Total Internal Reflection and Snell’s Law

Tip

Exam Focus

Near-certain items: derive the critical angle $\theta_c=\sin^{-1}(n_2/n_1)$, state the two TIR conditions ($n_1>n_2$ and $\theta_i>\theta_c$), draw the fibre cross-section ($n_1>n_2$) and the acceptance-cone ray diagram. Numericals on $\theta_c$ from a given $n_1,n_2$ are almost guaranteed — remember $\theta_c$ is large (near $80^\circ$) because $n_1\approx n_2$.

Optical Fibre Communication

Optical fibre communication transmits information as modulated light through a thin transparent dielectric waveguide of glass or plastic, guided by total internal reflection at the core–cladding boundary.

A basic link converts the electrical message to light, guides it through the fibre, and converts it back to electrical form at the receiver.

Optical fibre communication system: electrical input $\to$ source driver $\to$ optical source $\to$ fibre channel $\to$ photodetector $\to$ receiver electronics $\to$ recovered output.

Structure of an Optical Fibre

Fibre cross-section: high-index core ($n_1$), lower-index cladding ($n_2$) and a protective coating/jacket.

Part	Function
Core	Central region ($n_1$) where light propagates by TIR
Cladding	Surrounds core with lower index $n_2$ ($n_1>n_2$); confines light
Coating / jacket	Plastic layer giving mechanical and moisture protection

Three parts of an optical fibre.

Advantages and Limitations

Note

Key Point — Fibre versus metallic cable

• Advantages: very high bandwidth; low loss (long repeater spacing); immunity to EMI, lightning and crosstalk; small size and light weight; electrical isolation and no spark hazard; high tapping security; low bit-error rate.

• Limitations: brittle glass; skilled splicing/precision tools needed; costlier electro-optic transceivers; bending loss; cannot deliver electrical power to remote equipment.

Refraction and Refractive Index

The refractive index of a medium measures how much it slows light:

Caution

Key Formula — Refractive index

$$n=\frac{c}{v},$$

where $c$ is the speed of light in vacuum and $v$ its speed in the medium. A larger $n$ means slower light. When light crosses a boundary between media of different index, it changes direction (refraction).

Snell’s Law and the Critical Angle

Note

Law — Snell’s law of refraction

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

where angles are measured from the normal to the interface.

For light passing from a denser to a rarer medium ($n_1>n_2$) the refracted ray bends away from the normal. At the critical angle $\theta_c$ the refracted ray grazes the boundary ($\theta_2=90^\circ$):

Caution

Key Formula — Critical angle

$$n_1\sin\theta_c=n_2\sin90^\circ \;\Rightarrow\; \boxed{\;\theta_c=\sin^{-1}\!\frac{n_2}{n_1}\;}\quad(n_1>n_2).$$

Total Internal Reflection

Total internal reflection (TIR) is the complete reflection of light back into the denser medium, occurring when (i) light travels from higher to lower index ($n_1>n_2$) and (ii) the incidence angle exceeds the critical angle ($\theta_i>\theta_c$).

In a fibre the core index slightly exceeds the cladding index, so rays launched within the acceptance cone strike the boundary above $\theta_c$ and are guided along the core by repeated TIR.

Rays entering within the acceptance angle $\theta_a$ undergo repeated total internal reflection and stay confined to the core; steeper rays refract into the cladding and are lost.

Note

Example — Critical angle of a step-index fibre

A fibre has $n_1=1.48$ and $n_2=1.46$. Find the critical angle.

$$\sin\theta_c=\frac{n_2}{n_1}=\frac{1.46}{1.48}=0.9865 \;\Rightarrow\; \theta_c=\sin^{-1}(0.9865)\approx\boxed{80.6^\circ}.$$

Because $n_1\approx n_2$, $\theta_c$ is close to $90^\circ$: only rays that strike the wall at a very grazing angle are guided.

[!NOTE]

Key Point — Exam-ready model answer — guidance by Snell’s law and TIR

An optical fibre is a cylindrical dielectric waveguide (core $n_1$, cladding $n_2<n_1$, protective coating). Refractive index $n=c/v$. At any interface, Snell’s law $n_1\sin\theta_1=n_2\sin\theta_2$ holds. Going core$\to$cladding the ray bends away from the normal; at $\theta_c=\sin^{-1}(n_2/n_1)$ the refracted angle is $90^\circ$. TIR needs $n_1>n_2$ and $\theta_i>\theta_c$; rays inside the acceptance cone satisfy both and follow the core with negligible loss, even around gentle bends. Rays outside the cone refract into the cladding and are lost. For $n_1=1.48,\,n_2=1.46$, $\theta_c\approx80.6^\circ$.