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Why does light bend at a boundary, and when is it totally internally reflected?

Refraction and refractive index, Snell's law, the critical angle, total internal reflection, and optical fibres.

A focused answer to WJEC A-Level Physics Unit 2 refraction of light, 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. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to define refractive index, apply Snell's law at a boundary, find the critical angle, explain total internal reflection, and describe how optical fibres guide light. Refraction questions are reliable calculation marks, and the optical-fibre application connects the maths directly to the technology that carries modern communications.

The answer

Refractive index

The refraction happens because the wave slows in the denser medium: the part of the wavefront that enters first slows first, so the whole front swings round, changing the direction of travel.

Snell's law

Light entering a denser medium (higher nn) slows and bends towards the normal; entering a less dense medium it bends away from the normal. All angles are measured from the normal, not the surface.

Critical angle and total internal reflection

Optical fibres

An optical fibre has a core of high refractive index surrounded by cladding of lower index. Light hits the core-cladding boundary above the critical angle and is totally internally reflected repeatedly, so it travels along the fibre with little loss, the basis of high-speed communications and endoscopes.

Examples in context

Example 1. Why a straw looks bent in a glass of water
Light from the submerged part of the straw refracts away from the normal as it leaves the water into air, so it reaches your eye from a different direction than the light from the part above the surface. Your brain assumes straight-line travel, so the straw appears to bend at the waterline. Snell's law quantifies the offset.
Example 2. Endoscopes in medicine
A medical endoscope uses a bundle of optical fibres to carry an image out of the body. Light is totally internally reflected along each fibre, so the image survives the twists and turns of the instrument. The same total-internal-reflection physics that limits broadband loss lets doctors see inside a patient without surgery.
Example 3. A mirage on a hot road
On a hot day the air just above the tarmac is much warmer and therefore less dense, giving it a slightly lower refractive index than the cooler air above. Light from the sky bends gradually as it passes through this gradient and, near grazing incidence, is effectively totally internally reflected upward into your eye. Your brain interprets the shimmering reflected sky as a pool of water on the road. The effect is a continuous version of the same refraction and total-internal-reflection physics described above.

Try this

Q1. Light passes from glass (n=1.5n = 1.5) into air. Find the critical angle. [2 marks]

  • Cue. sin⁡θc=1.01.5=0.667\sin\theta_c = \frac{1.0}{1.5} = 0.667, so θc≈41.8∘\theta_c \approx 41.8^\circ.

Q2. State what happens to light meeting a boundary at an angle greater than the critical angle. [1 mark]

  • Cue. It is totally internally reflected.

Exam-style practice questions

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

WJEC 20195 marksA ray of light travels from water (n=1.33n = 1.33) into a glass block (n=1.52n = 1.52), striking the boundary at an angle of incidence of 40∘40^{\circ}. Calculate the angle of refraction, and determine 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 →

Apply Snell's law n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2 at the boundary.

1.33sin⁡40∘=1.52sin⁡θ21.33 \sin 40^{\circ} = 1.52 \sin\theta_2, so sin⁡θ2=1.33×0.6431.52=0.562\sin\theta_2 = \dfrac{1.33 \times 0.643}{1.52} = 0.562.

θ2=sin⁡−1(0.562)=34.2∘\theta_2 = \sin^{-1}(0.562) = 34.2^{\circ}. The ray bends toward the normal because it enters a denser medium.

Speed of light in the glass from n=c/vn = 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 the correct Snell's law substitution, the smaller refraction angle, and the speed from v=c/nv = c/n.

WJEC 20214 marksAn optical fibre has a core of refractive index 1.501.50 and cladding of refractive index 1.451.45. Calculate the critical angle at the core-cladding boundary and explain how total internal reflection keeps light within the core.
Show worked answer →

The critical angle for light going from the denser core to the less dense cladding is given by sin⁡θc=ncladdingncore\sin\theta_c = \dfrac{n_{\text{cladding}}}{n_{\text{core}}}.

sin⁡θc=1.451.50=0.967\sin\theta_c = \dfrac{1.45}{1.50} = 0.967, so θc=sin⁡−1(0.967)=75.2∘\theta_c = \sin^{-1}(0.967) = 75.2^{\circ}.

Light rays travelling nearly along the fibre meet the core-cladding boundary at angles of incidence greater than this critical angle. At any angle above the critical angle all the light is reflected back into the core (total internal reflection), so the ray bounces repeatedly along the fibre with negligible loss into the cladding. Markers reward the critical angle calculation and linking incidence above the critical angle to total internal reflection guiding the light.

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