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Refraction and total internal reflection

Light changes speed when it changes medium, and the bend that follows is completely predictable. Push the geometry far enough in one direction and the light stops leaving at all, which is the trick every optical fibre is built on.

Year 12AQA 3.3.2.3

IN THIS TOPIC

  • Use refractive index as a speed ratio, and Snell's law to predict which way and how far a ray bends.
  • State both conditions for total internal reflection and calculate a critical angle.
  • Describe a step-index optical fibre, the jobs of the cladding, and what causes material and modal dispersion.

WHAT YOU PROBABLY THINK

Total internal reflection happens at any boundary, as long as you hit it steeply enough.

Refractive index

Light travels at c = 3.00 × 108 m s−1 in a vacuum and more slowly in everything else. The refractive index n of a material compares the two speeds:

n = ccsON YOUR DATA SHEET

where cs is the speed of light in the material. Water has n ≈ 1.33 and typical glass n ≈ 1.5, and a larger n means a slower, optically denser material. Air slows light so little that its refractive index is taken as 1.

When light crosses into a new medium its frequency cannot change, because the wavefronts arrive at the boundary at a fixed rate and must leave at the same rate. With c = fλ, a smaller speed therefore means a proportionally smaller wavelength.

Snell's law

A ray crossing a boundary at an angle bends, because one side of each wavefront slows down before the other. All angles are measured between the ray and the normal, the line at right angles to the surface. The bend obeys Snell's law:

n1 sin θ1 = n2 sin θ2ON YOUR DATA SHEET
Refraction at an air-glass boundary: the ray bends towards the normal in the denser mediumnormalair n = 1.00glass n = 1.5048°29.7°
FIG. 1Air to glass at 48°. Snell's law gives a refraction angle of 29.7°: entering the denser medium, the ray bends towards the normal.

The direction of the bend follows from the equation. Going into a higher n, sin θ must shrink, so the ray bends towards the normal; going into a lower n it bends away from the normal. A ray along the normal itself passes straight through, slowed but unbent.

The critical angle and total internal reflection

Now send the light the other way, from glass towards air. It bends away from the normal, so the refracted ray is always at a larger angle than the ray inside the glass. Increase the angle of incidence and the refracted ray leans further and further towards the surface. At one particular incidence, the critical angle θc, the refracted ray runs exactly along the boundary. Setting θ2 = 90° in Snell's law gives

sin θc = n2n1ON YOUR DATA SHEET

valid when n1 > n2. For glass to air, sin θc = 1/1.5, so θc ≈ 42°.

Three rays in glass meeting the boundary below, at and above the critical angleairglass30°41.8°55°refractsat θc: along the boundarytotal internal reflection
FIG. 2The same boundary at three angles of incidence. Below the critical angle the ray escapes; at it, the ray skims the surface; beyond it, the boundary behaves as a perfect mirror.

Beyond the critical angle there is no angle that satisfies Snell's law, and the light cannot leave. All of it reflects back into the glass, obeying the ordinary law of reflection. This is total internal reflection, and it needs both conditions at once: the light must be travelling towards a lower refractive index, and it must meet the boundary beyond the critical angle. Light going from air into glass can never be totally internally reflected, however steep the angle.

Optical fibres

A step-index optical fibre is a thin glass core wrapped in cladding, a layer of glass with a slightly lower refractive index. Light entering the core meets the core-cladding wall beyond the critical angle and reflects, again and again, until it emerges at the far end.

A step-index optical fibre: the ray meets the core wall beyond the critical angle at every bouncecladding n = 1.40core n = 1.5075°
FIG. 3A ray guided along the core. With core n = 1.50 and cladding n = 1.40 the critical angle is 69°, and this ray meets the wall at 75° every time.

The cladding is doing several jobs. It provides the lower refractive index that makes TIR possible, it protects the core surface from scratches that would let light leak out, and it stops light crossing between fibres that are bundled together, which would mix up their signals.

Real signals are pulses, and a pulse can smear out as it travels, an effect called pulse broadening. It has two causes. Modal dispersion: rays bouncing at different angles travel different total distances, so parts of the pulse arrive at different times; making the core very narrow forces every ray onto nearly the same path. Material dispersion: different wavelengths travel at slightly different speeds in glass, so white light spreads out in time; using monochromatic light removes the spread. Broadened pulses can overlap their neighbours, corrupting the information, and absorption in the glass separately weakens the pulse, which is why long lines need repeaters.

THE EXAM BIT

  • Every angle in this topic is measured from the normal. If a question quotes an angle from the surface, subtract it from 90° before touching Snell's law.
  • For a critical angle, the smaller index goes on top: sin θc = n2/n1 with n1 > n2. If your calculator complains that sin θc > 1, the fraction is upside down.
  • State both TIR conditions: into a lower refractive index, and incidence beyond the critical angle. One without the other scores half.
  • “Explain the purpose of the cladding” has three creditable points: it gives the lower n needed for TIR, it protects the core from scratches, and it prevents signal crossover between adjacent fibres.
  • Pulse broadening questions want the cause named and the fix matched to it: narrow core for modal dispersion, monochromatic light for material dispersion.

CHECK YOURSELF

A glass block has a refractive index of 1.50. Calculate the critical angle for light travelling from this glass into air.

Show a hint

Which refractive index belongs on top of the fraction?

Show the answer

Going from glass (n1 = 1.50) into air (n2 = 1.00), the condition n1 > n2 holds, so a critical angle exists.

sin θc = n2/n1 = 1.00 / 1.50 = 0.667, so θc = sin−1(0.667) = 41.8°.

Any ray inside this glass that meets the surface at more than 41.8° to the normal cannot get out; it is totally internally reflected as if the surface were a mirror.

TIR needs both: into lower n,

and past the critical angle.

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