C.3 Wave phenomena

Define a wavefront

A wavefront is a line or surface joining points that are in phase. Adjacent wavefronts are separated by one wavelength. In a uniform medium, the wavefronts are perpendicular to the direction of propagation.

Use rays to show propagation

A ray is a line showing the direction in which the wave transfers energy. Draw rays perpendicular to the local wavefronts; straight, parallel wavefronts give parallel rays, while circular wavefronts from a point source give radial rays.

Read the geometry

When a wavefront diagram changes direction at a boundary, compare the ray direction and the spacing of wavefronts on each side. The wavefront construction helps distinguish a change in speed from a change in frequency.

Common trap

Do not draw rays parallel to wavefronts. A ray follows energy propagation and is normal to the wavefront at each point.

Write the boundary relationship

For a ray crossing from medium 1 to medium 2, Snell’s law is $n_1\sin\theta_1=n_2\sin\theta_2$. Each angle is measured between the ray and the normal, not between the ray and the surface.

Solve the geometry first

Draw or identify the normal at the boundary, label incident and refracted angles, then substitute the refractive indices and sines. If the question asks for speed, combine the result with $n=c/v$.

Check the bend

Entering a higher-index medium decreases speed and bends the ray toward the normal. Entering a lower-index medium increases speed and bends it away from the normal; the frequency remains fixed at a stationary boundary.

Common trap

Do not use the angle between a wavefront and the normal as if it were the ray angle. A wavefront is perpendicular to the ray, so convert the angle when the diagram labels wavefronts.

Define refractive index

The refractive index of a medium is $n=c/v$, where $c$ is the speed of light in vacuum and $v$ is its speed in the medium. A larger $n$ means a lower light speed in that medium.

Relate speed and wavelength

At a stationary boundary the frequency is unchanged. Since $v=f\lambda$, a lower speed means a shorter wavelength. For two media, $n_2/n_1=\lambda_1/\lambda_2$ when the frequency is common.

Use a graph or measurement

If a graph’s gradient represents $n$, state that interpretation before reading the value. For uncertainty, use the spread from suitable maximum and minimum lines or the specified uncertainty method.

Common trap

Do not use $n=v/c$, and do not assume wavelength stays fixed when light enters a different medium. Frequency is the quantity that remains fixed at a stationary boundary.

Check the two conditions

Total internal reflection can occur only when a wave travels from a higher-index medium to a lower-index medium, and the incidence angle is greater than the critical angle. At the critical angle, the refracted ray travels along the boundary: $\theta_2=90^\circ$.

Calculate the critical angle

From Snell’s law, $\sin\theta_c=n_2/n_1$ for $n_1>n_2$. For a dense medium to air, $n_2\approx1$, so $\sin\theta_c=1/n_1$.

Use the boundary picture

For incidence below $\theta_c$, there is a refracted ray. At $\theta_c$, it grazes the boundary. Above $\theta_c$, no refracted ray propagates into the lower-index medium and all the light is reflected back into the denser medium.

Common trap

Do not use the critical-angle equation when light travels from lower to higher refractive index, and do not measure the critical angle from the surface rather than the normal.

Add overlapping displacements

When waves overlap, the resultant displacement at a point is the algebraic sum of the individual displacements: $y_{\mathrm{resultant}}=y_1+y_2$. The waves then continue propagating after the overlap.

Keep the signs

Displacements on the same side of equilibrium add; opposite displacements partially or completely cancel. Equal opposite pulses can produce zero displacement at an instant without destroying either wave.

Connect to interference

Repeated superposition of coherent waves can create stable maxima and minima. A diffraction pattern extending beyond a geometrical shadow is evidence that wave overlap and interference are involved.

Common trap

Do not add amplitudes as positive magnitudes only, and do not treat destructive interference as permanent disappearance of the waves.

Define coherence

Two waves are coherent if they have the same frequency and a constant phase difference. The phase relationship does not drift with time.

Connect coherence to a pattern

When coherent waves overlap, the locations of constructive and destructive interference remain fixed, producing a stable interference pattern. An ordinary pair of independent light sources usually has a changing phase relationship and does not produce a stable pattern.

Use one source when needed

A single source split into two paths can provide a common frequency and phase relationship. The resulting secondary sources can then act coherently for a double-source interference experiment.

Common trap

Same frequency alone is not enough. The phase difference must also remain constant; otherwise bright and dark locations move or wash out over time.

Define path difference

Path difference is the difference between the distances travelled by two waves from their sources to the same observation point. For in-phase coherent sources, it determines whether the waves arrive in phase or out of phase.

Apply the conditions

Constructive interference occurs when path difference $=n\lambda$. Destructive interference occurs when path difference $=(n+\tfrac12)\lambda$, where $n$ is a whole number.

Count fringes carefully

Start from the central bright fringe when the path difference is zero. Each additional bright fringe changes the path difference by $\lambda$; dark fringes lie halfway between adjacent bright conditions.

Common trap

Do not assign $\lambda/2$ to every dark point. The first dark condition is $\lambda/2$, then $3\lambda/2$, $5\lambda/2$, and so on.

Set up two-source interference

Two coherent sources emit waves with the same frequency and a constant phase difference. At each observation point, compare the two source-to-point distances to find the path difference.

Map bright and dark regions

For in-phase sources, path difference $n\lambda$ gives constructive interference and a bright or high-amplitude region. Path difference $(n+\tfrac12)\lambda$ gives destructive interference and a dark or low-amplitude region.

Read the pattern

Points equidistant from the two sources have zero path difference and form a central constructive line. Further maxima and minima occur where the path difference changes by half-wavelength steps.

Common trap

Do not use source-to-source separation as the path difference. It is the difference between the two travel distances to the same observation point.

Relate fringe spacing to the apparatus

For Young’s double-slit interference, fringe separation is $s=\lambda D/d$, where $\lambda$ is wavelength, $D$ is slit-to-screen distance and $d$ is slit separation.

Rearrange before substituting

Use $\lambda=sd/D$, $d=\lambda D/s$, or $D=sd/\lambda$ as needed. Measure the separation between adjacent bright or dark fringe centres; if several fringes are measured, divide the total width by the number of intervals.

Check the trends

Fringes spread farther apart when wavelength or screen distance increases, and become closer when slit separation increases. The small-angle model assumes $D\gg d$ and approximately plane wavefronts normal to the slits.

Common trap

Do not use the total width across several fringes as s without dividing by the number of fringe spacings, and do not confuse slit separation d with screen distance D.

Compare the reflected rays

In thin-film interference, light reflects from the top and bottom surfaces of a film. The two reflected rays have a path difference from the extra distance travelled inside the film and may also acquire a phase change on reflection.

Account for phase changes

Determine whether each reflection introduces a phase reversal, then combine that phase difference with the optical path difference. For normal incidence, the geometric contribution is approximately $2nt$, where $n$ is film refractive index and $t$ is thickness.

Choose the correct interference condition

Use the phase-shift situation stated in the problem to decide whether the total condition for constructive reflection is an integer or half-integer number of wavelengths. For an air–film–air setup with one phase reversal, the minimum constructive thickness is $t=\lambda/(4n)$.

Common trap

Do not apply a single formula without checking the refractive indices at both reflecting surfaces. A phase reversal can swap the bright and dark conditions.

Understand the diffraction limit

A finite aperture produces a diffraction pattern rather than a perfect point image. Two nearby sources are resolved only when their diffraction patterns are sufficiently separated. Resolution is therefore limited by wavelength and aperture size.

State Rayleigh’s criterion

Two point sources are just resolved when the central maximum of one diffraction pattern lies on the first minimum of the other pattern. At smaller angular separation the patterns overlap too strongly to distinguish the sources.

Improve resolution

Use a shorter wavelength or a larger aperture. For visible light, violet light has a shorter wavelength than red light and can resolve closer sources under the same aperture conditions.

Common trap

Do not say that the two central maxima must coincide. Rayleigh’s limit is defined by one central maximum aligning with the other pattern’s first minimum.

Recognize the pattern

A monochromatic wave passing through a narrow rectangular slit spreads and forms a broad central maximum with weaker side maxima separated by minima. The pattern results from interference between contributions across the slit.

Use the first-minimum condition

For slit width $b$, the first minimum satisfies $\theta\approx\lambda/b$ for small angles. On a screen a distance $x$ away, the central maximum width is approximately $2x\lambda/b$.

Check the trends

A narrower slit or longer wavelength produces greater angular spreading and a wider central maximum. A wider slit or shorter wavelength produces a narrower pattern. The syllabus treatment is monochromatic light and rectangular slits at normal incidence.

Common trap

Do not confuse the distance from the central maximum to the first minimum with the full central-maximum width; the latter is twice the first-minimum distance on the screen.

Separate the two patterns

In a multiple-slit intensity pattern, the fine interference maxima are contained within a broad single-slit diffraction envelope. The envelope sets the overall intensity scale; the double-slit or multiple-slit interference determines the rapid fringe structure.

Relate the widths

The single-slit envelope depends on slit width $b$, while the separation of fine interference maxima depends on source or slit separation $d$. A smaller $b$ makes the envelope wider; a larger $d$ makes interference fringes closer together.

Explain missing or unequal maxima

Interference maxima at different angles can have different intensities because the envelope changes across the screen. Some interference maxima may fall at an envelope minimum and disappear.

Common trap

Do not treat every interference maximum as having the same height, and do not confuse the fine fringe spacing with the width of the single-slit envelope.

Core idea

A diffraction grating has many equally spaced parallel slits. Bright principal maxima occur when the path difference between adjacent slits is an integer number of wavelengths:

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

Here, $n$ is the order number $0,1,2,\ldots$, $\lambda$ is the wavelength, $d$ is the spacing between adjacent slits, and $\theta$ is measured from the central maximum to the chosen maximum.

Build the model

If a grating has $N$ lines per metre, the slit spacing is $d=1/N$. For a selected maximum, identify its order $n$, convert $d$ and $\lambda$ to consistent units, and solve for the unknown angle, wavelength or spacing. The central maximum is $n=0$; the first maxima on either side are $n=1$.

Check the allowed orders

Because $\lvert\sin\theta\rvert\le 1$, a wavelength can only produce orders satisfying $n\lambda\le d$. The largest possible order is therefore the greatest integer not exceeding $d/\lambda$. For overlapping wavelengths, equate their path-difference conditions: if the second-order maximum of $\lambda_1$ coincides with the third-order maximum of $\lambda_2$, then $2\lambda_1=3\lambda_2$.

Common trap

Do not use the number of lines per metre as $d$; invert it first. Do not count the central maximum as first order, and do not replace the grating equation with the small-angle approximation unless the question explicitly permits that approximation.