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EnglandPhysicsSyllabus dot point

Why do waves change direction when they cross a boundary, and when does light undergo total internal reflection?

Refraction and the refractive index of a substance, Snell's law at a boundary, the critical angle, total internal reflection, and the operation of optical fibres.

A focused answer to AQA A-Level Physics 3.3.2.4, covering refraction and refractive index, Snell's law, the critical angle, total internal reflection, and how optical fibres guide light.

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  1. What this dot point is asking
  2. Refraction and refractive index
  3. Snell's law
  4. The critical angle and total internal reflection
  5. Optical fibres
  6. Try this

What this dot point is asking

AQA specification point 3.3.2.4 wants you to define refractive index, apply Snell's law at a boundary, calculate the critical angle, explain total internal reflection, and describe how optical fibres use it to guide light.

Refraction and refractive index

The frequency of the wave stays the same on crossing the boundary, but the speed changes, so the wavelength changes too, and it is this change in speed that bends the ray.

Snell's law

Light slowing down (entering a denser medium) bends towards the normal; light speeding up (entering a less dense medium) bends away from the normal. A useful special case is light passing from a vacuum or air (n1=1n_1 = 1) into a medium, where Snell's law simplifies to sin⁡θ1=n2sin⁡θ2\sin\theta_1 = n_2 \sin\theta_2. Snell's law also explains everyday observations such as a straw appearing bent at the surface of a glass of water and the apparent shallowing of a swimming pool, both caused by the rays bending at the water-air boundary.

The critical angle and total internal reflection

Optical fibres

An optical fibre has a high-refractive-index core surrounded by a lower-index cladding. Light entering at a shallow enough angle repeatedly undergoes total internal reflection at the core-cladding boundary, so it is guided along the fibre with little loss over long distances. The cladding protects the surface, keeps the critical angle well defined, and reduces signal loss and crossover between adjacent fibres. This is the basis of high-speed communications and medical endoscopes.

Try this

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

  • Cue. The ratio of the speed of light in a vacuum to its speed in the substance, n=ccsn = \dfrac{c}{c_s}.

Q2. Explain why light is totally internally reflected in an optical fibre. [3 marks]

  • Cue. Light hits the core-cladding boundary at an angle greater than the critical angle, so no light refracts out and all of it reflects back into the core.

Q3. State which way a ray bends when it enters a denser medium. [1 mark]

  • Cue. Towards the normal.

Exam-style practice questions

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

AQA 20184 marksA ray of light travels from water (refractive index 1.331.33) into glass (refractive index 1.501.50), striking the boundary at an angle of incidence of 40∘40^{\circ} to the normal. Calculate the angle of refraction in the glass.
Show worked answer →

Apply Snell's law n1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2, with n1=1.33n_1 = 1.33, θ1=40∘\theta_1 = 40^{\circ} and n2=1.50n_2 = 1.50.

sin⁡θ2=n1sin⁡θ1n2=1.33×sin⁡40∘1.50=1.33×0.6431.50=0.570\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{1.33 \times \sin 40^{\circ}}{1.50} = \dfrac{1.33 \times 0.643}{1.50} = 0.570.

θ2=sin⁡−1(0.570)=34.8∘\theta_2 = \sin^{-1}(0.570) = 34.8^{\circ}, smaller than the incident angle because light bends towards the normal entering the denser glass.

Markers reward correct use of Snell's law, identifying which index goes with which angle, and the ray bending towards the normal.

AQA 20214 marksExplain how light is transmitted along an optical fibre, referring to total internal reflection and the role of the cladding.
Show worked answer →

The fibre has a core of high refractive index surrounded by a cladding of lower refractive index. Light enters the core at a shallow angle and strikes the core-cladding boundary at an angle greater than the critical angle, so it undergoes total internal reflection rather than refracting out.

The light reflects repeatedly along the fibre, staying within the core with little loss. The cladding keeps the critical angle well defined, protects the core surface, and prevents light crossing between adjacent fibres, which would blur the signal.

Markers reward total internal reflection at the core-cladding boundary, the angle exceeding the critical angle, and a correct role for the cladding.

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