C.3 Wave phenomena

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.

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.

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.

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.

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.

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.

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.

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.

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.

A boundary can split wave energy

When a travelling wave reaches a boundary, part of its energy may be reflected back into the first medium and part may be transmitted into the second. The transmitted wave is refracted when its speed changes and it crosses the boundary at a non-zero angle.

Read rays and wavefronts together

Angles are measured from the normal. Rays show energy-transfer direction and remain perpendicular to wavefronts. Frequency stays fixed at a stationary boundary; a change in speed therefore changes wavelength and wavefront spacing.

Diffraction is wave spreading

A wave diffracts when its wavefront bends into the region behind an obstacle or spreads after passing through an aperture. The wave remains in the same medium, so diffraction itself does not require a change of speed or frequency.

Compare wavelength with the opening or body

Spreading is most noticeable when the aperture width or obstacle size is comparable to the wavelength. An aperture much wider than the wavelength gives a broad central region that travels nearly straight with limited edge spreading; narrowing the aperture increases the angular spread.

Wavefront–ray representation

Before a straight aperture, incident wavefronts are parallel. Beyond a narrow aperture, draw curved outgoing wavefronts; rays stay perpendicular to them and fan outward. Around a body, wavefronts curve into the geometrical shadow.

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.

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.

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 $2\lambda_1=3\lambda_2$.