How Two Beams of Light Can Add Up to Darkness

A thin film of oil on a wet road shimmers with colour, and a soap bubble glows in bands of pink and green. These patterns come from light waves adding and cancelling — interference — and from light spreading as it passes narrow gaps — diffraction.

This part covers coherent and incoherent addition, Young's double-slit experiment, and single-slit diffraction.

Waves from coherent sources keep a constant phase difference, so their displacements add to give a steady pattern of bright and dark regions, while waves from incoherent sources change phase randomly, so only their intensities add and no pattern forms.

Coherent addition. Two waves of intensity $I_0$ with phase difference $\varphi$ give

$$I=4I_0{\cos }^2\frac{\varphi }{2}$$

- Constructive — $\varphi =0,2\pi ,4\pi ,\dots$ or path difference $n\lambda$: $I=4I_0$
- Destructive — $\varphi =\pi ,3\pi ,\dots$ or path difference $\left(n+\frac{1}{2}\right)\lambda$: $I=0$

Incoherent addition. The phase difference changes rapidly and randomly, the average of ${\cos }^2\frac{\varphi }{2}$ is $\frac{1}{2}$, and $I=2I_0$ everywhere.

Conditions for sustained interference:

- Sources must be coherent — same frequency and constant phase difference
- Amplitudes should be nearly equal, so dark fringes are truly dark
- Sources should be narrow and close together

Worked example. Coherent waves of intensities $4$ units and $1$ unit give

$$I_{max}={\left(\sqrt{4}+\sqrt{1}\right)}^2=9,I_{min}={\left(\sqrt{4}-\sqrt{1}\right)}^2=1$$

Incoherent, they give a uniform $4+1=5$ units.

An everyday example. Two bulbs lighting a room never produce fringes on the wall, because atoms in separate bulbs emit light with random, constantly changing phases.

The substance. Interference does not destroy energy — light missing from dark fringes reappears in bright ones, keeping the average at $2I_0$.

**Two narrow slits lit by the same source act as coherent sources, and at a point a distance $y$ from the centre of a screen the path difference is $\frac{yd}{D}$, giving equally spaced bright and dark fringes of width $\beta =\frac{\lambda D}{d}$.

Set-up:** slits $S_1$ and $S_2$ a distance $d$ apart, with the screen a distance $D$ away, where $D$ is much larger than $d$.

Fringe positions:

- Bright fringes — $y_n=\frac{n\lambda D}{d}$
- Dark fringes — $y_n=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$
- Fringe width — $\beta =\frac{\lambda D}{d}$, the same for bright and dark fringes

Worked example. Light of wavelength $600$ nm, slits $0.2$ mm apart and a screen $1$ m away:

$$\beta =\frac{600\times 10^{-9}\times 1}{0.2\times 10^{-3}}=3\times 10^{-3}\text{ m}=3\text{ mm}$$

The third bright fringe lies $9$ mm from the centre. With the apparatus immersed in water ($n=\frac{4}{3}$), the wavelength and fringe width shrink to $\frac{3}{4}$ of their values, giving $2.25$ mm.

Changing the set-up: a larger $D$ or longer $\lambda$ widens the fringes, while a larger $d$ narrows them; white light gives a white central fringe with coloured fringes on either side.

An everyday example. A red laser pointer shone through two fine scratches on a blackened glass slide produces a row of evenly spaced red spots on a far wall.

The substance. The central fringe is always bright, because the path difference there is zero for every wavelength.

**Light passing through a single narrow slit of width $a$ spreads out into a broad bright central maximum with fainter bands on either side; the first minima occur at $\sin \theta =\frac{\lambda }{a}$, so the central maximum has angular width $\frac{2\lambda }{a}$ and linear width $\frac{2\lambda D}{a}$ on a screen at distance $D$.

Why minima form.** Divide the slit into two halves: when the path difference between the edges is $\lambda$, each point in one half cancels a matching point in the other.

- Minima — $a\sin \theta =n\lambda$, with $n=\pm 1,\pm 2,\dots$
- Secondary maxima — roughly at $a\sin \theta =\left(n+\frac{1}{2}\right)\lambda$, growing fainter
- Central maximum — twice as wide as the other bands

Worked example. Light of wavelength $500$ nm passes through a slit $0.1$ mm wide onto a screen $1$ m away:

$$\theta =\frac{\lambda }{a}=\frac{500\times 10^{-9}}{0.1\times 10^{-3}}=5\times 10^{-3}\text{ rad},\text{width}=\frac{2\lambda D}{a}=10\text{ mm}$$

Halving the slit width doubles the central maximum to $20$ mm.

Interference versus diffraction: interference fringes are equally wide and nearly equally bright, while diffraction bands have a wide central maximum and rapidly fading sides.

An everyday example. Looking at a distant streetlight through a narrow gap between two fingers shows faint dark and bright bands along the gap.

The substance. Diffraction limits how finely any microscope or telescope can resolve detail, linking this part back to optical instruments.