Refraction, total internal reflection and optical fibresAQA A-Level Physics: Revision notes

Light travels more slowly in a transparent material than in a vacuum. The absolute refractive index of a substance is

$n = \frac{c}{c_s}$

where $c$ is the speed of light in a vacuum ($3.00 \times 10^8$ m s⁻¹) and $c_s$ is the speed of light in the substance. Since $c_s < c$, $n > 1$ for every material, and the refractive index of air is approximately 1.

Worked example. Glass has $n = 1.50$, so $c_s = 3.00 \times 10^8 / 1.50 = 2.00 \times 10^8$ m s⁻¹.

Refraction is the change in direction of a wave when its speed changes as it crosses a boundary between two media. For a boundary between media of refractive index $n_1$ and $n_2$:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

Angles are measured from the normal. Going into a medium of higher $n$ the light slows down and bends towards the normal; going into a medium of lower $n$ it speeds up and bends away from the normal. A ray along the normal is not deviated.

Worked example. Light in water ($n_1 = 1.33$) meets oil ($n_2 = 1.47$) at 30°. $\sin\theta_2 = 1.33 \sin 30° / 1.47 = 0.452$, so $\theta_2 = 27°$.

When light travels from a medium of higher $n_1$ towards one of lower $n_2$ it bends away from the normal. As the angle of incidence increases, the refracted ray approaches 90° to the normal. The critical angle $c$ is the angle of incidence in the denser medium for which the angle of refraction is 90°:

$\sin c = \frac{n_2}{n_1}$

If the angle of incidence is greater than the critical angle, there is no refracted ray and all the light is reflected: total internal reflection (TIR).

Two conditions are needed: light must travel from a higher to a lower refractive index, and the angle of incidence must exceed $c$.

Worked example. For glass to air, $\sin c = 1/1.50$, so $c = 41.8°$. A ray at 50° inside the glass is totally internally reflected.

A step-index optical fibre has a central core of glass surrounded by cladding, a thin layer of glass of lower refractive index. Light entering the core at a small angle to the axis meets the core–cladding boundary at an angle above the critical angle and is totally internally reflected again and again along the fibre.

The cladding:

• has a lower refractive index than the core, which makes TIR possible at the boundary
• protects the core from scratches, which would let light escape
• prevents light leaking between neighbouring fibres (cross-talk)

A ray at angle $\alpha$ to the axis meets the boundary at an angle of incidence of $90° - \alpha$. TIR only occurs if $90° - \alpha > c$.

Digital data is sent as short pulses of light. As a pulse travels it becomes longer: pulse broadening.

• Modal dispersion: rays enter at different angles, so they take paths of different length. Rays that zigzag more travel further and arrive later. A very narrow core reduces it.
• Material dispersion: light contains a range of wavelengths, and the refractive index of glass depends on wavelength, so each wavelength travels at a different speed. Using monochromatic light (a laser) reduces it.

Consequences: broadened pulses overlap, so the receiver cannot separate them and errors occur. Pulses must be sent further apart, so the data rate falls, and the problem grows with fibre length.

Glass absorbs some of the light energy as the pulse travels, so the amplitude (power) of the pulse falls with distance. The signal becomes weaker and may fall below what the receiver can detect. On long links the signal is boosted by repeaters (regenerators), which also restore the pulse shape. Absorption reduces the signal strength; it does not change the pulse length, which is why it is treated separately from dispersion.