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How do waves bend, spread and reflect at boundaries?

Refraction and Snell's law, refractive index, total internal reflection and the critical angle, diffraction at a single slit, and the diffraction grating equation.

A focused answer to the Edexcel 9PH0 refraction and diffraction content, covering Snell's law, refractive index, total internal reflection and the critical angle, single-slit diffraction, and the diffraction grating equation.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Edexcel wants you to apply Snell's law and refractive index, analyse total internal reflection and the critical angle, describe single-slit diffraction, and use the diffraction grating equation dsin⁡θ=nλd\sin\theta = n\lambda.

The answer

Refraction and Snell's law

Snell's law relates the angles to the normal on each side: n1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2. Light slows and bends towards the normal entering a denser (higher nn) medium, and speeds up and bends away from the normal leaving it. The frequency stays the same; the wavelength changes in proportion to the speed.

Total internal reflection

At the critical angle the refracted ray grazes along the boundary (θ2=90\theta_2 = 90 degrees). Beyond it, no light escapes and the boundary acts as a perfect mirror. This is the basis of optical fibres, which guide light along their length by repeated total internal reflection.

Diffraction

The narrower the slit, the more the central maximum spreads. Single-slit diffraction is direct evidence of the wave nature of light, since particles would simply pass straight through.

The diffraction grating

A diffraction grating has many thousands of fine, equally spaced slits, producing very sharp, bright maxima:

Because the maxima are sharp and well separated, gratings measure wavelengths far more accurately than a double slit, and they are central to spectroscopy.

Examples in context

Optical fibres carry internet and phone data as pulses of light trapped by total internal reflection, with almost no loss. Endoscopes use fibre bundles to see inside the body. Diffraction gratings in spectrometers split light into its component wavelengths to identify chemical elements in stars and laboratory samples. Refraction explains lenses, prisms, mirages, and why a straw looks bent in a glass of water. The colours on a CD or DVD are grating diffraction from its closely spaced tracks.

Try this

Q1. Define the refractive index of a medium. [1 mark]

  • Cue. The ratio of the speed of light in vacuum to its speed in the medium, n=cvn = \frac{c}{v}.

Q2. A glass has refractive index 1.51.5. Find its critical angle in air. [2 marks]

  • Cue. sin⁡θc=1n=11.5=0.667\sin\theta_c = \frac{1}{n} = \frac{1}{1.5} = 0.667, so θc=42\theta_c = 42 degrees.

Q3. Explain why a diffraction grating gives sharper maxima than a double slit. [2 marks]

  • Cue. The many slits reinforce only at the exact maxima angles and cancel elsewhere, producing narrow, bright, well-separated fringes.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20184 marksLight passes from air into glass of refractive index 1.501.50. The angle of incidence is 4040 degrees. Calculate the angle of refraction.
Show worked answer →

Snell's law: n1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2, with n1=1.00n_1 = 1.00 for air.

1.00×sin⁡40=1.50×sin⁡θ21.00 \times \sin 40 = 1.50 \times \sin\theta_2, so sin⁡θ2=0.6431.50=0.428\sin\theta_2 = \frac{0.643}{1.50} = 0.428.

θ2=sin⁡−1(0.428)=25\theta_2 = \sin^{-1}(0.428) = 25 degrees.

Markers reward Snell's law, the correct rearrangement, and the angle about 2525 degrees (bending towards the normal in the denser medium).

Edexcel 20215 marksA diffraction grating has 500500 lines per mm. Monochromatic light of wavelength 600600 nm is shone on it normally. Determine the angle of the first-order maximum and the highest order that can be observed.
Show worked answer →

Slit spacing: d=1500×103=2.0×10−6d = \frac{1}{500 \times 10^{3}} = 2.0 \times 10^{-6} m.

First order (n=1n = 1): dsin⁡θ=nλd\sin\theta = n\lambda, so sin⁡θ=1×600×10−92.0×10−6=0.30\sin\theta = \frac{1 \times 600 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.30, giving θ=17\theta = 17 degrees.

Highest order: sin⁡θ\sin\theta cannot exceed 11, so nmax⁡=dλ=2.0×10−6600×10−9=3.3n_{\max} = \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}}{600 \times 10^{-9}} = 3.3, so the highest complete order is n=3n = 3.

Markers reward dd from lines per mm, the first-order angle about 1717 degrees, and the maximum order n=3n = 3.

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