination with no fringes.
Breaks when
  • Incoherent sources. If the phase difference between the two slits fluctuates randomly in time (e.g., two independent light bulbs), the interference term \(\cos^2(\frac{\pi d \sin\theta}{\lambda})\) averages to \(1/2\), and the intensity is simply \(2I_0\), with no fringes.
  • Near-field (Fresnel) regime. If the screen distance \(L\) is not much greater than \(d\), the parallel-ray approximation fails, and the path difference is not \(d\sin\theta\). The Fresnel diffraction integrals must be used instead.
  • Polychromatic light. If the light contains a range of wavelengths, each wavelength produces a fringe pattern with a different spacing. The patterns overlap and wash out, especially at higher orders, leading to a blurred or colored fringe pattern.
Failure modes
  • Confusing path difference with phase difference. The path difference is \(d\sin\theta\), while the phase difference is \(\frac{2\pi}{\lambda} d\sin\theta\). The intensity depends on the phase difference, not the path difference directly.
  • Forgetting the small-angle approximation. Using \(y = L\tan\theta\) instead of \(y \approx L\theta\) when \(\theta\) is not small. This leads to errors in the fringe position, especially for higher orders.
  • Assuming the intensity is \(I_1 + I_2\). The intensity is not simply the sum of the individual intensities; it includes an interference term. The total intensity is \(I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta\), where \(\delta\) is the phase difference.
  • Ignoring the single-slit diffraction envelope. In real experiments, the slits have a finite width, and the double-slit pattern is modulated by a single-slit diffraction envelope. Some interference orders may be missing if they coincide with the minima of the envelope.
Discussion

The derivation of Young's double-slit interference is a beautiful example of how the principle of superposition leads to observable patterns. The key insight is that the interference pattern arises from the phase difference between the two waves, which is determined by the path difference. The intensity distribution is a direct consequence of the constructive and destructive interference of the electric fields.

The result also highlights the importance of coherence. Without a stable phase relationship between the two sources, the interference pattern would wash out. This is why Young's experiment requires a single source split into two coherent beams, rather than two independent sources. The coherence length of the light source must be greater than the path difference for the interference to be observable.

The energy conservation aspect is also crucial. Averaged over one fringe period the intensity is \(4I_0\langle\cos^2\rangle = 2I_0\), exactly the sum of the two single-slit intensities, even though the peaks reach \(4I_0\). This means that the interference pattern redistributes energy, with bright fringes at the expense of dark fringes. No energy is created or destroyed; it is merely redistributed in space.

At a deeper level, the double-slit experiment is a cornerstone of quantum mechanics. When performed with single photons or electrons, the interference pattern builds up one particle at a time, demonstrating the wave-particle duality. Each particle interferes with itself, and the probability distribution is given by the same intensity formula. This is a profound illustration of the probabilistic nature of quantum mechanics.

Common misconceptions. One common misconception is that the two waves "cancel" each other in the dark fringes, destroying energy. In reality, the energy is redistributed to the bright fringes. Another misconception is that the interference pattern is due to the interaction of the two waves with each other. In fact, the pattern is a result of the superposition of the fields, and each wave contributes independently to the total field.

Related pages. Fraunhofer Diffraction as a Fourier Transform · Diffraction Grating: Grating Equation and Resolving Power · The van Cittert-Zernike Coherence Theorem · The de Broglie Wavelength of a Material Particle · Optical Wave and Helmholtz Equations from Maxwell.

Worked examples

Example 1. A double-slit experiment uses light of wavelength \(\lambda = 550\ \mathrm{nm}\). The slit separation is \(d = 0.10\ \mathrm{mm}\), and the screen is \(L = 1.0\ \mathrm{m}\) away. Find the fringe spacing \(\Delta y\) and the position of the third-order bright fringe (\(m=3\)).

1
\[ \Delta y = \frac{\lambda L}{d} \]
The fringe spacing in the small-angle approximation. A
2
\[ \Delta y = \frac{(550 \times 10^{-9}\ \mathrm{m})(1.0\ \mathrm{m})}{0.10 \times 10^{-3}\ \mathrm{m}} = 5.5 \times 10^{-3}\ \mathrm{m} = 5.5\ \mathrm{mm} \]
Substitute the given values. A
3
\[ y_3 = 3 \Delta y = 3(5.5\ \mathrm{mm}) = 16.5\ \mathrm{mm} \]
The position of the third-order bright fringe is three times the fringe spacing. A
\[ \Delta y = 5.5\ \mathrm{mm}, \qquad y_3 = 16.5\ \mathrm{mm} \]

Reading. The fringes are spaced 5.5 mm apart, and the third-order bright fringe is 16.5 mm from the central maximum.

Units check. \(\lambda L / d\) has units \(\mathrm{m \cdot m / m = m}\), consistent with a length.

Example 2. In a double-slit experiment, the slit separation is \(d = 0.25\ \mathrm{mm}\), and the screen is \(L = 2.0\ \mathrm{m}\) away. The distance between the central maximum and the second-order bright fringe (\(m=2\)) is measured to be \(12.0\ \mathrm{mm}\). Find the wavelength of the light.

1
\[ y_2 = 2 \frac{\lambda L}{d} \]
The position of the second-order bright fringe. A
2
\[ \lambda = \frac{y_2 d}{2 L} \]
Solve for the wavelength. B
3
\[ \lambda = \frac{(12.0 \times 10^{-3}\ \mathrm{m})(0.25 \times 10^{-3}\ \mathrm{m})}{2(2.0\ \mathrm{m})} = \frac{3.0 \times 10^{-6}\ \mathrm{m^2}}{4.0\ \mathrm{m}} = 7.5 \times 10^{-7}\ \mathrm{m} = 750\ \mathrm{nm} \]
Substitute the given values. A
\[ \lambda = 750\ \mathrm{nm} \]

Reading. The wavelength of the light is 750 nm, at the deep-red edge of the visible spectrum.

Units check. \(y d / L\) has units \(\mathrm{m \cdot m / m = m}\), consistent with a wavelength.

Problems
  1. A double-slit experiment uses light of wavelength \(\lambda = 600\ \mathrm{nm}\). The slit separation is \(d = 0.15\ \mathrm{mm}\), and the screen is \(L = 1.5\ \mathrm{m}\) away. Find the fringe spacing \(\Delta y\).
    Solution \(\Delta y = \frac{\lambda L}{d} = \frac{(600 \times 10^{-9}\ \mathrm{m})(1.5\ \mathrm{m})}{0.15 \times 10^{-3}\ \mathrm{m}} = 6.0 \times 10^{-3}\ \mathrm{m} = 6.0\ \mathrm{mm}\).
  2. In a double-slit experiment, the slit separation is \(d = 0.20\ \mathrm{mm}\), and the screen is \(L = 1.0\ \mathrm{m}\) away. The distance between the central maximum and the first-order bright fringe is measured to be \(2.5\ \mathrm{mm}\). Find the wavelength of the light.
    Solution \(y_1 = \frac{\lambda L}{d} \Rightarrow \lambda = \frac{y_1 d}{L} = \frac{(2.5 \times 10^{-3}\ \mathrm{m})(0.20 \times 10^{-3}\ \mathrm{m})}{1.0\ \mathrm{m}} = 5.0 \times 10^{-7}\ \mathrm{m} = 500\ \mathrm{nm}\) (blue-green light).
  3. A double-slit experiment uses light of wavelength \(\lambda = 500\ \mathrm{nm}\). The slit separation is \(d = 0.10\ \mathrm{mm}\), and the screen is \(L = 2.0\ \mathrm{m}\) away. Find the position of the dark fringe with \(m=4\) in the condition \(d\sin\theta = (m+\tfrac12)\lambda\) (the fifth dark fringe counting out from the centre).
    Solution The position of the \(m\)-th dark fringe is \(y = (m + \tfrac{1}{2}) \frac{\lambda L}{d}\). For \(m=4\), \(y = (4.5) \frac{(500 \times 10^{-9}\ \mathrm{m})(2.0\ \mathrm{m})}{0.10 \times 10^{-3}\ \mathrm{m}} = 4.5 \times 10^{-2}\ \mathrm{m} = 45\ \mathrm{mm}\).
  4. In a double-slit experiment, the slit separation is \(d = 0.30\ \mathrm{mm}\), and the screen is \(L = 1.5\ \mathrm{m}\) away. The distance between the central maximum and the third-order bright fringe is measured to be \(9.0\ \mathrm{mm}\). Find the wavelength of the light.
    Solution \(y_3 = 3 \frac{\lambda L}{d} \Rightarrow \lambda = \frac{y_3 d}{3 L} = \frac{(9.0 \times 10^{-3}\ \mathrm{m})(0.30 \times 10^{-3}\ \mathrm{m})}{3(1.5\ \mathrm{m})} = \frac{2.7 \times 10^{-6}\ \mathrm{m^2}}{4.5\ \mathrm{m}} = 6.0 \times 10^{-7}\ \mathrm{m} = 600\ \mathrm{nm}\) (orange light).
  5. A double-slit experiment uses light of wavelength \(\lambda = 650\ \mathrm{nm}\). The slit separation is \(d = 0.12\ \mathrm{mm}\), and the screen is \(L = 1.2\ \mathrm{m}\) away. Find the intensity at a point on the screen that is \(10.0\ \mathrm{mm}\) from the central maximum, given that the intensity at the central maximum is \(I_{\text{max}} = 100\ \mathrm{W/m^2}\).
    Solution The intensity is \(I = I_{\text{max}} \cos^2\left(\frac{\pi d y}{\lambda L}\right)\). First, find the argument: \(\frac{\pi d y}{\lambda L} = \frac{\pi (0.12 \times 10^{-3}\ \mathrm{m})(10.0 \times 10^{-3}\ \mathrm{m})}{(650 \times 10^{-9}\ \mathrm{m})(1.2\ \mathrm{m})} = \frac{3.77 \times 10^{-6}}{7.80 \times 10^{-7}} = 4.83\ \mathrm{rad}\). Then, \(I = 100 \cos^2(4.83) = 100 \cos^2(276.9^\circ) = 100 (0.121)^2 = 1.45\ \mathrm{W/m^2}\): the point lies close to a dark fringe (the \(m=1\) minimum, at \(\pi d y/\lambda L = 3\pi/2 \approx 4.71\)).