IB Physics HL C.3 Wave phenomena Notes

Use wavefronts and rays as two linked representations of the same travelling wave. The wavefront marks equal phase; the ray points in the propagation direction and must meet the wavefront at 90 degrees. When a wave enters a slower medium, frequency remains set by the source, so wavelength and wavefront spacing decrease. A correct IB refraction sketch therefore shows the refracted ray bent toward the normal and the refracted wavefronts closer together.

A plane wave enters a slower medium at an angle. Describe how to sketch the transmitted ray and transmitted wavefronts.

Snell’s law is a direction rule for refraction. First label the incident medium as 1 and the refracting medium as 2, then measure each angle from the normal. If the second medium has a larger refractive index, light travels more slowly there and the refracted ray is closer to the normal. The wave frequency does not change at the boundary; the wavelength changes because the wave speed changes.

Light passes from air into glass. Explain why the ray bends toward the normal and write the equation needed to calculate the refracted angle.

Refractive index is a speed ratio, not a new unit. In vacuum light travels at c, so any material with n greater than 1 gives a lower speed v = c/n. When light enters the material, the source still drives the same frequency. Since v = f lambda, the wavelength decreases when the speed decreases. This is the physical reason a ray bends toward the normal when it enters a higher-index medium.

A transparent material has refractive index 1.60. Calculate the speed of light in the material and state what happens to wavelength when light enters it from air.

TIR is a limiting case of refraction. As light travels from a higher refractive index medium into a lower refractive index medium, the refracted ray bends away from the normal. Increasing the incident angle eventually makes the refracted angle 90 degrees; this defines the critical angle. Any larger incident angle gives total internal reflection, so no refracted ray carries energy away into the second medium in the ray model.

Light travels from glass, n = 1.50, into air. Calculate the critical angle and state what happens for an incident angle of 50 degrees.

Superposition is a local rule. At one instant and one position, read each wave displacement from equilibrium and add the signed values. The rule explains both pulse diagrams and later interference patterns: maxima occur where waves arrive in phase and minima occur where opposite displacements reduce the resultant. The waves are not used up by the overlap in the ideal model.

Two pulses overlap at a point with displacements +4.0 cm and -1.5 cm. State the resultant displacement and identify whether the overlap is constructive or destructive.

Coherence is the condition that makes interference observable as a stable pattern. If the phase difference between two sources drifts, a point on the screen alternates between constructive and destructive conditions and the pattern washes out over time. IB answers should state both parts of the condition: same frequency and constant phase difference.

State the condition for two sources to be coherent and explain why coherence is needed in a double-source interference experiment.

Path difference turns geometry into interference. If one wave travels an extra whole number of wavelengths, crests still meet crests and the result is constructive. If one wave travels an extra half-integer number of wavelengths, crests meet troughs and the result is destructive. In diagrams, measure both paths to the same observation point before subtracting.

Two coherent waves reach a point with path lengths 4.20 m and 3.90 m. The wavelength is 0.20 m. Determine the type of interference.

Two-source interference combines the previous three ideas. Coherence keeps the relative phase fixed, path difference tells whether a point receives waves in phase or antiphase, and superposition gives the resultant displacement. The visible pattern is a map of maxima and minima, not a set of new waves travelling along the nodal lines.

Describe the conditions needed to observe a stable two-source interference pattern and explain why the central line is bright for two in-phase sources.

Young’s experiment converts two-source interference into a measurable pattern. The two slits act as coherent secondary sources. Bright fringes occur where path difference is a whole number of wavelengths, and the small-angle geometry gives an approximately constant fringe spacing s = lambda D / d. Use the formula only after identifying which distance is on the screen and which is between the slits.

In a Young double-slit experiment, lambda = 6.0 × 10^-7 m, D = 1.8 m, and d = 0.30 mm. Calculate the fringe spacing and state what happens if d is doubled.

Diffraction is a wavefront effect. When a wave meets an edge or passes through a gap, the wavefront spreads into the region beyond the edge. The comparison that matters is wavelength against the size of the aperture or obstacle. If the gap is much larger than the wavelength, spreading is small. If the gap is about one wavelength wide, the outgoing wavefronts spread strongly.

Plane water waves pass through a gap. Sketch and explain the difference between the pattern when the gap is about one wavelength wide and when the gap is much wider than the wavelength.

In an HL double-slit graph, there are two scales to read. The repeating bright and dark fringes are the two-slit interference pattern. Their separation depends on lambda D / d, so larger slit separation gives closer fringes. The slow rise and fall of the fringe heights is the diffraction envelope from the finite width of each slit. A narrower slit produces a wider envelope, so more interference fringes fit under the central envelope.

Two double-slit intensity graphs use the same wavelength and screen distance. In graph B the fringes are closer together, but the broad envelope has the same width. Explain what changed in the apparatus.

Single-slit diffraction is not just “light spreading”; HL expects the intensity pattern. The strongest feature is the broad central maximum. The first dark minima occur where the path difference across the slit produces cancellation, summarized by sin theta = lambda/a for m = 1. If a decreases, lambda/a increases, so the first minima move farther from the centre and the central maximum widens.

Monochromatic light of wavelength 5.0 × 10^-7 m passes through a single slit of width 2.0 × 10^-5 m. Estimate the angle to the first minimum and state how the pattern changes if the slit is narrowed.

A double-slit pattern has two layers. The separation between the two slits controls the closely spaced interference maxima. The finite width of each slit controls the broad diffraction envelope that changes the heights of those maxima. Where the envelope is zero, a possible interference maximum is missing because the single-slit diffraction from each slit contributes no intensity at that angle.

A double-slit intensity graph shows regularly spaced bright fringes, but one expected bright fringe is absent at a position where the broad envelope is zero. Explain this observation.

A grating is a many-source interference system. For a bright principal maximum, light from neighbouring slits must arrive in phase, so their path difference d sin theta must equal an integer number of wavelengths. More slits narrow the maxima because off-condition angles cancel more strongly. Before calculating an angle, convert line density into slit spacing if necessary and check that the requested order is possible.

A grating has 500 lines per mm and is illuminated with light of wavelength 600 nm. Calculate the angle of the first-order maximum and determine whether a fourth order is possible.

Core C.3 is secure when the student can move from a wave diagram to the correct physical rule. Wavefronts and rays set the geometry. Refraction changes speed and wavelength but not frequency. TIR requires a direction condition before a critical-angle calculation. Diffraction depends on size compared with wavelength. Interference then builds from superposition, coherence, and path difference, ending with Young’s measurable fringe spacing.

HL C.3 is mostly about choosing the right scale. Single-slit diffraction uses slit width a to find minima and envelope width. Double-slit interference uses slit separation d to find fine fringe spacing, while the finite slit width creates the broad envelope. A grating uses adjacent-line spacing d and integer order n to locate sharp principal maxima. The same symbols can appear in different models, so the diagram or graph feature must be identified before choosing an equation.