Wave Phenomena

When a wave meets a boundary between two media, part of it reflects and part may pass through. Reflection follows a single rule: the angle of incidence equals the angle of reflection, with both angles measured from the normal (the line perpendicular to the surface).

Refraction happens because the wave changes speed in the new medium. The frequency is fixed by the source and never changes, so the wavelength must change in proportion to the speed ($v = f\lambda$).

The refractive index of a medium, $n = \frac{c}{v}$, measures how much it slows light. Snell's law links the angles and indices on the two sides of a boundary:

$$n_1 \sin\theta_1 = n_2 \sin\theta_2$$

Light entering a denser medium (higher $n$) slows down and bends towards the normal; on leaving, it speeds up and bends away. For example, a ray passing from air into a glass block ($n = 1.5$) at $40^\circ$ refracts to only about $25^\circ$ from the normal.

You should be able to calculate any one of $n$, $\theta$, $v$ or $\lambda$ from the others, and draw wavefront diagrams with the wavefronts closer together in the slower (denser) medium and the ray always perpendicular to the wavefronts.

Total internal reflection (TIR) is the special case in which no light escapes a boundary at all: the entire wave reflects back into the first medium. Two conditions must both hold: the light must be travelling from an optically denser medium into a less dense one ($n_1 > n_2$), and the angle of incidence must exceed the critical angle $\theta_c$, found from:

$$\sin\theta_c = \frac{n_2}{n_1}$$

• At exactly the critical angle, the refracted ray skims along the boundary at $90^\circ$.
• Below it, the ray refracts out (with some partial reflection).
• Above it, refraction becomes impossible and the reflection is total.

For glass ($n = 1.50$) in air the critical angle is about $42^\circ$, which is why $45^\circ$ prisms can redirect light through $90^\circ$ or $180^\circ$ more efficiently than mirrors.

Optical fibres exploit TIR: light entering the core at a shallow enough angle strikes the core-cladding boundary beyond the critical angle and reflects repeatedly, staying trapped along kilometres of fibre with very little loss.

You should be able to calculate $\theta_c$ for any pair of media, decide whether a given ray will totally internally reflect, and explain why TIR cannot occur when light travels from a less dense to a denser medium (the refracted ray always bends towards the normal, so a $90^\circ$ angle of refraction can never be reached).

Diffraction is the spreading of a wave after it passes through an aperture or around the edge of an obstacle. It is a property of all waves (sound, water waves, light, even electrons) and cannot be explained by rays travelling in straight lines.

The amount of spreading depends on the ratio of wavelength to gap size: diffraction is most pronounced when the wavelength is comparable to the aperture width, that is, when:

$$\lambda \approx b$$

A wide gap, many wavelengths across, lets the wave pass almost straight through with only slight curving at the edges; a gap of about one wavelength turns the opening into what looks like a point source of circular wavefronts.

Crucially, diffraction changes only the direction of travel: the wave stays in the same medium, so its speed, wavelength and frequency are all unchanged. This is what distinguishes a diffraction wavefront diagram from a refraction one.

Everyday examples make the scale-dependence clear. Sound with wavelengths around a metre diffracts strongly through a doorway, which is why you can hear a conversation from around a corner; light, with wavelengths around $5 \times 10^{-7}\ \text{m}$, barely spreads at the same doorway, so you cannot see around the corner.

You should be able to sketch wavefronts before and after wide and narrow gaps and around obstacles, and predict how the pattern changes if the wavelength or gap width is altered.

When two waves occupy the same point in space, they do not bounce off each other; they pass straight through, and while they overlap the principle of superposition applies: the resultant displacement at any point is the vector sum of the individual displacements.

If two crests arrive together (path difference $n\lambda$, phase difference zero) the waves interfere constructively, giving a larger amplitude. If a crest meets a trough (path difference $(n + \frac{1}{2})\lambda$, phase difference $\pi$ rad) they interfere destructively and can cancel completely if the amplitudes are equal.

For an interference pattern to be stable and observable over time, the interfering waves must be coherent: they must share the exact same frequency and keep a constant phase difference.

Independent light sources, such as two ordinary lamp filaments, fail this test. Their phase relationship changes randomly every few nanoseconds as atoms spontaneously emit light, so the constructive and destructive regions shift far too rapidly to be resolved and the pattern washes out into a uniform, average background intensity. This is why laboratory experiments use a single laser or a single source split in two.

Two loudspeakers driven by the same signal generator give an audible example: walking across the room, you pass through loud points and near-silent points.

You should be able to state the superposition principle, give the path- and phase-difference conditions for each type of interference, and explain why coherence is required.

Young's double-slit experiment is the standard demonstration that light behaves as a wave. A single source illuminates two narrow slits separated by a small distance $d$; the slits act as coherent sources because each wavefront reaches both together.

The diffracted waves overlap and interfere, producing on a screen a distance $D$ away a series of equally spaced bright and dark fringes. Bright fringes mark directions where the path difference from the two slits is a whole number of wavelengths; dark fringes mark half-integer path differences.

Provided $D \gg d$, the fringe spacing is:

$$s = \frac{\lambda D}{d}$$

The formula tells you how the pattern responds to changes. The fringes widen with:

• a longer wavelength (red light)
• a more distant screen
• closer slits

A shorter wavelength (blue light) or wider slit separation squeezes them together. With white light the central fringe is white but the others show coloured edges, because each wavelength has its own spacing.

Typical numbers make the scales clear: with $\lambda \approx 600\ \text{nm}$, $d \approx 0.1\ \text{mm}$ and $D \approx 2\ \text{m}$, the fringes are roughly a centimetre apart and easily visible.

You should be able to use the fringe equation, explain the pattern in terms of path difference, and describe what happens when each variable is changed.

A single slit of finite width $b$ produces its own diffraction pattern: a broad, bright central maximum flanked by much weaker secondary maxima. The first minimum occurs (for small angles) at the angle:

$$\theta = \frac{\lambda}{b}$$

So the central maximum is twice the width of the secondary maxima and narrows as the slit widens or the wavelength shortens. Treating the slit as a row of tiny wave sources explains the minima: at these angles the contributions cancel in pairs.

A diffraction grating takes interference to the extreme by using hundreds of slits per millimetre. Constructive interference then survives only at sharply defined angles given by:

$$d\sin\theta = n\lambda$$

Here $d$ is the spacing between slit centres and $n$ is the order. The many slits make the maxima far sharper and brighter than in a double-slit pattern, and because grating slit spacings are much smaller, the maxima are also far more widely separated, which is why gratings are used in spectrometers to measure wavelengths precisely; white light is spread into a spectrum in each non-zero order.

In realistic double-slit and grating setups the two effects combine. Each slit has a finite width $b$ which causes its own single-slit diffraction, so the overall pattern is the product of the two effects: the sharp interference maxima are modulated in amplitude by the broader single-slit envelope, and bright orders fade out where they coincide with single-slit minima, producing characteristic missing orders.

You should be able to locate single-slit minima and grating maxima, find the highest visible order (set $\sin\theta \le 1$), and sketch the combined intensity pattern.