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Why does light bend when it changes medium, and when is it totally internally reflected?

Refraction of light: the refractive index and Snell's law, the change of speed and wavelength, total internal reflection and the critical angle, and optical fibres.

A focused answer to the Eduqas A-Level Physics Component 3 refraction content, covering the refractive index and Snell's law, the change of speed and wavelength on refraction, total internal reflection and the critical angle, and the operation of optical fibres.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

Eduqas wants you to define the refractive index, apply Snell's law, explain how the speed and wavelength change on refraction while the frequency stays the same, define the critical angle and total internal reflection, and describe how an optical fibre guides light.

The answer

The refractive index and Snell's law

Speed, wavelength and frequency

Total internal reflection and the critical angle

Optical fibres

Examples in context

Refraction explains lenses, prisms, the apparent bending of a straw in water and the dispersion of white light into a spectrum. Total internal reflection makes diamonds sparkle (their very high index gives a small critical angle, trapping light), and is the working principle of optical fibres that carry the world's internet traffic and of the reflecting prisms in binoculars and periscopes. Medical endoscopes use fibre bundles to see inside the body.

Try this

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

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

Q2. Light passes from air into a medium of refractive index 1.331.33 at an angle of incidence of 30∘30^\circ. Find the angle of refraction. [2 marks]

  • Cue. sin⁡θ2=sin⁡30∘1.33=0.376\sin\theta_2 = \frac{\sin 30^\circ}{1.33} = 0.376, so θ2=22∘\theta_2 = 22^\circ.

Q3. State the condition for total internal reflection to occur. [2 marks]

  • Cue. The light must travel from a denser to a less dense medium and meet the boundary at an angle greater than the critical angle.

Exam-style practice questions

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

Eduqas 20184 marksA ray of light passes from air into glass of refractive index 1.521.52, striking the surface at an angle of incidence of 40∘40^\circ. Calculate the angle of refraction and the speed of light in the glass. Take c=3.0×108 m s−1c = 3.0 \times 10^{8}\ \text{m s}^{-1}.
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 (air): 1.00sin⁡40∘=1.52sin⁡θ21.00\sin 40^\circ = 1.52\sin\theta_2.

sin⁡θ2=sin⁡40∘1.52=0.6431.52=0.423\sin\theta_2 = \dfrac{\sin 40^\circ}{1.52} = \dfrac{0.643}{1.52} = 0.423, so θ2=sin⁡−1(0.423)=25∘\theta_2 = \sin^{-1}(0.423) = 25^\circ.

Speed in glass from n=cvn = \dfrac{c}{v}: v=cn=3.0×1081.52=1.97×108 m s−1v = \dfrac{c}{n} = \dfrac{3.0 \times 10^{8}}{1.52} = 1.97 \times 10^{8}\ \text{m s}^{-1}.

Markers reward Snell's law giving θ2≈25∘\theta_2 \approx 25^\circ and the speed about 2.0×108 m s−12.0 \times 10^{8}\ \text{m s}^{-1}.

Eduqas 20214 marksThe core of an optical fibre has refractive index 1.501.50 and the cladding has refractive index 1.461.46. Calculate the critical angle at the core-cladding boundary and explain how this allows the fibre to guide light.
Show worked answer →

The critical angle at a boundary between a denser and a less dense medium is given by sin⁡θc=n2n1\sin\theta_c = \dfrac{n_2}{n_1}, where n1n_1 is the core and n2n_2 the cladding.

sin⁡θc=1.461.50=0.973\sin\theta_c = \dfrac{1.46}{1.50} = 0.973, so θc=sin⁡−1(0.973)=76.7∘\theta_c = \sin^{-1}(0.973) = 76.7^\circ.

Light striking the core-cladding boundary at an angle greater than 76.7∘76.7^\circ is totally internally reflected, so it stays within the core and is guided along the fibre by repeated total internal reflection. Markers reward sin⁡θc=n2n1\sin\theta_c = \frac{n_2}{n_1}, the critical angle about 77∘77^\circ, and explaining the guiding by total internal reflection.

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