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C.3

Wave phenomena

Theme C · Wave behaviour · SL and HL, with additional HL material marked

🎯What you need to be able to do

  • Apply the laws of reflection and refraction, including Snell’s law and refractive index.
  • Find the critical angle and describe total internal reflection.
  • Describe diffraction and the conditions under which it is noticeable.
  • Apply the principle of superposition, and use path difference to predict interference.
  • Use the double-slit formula.
  • HLSingle-slit diffraction, diffraction gratings, resolution, and polarisation with Malus’s law.

📚The physics

Refraction happens because the wave changes speed. Snell’s law follows from that, with the refractive index \( n = c/v \) telling you how much a medium slows light.

\[ n_1\sin\theta_1 = n_2\sin\theta_2 \]

All angles are measured from the normal, never from the surface — a convention that costs marks every year. Going into a denser medium the wave slows and bends towards the normal; coming out it speeds up and bends away.

Total internal reflection is what happens when “bending away” runs out of room. Beyond the critical angle, given by \( \sin\theta_c = n_2/n_1 \), no refracted ray can exist and all the light reflects. It requires travel from denser to less dense — there is no critical angle the other way. This is the physics of optical fibres and of the sparkle of a cut diamond, whose high refractive index gives it a critical angle of only about 24°.

Diffraction is the spreading of a wave as it passes an obstacle or aperture. It becomes noticeable when the gap is comparable to the wavelength — which is why you can hear someone around a corner but not see them. Sound has wavelengths of order a metre and diffracts round doorways easily; visible light, at a fraction of a micrometre, effectively does not.

Superposition is the rule for what happens when waves meet: displacements add, with sign. Where they arrive in phase you get constructive interference; where they arrive antiphase, destructive. For two coherent sources, that reduces to a statement about path difference: constructive when the path difference is a whole number of wavelengths, \( n\lambda \); destructive when it is an odd number of half-wavelengths, \( (n + \tfrac{1}{2})\lambda \). Almost every interference question is this one idea in a costume.

Coherence matters: the sources must maintain a constant phase relationship, which in practice means deriving them from a single source. Two independent lamps produce no visible fringes.

Young’s double slit gives evenly spaced fringes with separation

\[ s = \frac{\lambda D}{d} \]

where \(d\) is the slit separation and \(D\) the distance to the screen. Note that a smaller slit separation gives wider fringes — the relationship is inverse, and it is worth checking your intuition against that.

HLSingle-slit diffraction

A single slit of width \(b\) produces its first minimum at \( \theta = \lambda/b \). This envelope modulates the double-slit pattern: the fringes you see are double-slit interference sitting underneath a single-slit diffraction envelope, which is why the central fringes are brighter than the outer ones.

HLDiffraction gratings

With many slits, \( n\lambda = d\sin\theta \), and the maxima become very sharp because light from thousands of slits must all agree. That sharpness is what makes gratings the instrument of choice for spectroscopy.

HLResolution

The Rayleigh criterion says two sources are just resolved when the central maximum of one falls on the first minimum of the other, at \( \theta = 1.22\lambda/b \) for a circular aperture. Bigger apertures and shorter wavelengths resolve better — the reason telescopes are built large and electron microscopes beat optical ones.

HLPolarisation

Polarisation restricts the oscillation to one plane, and is possible only for transverse waves. Through a polariser, Malus’s law gives \( I = I_0\cos^{2}\theta \). Two crossed polarisers at 90° block everything — and yet inserting a third at 45° between them lets light through again, which is a genuinely surprising result worth thinking about.

✏️Worked example

Light of wavelength 590 nm falls on two slits 0.25 mm apart. Fringes are observed on a screen 1.8 m away.

(a) Fringe separation.

\[ s = \frac{\lambda D}{d} = \frac{590 \times 10^{-9} \times 1.8}{0.25 \times 10^{-3}} = 4.25 \times 10^{-3}\ \text{m} = 4.2\ \text{mm} \]

(b) The apparatus is submerged in water, \( n = 1.33 \). What happens to the fringes? The frequency is fixed by the source, but the speed falls, so the wavelength in water is \( 590/1.33 = 444 \) nm. The fringe separation becomes \( 4.25 \times (1/1.33) = 3.2 \) mm — the pattern contracts.

(c) The screen is moved to 2.5 m. New separation? \(s\) scales with \(D\): \( 3.2 \times (2.5/1.8) = 4.4 \) mm.

What to notice. Part (b) is the \( c = f\lambda \) idea from C.2 doing real work. Students who assume the frequency changes in water get a wavelength of 785 nm and fringes that spread instead of contracting — the opposite of what happens.

🔭See it happen

PhET, Wave Interference. Switch to two slits, then vary the slit separation and watch the fringes move the “wrong” way — closer slits, wider fringes. Seeing the inverse relationship is worth more than memorising the formula.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Light travelling in air strikes a glass block of refractive index 1.52 at 40° to the normal. Find the angle of refraction.
\( \sin\theta_2 = \dfrac{\sin 40^\circ}{1.52} = \dfrac{0.643}{1.52} = 0.423 \), so \( \theta_2 = 25^\circ \). It bends towards the normal on entering the denser medium.
2. Find the critical angle for light passing from that glass (\( n = 1.52 \)) into air.
\( \sin\theta_c = n_2/n_1 = 1/1.52 = 0.658 \), so \( \theta_c = 41^\circ \). There is no critical angle going the other way — it requires travel from denser to less dense.
3. Light of wavelength 633 nm falls on two slits 0.30 mm apart, with a screen 2.0 m away. Find the fringe separation.
\( s = \dfrac{\lambda D}{d} = \dfrac{633 \times 10^{-9} \times 2.0}{0.30 \times 10^{-3}} = 4.2 \times 10^{-3} \) m, or 4.2 mm.
4. HLA diffraction grating has 600 lines per mm. Find the angle of the first-order maximum for light of wavelength 589 nm.
The slit spacing is \( d = 1/(600 \times 10^{3}) = 1.67 \times 10^{-6} \) m. From \( n\lambda = d\sin\theta \) with \( n = 1 \): \( \sin\theta = \dfrac{589 \times 10^{-9}}{1.67 \times 10^{-6}} = 0.353 \), so \( \theta = 21^\circ \).
5. HLUnpolarised light passes through a polariser and then a second polariser at 30° to the first. What fraction of the light emerging from the first polariser gets through the second?
Malus’s law: \( I = I_0\cos^{2}\theta = I_0\cos^{2}30^\circ = 0.75 I_0 \), so 75% passes. At 90° it would be zero.
6. HLA telescope has an aperture of 0.10 m and observes at 550 nm. Find the smallest angular separation it can resolve.
Rayleigh criterion: \( \theta = \dfrac{1.22\lambda}{b} = \dfrac{1.22 \times 550 \times 10^{-9}}{0.10} = 6.7 \times 10^{-6} \) rad. A larger aperture or a shorter wavelength resolves finer detail.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Refraction and the Ray Model of Light
  • The Physics Hypertextbook — interference, diffraction and polarisation