Why Light Spreads Out More the Narrower the Slit It Passes Through

Ray optics treats light as straight lines, but it cannot explain dark and bright bands formed by two overlapping beams, or light spreading into the shadow of a narrow slit. Treating light as a wave explains both, and also rebuilds the familiar laws of reflection and refraction.

This lesson covers Huygens' principle, Young's double-slit experiment, and single-slit diffraction.

**Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the surface touching all these wavelets; applying it at a boundary gives $i=r$ for reflection and $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$ for refraction.

Refraction.** A plane wavefront AB meets a surface at A. While its far end B travels a distance $BC=v_{1}t$ to the surface, the wavelet from A travels $AE=v_{2}t$ in the second medium. The new wavefront is CE, and from the two right-angled triangles on the common side AC,

$$\sin i=\frac{v_{1}t}{AC}\sin r=\frac{v_{2}t}{AC}\Rightarrow \frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{n_2}{n_1}$$

Reflection. The wavelet returns into the same medium at the same speed, so the two triangles are congruent and $i=r$.

Worked example. Light at $3.0\times 10^8$ m s${}^{-1}$ enters glass of $n=1.5$ at $30^{\circ }$:

$$v=\frac{3.0\times 10^8}{1.5}=2.0\times 10^8{\text{ m s}}^{-1}\sin r=\frac{0.50}{1.5}=0.333\Rightarrow r=19.5^{\circ }$$

A 600 nm wavelength becomes 400 nm in the glass, while the frequency stays the same.

An everyday example. Sea waves approaching a beach at an angle slow down in shallow water and swing round to arrive nearly parallel to the shore.

The substance. Refraction towards the normal means light is slower in glass — the wave picture predicts this, which is how experiments could test it.

**Two coherent slits a distance d apart, with a screen at distance D, give bright fringes at $y_n=\frac{n\lambda D}{d}$ and a fringe width of $\beta =\frac{\lambda D}{d}$.

The derivation.** For a point P at height y on the screen, with $D\gg d$, the path difference between the waves from the two slits is $\Delta =\frac{yd}{D}$.

- Bright fringe: $\Delta =n\lambda$, so $y_n=n\lambda D/d$
- Dark fringe: $\Delta =(2n-1)\lambda /2$, so $y_n=(2n-1)\lambda D/2d$

The gap between neighbouring bright fringes is $\beta =\lambda D/d$.

Worked example. With $\lambda =600$ nm, $d=0.50$ mm and $D=1.0$ m:

$$\beta =\frac{6.0\times 10^{-7}\times 1.0}{5.0\times 10^{-4}}=1.2\times 10^{-3}\text{ m}=1.2\text{ mm}$$

In water, with $n=1.33$, the wavelength shrinks and $\beta =1.2/1.33=0.90$ mm.

Conditions for sustained interference:

- The sources must be coherent, with a constant phase difference, so both slits are lit from one source
- They must have the same frequency, and nearly equal amplitudes for good contrast
- The slits must be narrow and close together, with $D\gg d$

An everyday example. A thin film of oil on a wet road shows coloured bands, because light reflected from its top and bottom surfaces interferes.

The substance. Two separate bulbs never produce fringes — their phases change randomly, so the pattern shifts too fast to see.

**Light passing through a single slit of width a spreads out, with minima where $a\sin \theta =n\lambda$, so the central maximum has an angular width of $\frac{2\lambda }{a}$ and a linear width of $\frac{2\lambda D}{a}$ on a screen at distance D.

The derivation.** Divide the slit into two equal halves. When $a\sin \theta =\lambda$, every point in the upper half has a partner in the lower half whose light arrives exactly $\lambda /2$ later, so each pair cancels and the first minimum appears. Dividing the slit into 4, 6, and more parts gives the higher minima.

Worked example. A slit of width 0.20 mm is lit by 600 nm light, with the screen at 2.0 m:

$${\theta }_1=\frac{\lambda }{a}=\frac{6.0\times 10^{-7}}{2.0\times 10^{-4}}=3.0\times 10^{-3}\text{ rad}\text{width}=\frac{2\lambda D}{a}=\frac{2\times 6.0\times 10^{-7}\times 2.0}{2.0\times 10^{-4}}=12\text{ mm}$$

Halving the slit width to 0.10 mm doubles the central maximum to 24 mm.

How it differs from interference. In a double-slit pattern the fringes have equal widths and similar brightness; in single-slit diffraction the central maximum is twice as wide as the others, and the secondary maxima grow rapidly fainter.

An everyday example. Look at a tube light through the narrow gap between two fingers and you will see faint dark lines running parallel to the gap.

The substance. **The condition $a\sin \theta =n\lambda$ gives dark bands here**, while the same form gives bright fringes in the double-slit experiment.