3.3.2.3 - Refraction at a Plane Surface

Refractive index

The refractive index of a substance tells you how much light slows down in that substance compared with a vacuum. A larger refractive index means light travels more slowly and the substance is more optically dense.

Here, $c$ is the speed of light in a vacuum and $c_s$ is the speed of light in the substance. For the transparent materials in this lesson, $n>1$ and light travels more slowly than in a vacuum. The refractive index of air is very close to 1, so for A-level work you should recall that the refractive index of air is approximately 1.

When light enters a substance and slows down, its frequency stays constant. The source fixes the oscillation frequency, and the boundary must be driven at that same frequency. Hence wavelength decreases with speed: $\lambda_{\mathrm{medium}}=\lambda_{\mathrm{vacuum}}/n$. Refractive index is a dimensionless ratio, not a speed or a measure of mass density.

Snell's law

Refraction happens because one side of a wavefront reaches the new substance before the other side. If the new substance has a lower light speed, that first part slows down while the rest of the wavefront is still moving faster. The wavefront pivots, and the ray changes direction.

In the figure below, notice that both angles are measured from the normal and that the ray bends towards the normal when it enters the more optically dense substance.

[DIAGRAM: asset_name: 3.2.3 - Refraction at a Plane Surface - Diagram 1; asset_slug: 3.2.3 - Refraction at a Plane Surface - Diagram 1; recommended_method: retained_png; caption: Refraction angles are measured from the normal; description: A plane boundary between substance 1 and substance 2. A normal is drawn at the boundary. An incident ray in substance 1 makes angle theta_1 to the normal and a refracted ray in substance 2 makes angle theta_2. The refracted ray bends towards the normal when substance 2 is more optically dense.]

Light bends towards the normal when it enters a more optically dense substance and away from the normal when it enters a less optically dense substance. If it travels along the normal, there is no change in direction.

Using the wavefront geometry for a boundary between two substances gives the relationship
$\frac{\sin \theta_1}{\sin \theta_2} = \frac{c_1}{c_2}$.
Replacing the speeds with refractive indices gives Snell's law.

Here, $n_1$ and $n_2$ are the refractive indices of the two substances, and $\theta_1$ and $\theta_2$ are the angles to the normal. If the first substance is air, you can often take $n_1 \approx 1$.

Total internal reflection

When light travels from a more optically dense substance into a less optically dense one, it bends away from the normal. As the angle of incidence increases, the angle of refraction also increases. Eventually the refracted ray reaches $90^\circ$ and runs along the boundary.

The figure below compares the three key cases at a glass-air boundary, so focus on how the refracted ray moves from leaving the glass to skimming the surface and then disappearing when total internal reflection begins.

[DIAGRAM: asset_name: 3.2.3 - Refraction at a Plane Surface - Diagram 2; asset_slug: 3.2.3 - Refraction at a Plane Surface - Diagram 2; caption: Increasing the incident angle from glass to air reaches the critical angle, then total internal reflection; description: True-angle geometry for glass of index 1.50 to air of index 1.00. Incidence 30.0 degrees gives refraction 48.6 degrees. At incidence 41.8 degrees the refracted ray follows the boundary. At 55.0 degrees total internal reflection has equal incident and reflected angles.]

Applying Snell's law at the critical angle gives
$n_1 \sin \theta_c = n_2 \sin 90^\circ$.
Since $\sin 90^\circ = 1$, the critical-angle equation is:

Total internal reflection happens only when two conditions are both true:

• light is travelling from the more optically dense substance to the less optically dense substance
• the angle of incidence is greater than the critical angle

For example, at a glass-air boundary with $n_1 = 1.50$ and $n_2 = 1.00$,
$\sin \theta_c = 1.00/1.50 = 0.667$, so $\theta_c = 41.8^\circ$.

Step-index optical fibres

Optical fibres use total internal reflection to guide light along a thin flexible core, even when the fibre bends gently.

A step-index optical fibre has a core with a higher refractive index surrounded by cladding with a slightly lower refractive index. For this specification, you only need the step-index case, where the refractive index changes sharply at the core-cladding boundary.

In the figure below, notice that the core has the higher refractive index and that the light stays trapped by repeated total internal reflection at the core-cladding boundary.

[DIAGRAM: asset_name: 3.2.3 - Refraction at a Plane Surface - Diagram 3; asset_slug: 3.2.3 - Refraction at a Plane Surface - Diagram 3; caption: A continuous ray reflects within the higher-index fibre core; description: Equal-angle zigzag ray confined to a straight core of index 1.50 surrounded by cladding of index 1.45. A ray at 10 degrees to the fibre axis meets each wall at 80 degrees to its normal, above the 75.2 degree critical angle.]

The cladding matters for three main reasons:

• it provides the lower refractive index needed for total internal reflection at the core-cladding boundary
• it reduces light leakage and crossover between neighbouring fibres
• it protects the core from surface damage that would let light escape

For the drawn fibre, $\theta_c=\sin^{-1}(1.45/1.50)=75.2^\circ$. A ray making $10^\circ$ to the fibre axis strikes the wall at $80^\circ$ to its normal, so it undergoes total internal reflection. The two angles are complementary. Bending the fibre too sharply can reduce the incidence angle below the critical angle and cause leakage; a higher-index core alone does not trap every incident ray.

Absorption and dispersion

Real fibres do not transmit signals perfectly. Over long distances, two broad problems matter: the signal can get weaker, and the pulses can spread out.

Absorption reduces the amplitude of a pulse because some of the light energy is absorbed by the glass. If the pulse becomes too weak, the receiver may not be able to distinguish it from noise. Low-loss glass reduces absorption. Repeaters restore the weakened signal after transmission; they do not prevent absorption in the glass.

Two important causes of pulse broadening are:

• modal dispersion: different rays take different paths through the core, so they take different times to reach the end; this is reduced by using a sufficiently narrow core that supports fewer propagation modes
• material dispersion: different wavelengths travel at slightly different speeds in the glass, so parts of the same pulse separate; this is reduced by using monochromatic light, usually from a laser

These effects matter because a communication system relies on clear, separate pulses. If pulses merge, the receiver cannot distinguish successive bits. For example, pulses initially $2\,\mathrm{ns}$ wide and separated by $5\,\mathrm{ns}$ between centres can become distinguishable less reliably when broadened beyond $5\,\mathrm{ns}$. Reducing the pulse rate gives more separation but reduces the information rate. Raising the source brightness can help against attenuation; it does not undo pulse overlap.