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Refraction and total internal reflection

Light changes speed when it crosses into a new medium, and Snell's law relates the change of direction to the two refractive indices. Beyond the critical angle it does not cross at all but reflects back inside, which is how an optical fibre carries light along its length.

IN THIS TOPIC

  • Use refractive index as a ratio of speeds, and say what happens to frequency and to wavelength at a boundary.
  • Apply Snell's law, predicting which way the ray bends before calculating how far.
  • Measure the refractive index of a solid by tracing a ray through a block and taking the gradient of sin θ₁ against sin θ₂.
  • State both conditions for total internal reflection, and calculate a critical angle.
  • Describe a step-index optical fibre and the three jobs the cladding does.
  • Separate modal from material dispersion, and match each one to its fix.

COMMON MISCONCEPTION

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

Refractive index

Light travels at c=3.00×108c = 3.00 \times 10^8 m s−1 in a vacuum and more slowly in everything else. The refractive index n of a material compares the two speeds.

n=ccsn = \frac{c}{c_s}ON THE AQA DATA SHEET

where csc_s 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=fλc = f\lambda, a smaller speed therefore means a proportionally smaller wavelength.

WORKED EXAMPLE

Light slowed down, measurably

Water has a refractive index of 1.33. Find the speed of light in water.

n = c/v is a ratio of two speeds and nothing more mysterious, so rearrange it for v.

v = c/n = 3.0 × 108/1.33 = 2.3 × 108 m s−1.

Sense check. Any n above 1 must give a speed below c, and a quarter slower is the right size for water. The frequency has not moved, so the wavelength absorbed the whole cut.

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.

n1sinθ1=n2sinθ2n_1 \sin\theta_1 = n_2 \sin\theta_2ON THE AQA 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°, the ray bending towards the normal as it enters the denser medium.

The direction of the bend follows from the equation. Going into a higher n, sinθ\sin\theta 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.

GUIDED PRACTICE

Into the glass, towards the normal

Light strikes a glass block (n = 1.52) at 40° to the normal. Predict which way it bends, then find the angle of refraction.

Show the working

Going into a higher n bends the ray towards the normal, so the refracted angle has to come out below 40°.

Snell gives sin θ2 = sin 40°/1.52 = 0.423, so θ2 = 25°. Smaller, as predicted. A refraction answer that bends the wrong way should never survive its own first line.

Measuring the refractive index of a solid

Snell's law is also a recipe for measuring n, and a rectangular glass block, a protractor and four pins are the whole apparatus. Air has n = 1.00 to three figures, so for a ray going from air into a solid the law reduces to a ratio of two sines,

n=sinθ1sinθ2n = \frac{\sin\theta_1}{\sin\theta_2}

with θ1\theta_1 the angle in air and θ2\theta_2 the angle in the solid, both measured from the normal.

The difficulty is that you cannot see the ray inside the glass, so the method reconstructs it. Lay the block on paper and draw round it, then mark a point on one long face and rule the normal there. Send a narrow ray in at that point, either from a ray box or by sighting along two optical pins pushed into the paper, and mark the incoming ray. Walk round to the far side, look through the block, and push in two more pins so that all four appear in a line through the glass. Now lift the block off. Joining the entry point to the exit point draws the path the light took inside, the one segment you were never able to watch, and the two angles can be measured with a protractor.

Measuring the refractive index of a glass block by tracing a ray, and the graph of sine of the angle of incidence against sine of the angle of refractionglassθ₁θ₂trace the ray with pinssin θ₁sin θ₂gradient = none line, through the origin
FIG. 2The tracing on the left and the graph it feeds on the right. The ray bends towards the normal on the way in and away from it on the way out, so it leaves the block travelling in its original direction, shifted sideways. Plotting the sine of the angle in air against the sine of the angle in the glass gives a straight line through the origin, and its gradient is the refractive index.

One reading would give you an answer, and a poor one. Instead repeat the tracing for perhaps six angles of incidence spread between 20° and 70°, and plot sin θ1\theta_1 against sin θ2\theta_2. Snell's law says the points should lie on a straight line through the origin, so the graph tests the law and measures the constant in the same stroke: the gradient is n. Averaging six points through a line of best fit beats any single protractor reading, and an intercept away from the origin indicates that the block outline moved between runs.

Three habits carry most of the precision. Use a sharp pencil and the thinnest ray the box will give, because a fat ray has no single edge to measure from. Space the pins as far apart as the paper allows, since the same one-millimetre slip in a pin position turns into a smaller angular error over a longer baseline. And keep off the very small angles, where sinθ\sin\theta is small and a half-degree protractor error is a large fraction of the reading.

The tracing checks itself. A block with parallel faces bends the ray towards the normal going in and away from it by the same angle coming out, so the emergent ray must be parallel to the incident ray and merely displaced sideways. Lay a ruler along both and if they are not parallel, something in the tracing is wrong and no amount of arithmetic afterwards will fix it.

WORKED EXAMPLE

n from the gradient

A student's line of best fit on a plot of sin θ1 against sin θ2 passes through the origin and through the point (0.42, 0.64). Find the refractive index, and the speed of light in the solid.

The gradient is n, and a line through the origin makes the gradient a single division: n = 0.64/0.42 = 1.5.

Then n = c/cs gives cs = c/n = 3.00 × 108/1.5 = 2.0 × 108 m s−1.

Read the gradient off the line, never off one plotted point. The point above happens to lie on the line, but a question that gives you a scattered set expects the fit, and quoting a single reading throws away the whole reason for plotting.

There is a second route to the same number, and it is worth knowing because it needs no protractor work in air. Find the angle at which light inside the solid stops escaping, the critical angle of the next section, and n follows from sinθc=1/n\sin\theta_c = 1/n. It suits a semicircular block, where the ray can be aimed at the curved face along a radius and enters without bending at all.

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\theta_c, the refracted ray runs exactly along the boundary. Setting θ2=90°\theta_2 = 90° in Snell's law gives

sinθc=n2n1\sin\theta_c = \frac{n_2}{n_1}ON THE AQA DATA SHEET

valid when n1>n2n_1 > n_2. For glass to air, sinθc=1/1.5\sin\theta_c = 1/1.5, so θc42°\theta_c \approx 42°.

Refraction, and the angle where it stops (animated figure)bending away, then not getting out at allpast 41.8° the boundary turns mirror: total internal reflectionairglass
FIG. 3The incidence angle sweeps up and back. Below the critical angle the refracted ray escapes, bending ever further from the normal while a faint reflection lingers inside; at 41.8 degrees for this glass the escape route closes, and past it every bit of the light reflects internally, the boundary a perfect mirror. The dashed ray marks the critical direction the sweep crosses twice per loop.

Beyond the critical angle no angle satisfies Snell's law, and the light cannot leave. All of it reflects back into the glass, obeying the ordinary law of reflection. That is total internal reflection, and it needs two 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 you choose.

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. 4A 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 earns its place three times over: it provides the lower refractive index that makes TIR possible, it protects the core surface from scratches that would let light leak away, and it stops light crossing between fibres bundled side by side, which would scramble their signals.

Real signals are pulses, and a pulse smears out as it travels, an effect called pulse broadening. Two things cause it. Modal dispersion comes from geometry, because rays bouncing at different angles cover different total distances and so arrive at different times. Make the core very narrow and every ray is forced onto much the same path. Material dispersion comes from the glass, in which different wavelengths travel at slightly different speeds, so a pulse of white light spreads in time. Monochromatic light removes that spread.

A broadened pulse can overlap its neighbour and corrupt the information. Separately, absorption in the glass weakens the pulse, and long lines therefore need repeaters.

INDEPENDENT PRACTICE

The critical angle inside a fibre

A fibre's core has n = 1.52 and its cladding n = 1.43. Find the critical angle at the core-cladding boundary, and state what becomes of rays that meet the wall at a larger angle to the normal.

Show the working

At a boundary between two media, sin C = n2/n1 = 1.43/1.52 = 0.941, so C = 70°.

Rays meeting the wall beyond 70° from the normal are skimming along the fibre, and they are totally internally reflected back into the core, which is how light is meant to travel down a fibre. Rays that strike the wall inside the critical angle leak away into the cladding, so the steep zigzag paths, the ones that smear pulses worst, are shed early.

ASSESSMENT FOCUS

  • Every angle in this topic is measured from the normal. When a question quotes an angle from the surface, subtract it from 90° before you touch Snell's law.
  • In a critical angle the smaller index goes on top, sinθc=n2/n1\sin\theta_c = n_2/n_1 with n1>n2n_1 > n_2. A calculator complaining about sinθc>1\sin\theta_c > 1 is telling you the fraction is upside down.
  • Asked how you would measure the refractive index of a block, describe the graph as well as the tracing. Several angles of incidence, sin θ1\theta_1 plotted against sin θ2\theta_2, a straight line through the origin, and n read off as the gradient. Edexcel names this measurement as a statement in its own right; the practical is standard on every board.
  • State both TIR conditions, into a lower refractive index and beyond the critical angle. One without the other scores half.
  • “Explain the purpose of the cladding” has three creditable points. It supplies the lower n that makes TIR possible, it protects the core surface from scratches, and it stops signals crossing between fibres in a bundle.
  • 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=1.50n_1 = 1.50) into air (n2=1.00n_2 = 1.00), the condition n1>n2n_1 > n_2 holds, so a critical angle exists.

sinθc=n2/n1\sin\theta_c = n_2/n_1 = 1.00 / 1.50 = 0.667, so θc\theta_c = 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.

Angles are measured from the normal.

TIR needs a lower n on the far side,

and an angle past the critical one.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the refraction and total internal reflection questions page.

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  • Use refractive index as a ratio of speeds, and say what happens to frequency and to wavelength at a boundary.
  • Apply Snell's law, predicting which way the ray bends before calculating how far.
  • Measure the refractive index of a solid by tracing a ray through a block and taking the gradient of sin θ₁ against sin θ₂.
  • State both conditions for total internal reflection, and calculate a critical angle.
  • Describe a step-index optical fibre and the three jobs the cladding does.
  • Separate modal from material dispersion, and match each one to its fix.

Open the full revision checklist to track your progress across the whole unit.