6.4.6 Total reflection and the critical angle

Write Snell's law as

If the wave enters a denser medium, ${\beta }_2>{\beta }_1$, the right-hand side is always smaller than one and there is a transmitted wave at every angle of incidence. If the wave goes into a less dense medium, ${\beta }_1>{\beta }_2$, as light that tries to leave glass for air, the transmitted wave is bent away from the normal, and ${\theta }_t$ reaches $90^{\circ }$ at a finite angle of incidence. Beyond that angle the right-hand side of Eq. (6.4.16) exceeds one, there is no real angle ${\theta }_t$, and, as we show below, the incident wave is totally reflected.

Figure 6.4.6 follows three rays from a source inside glass. At $25^{\circ }$ the ray leaves into the air at $39.3^{\circ }$ and a weak part is reflected. At ${\theta }_c=41.8^{\circ }$ the refracted ray grazes the surface. At $55^{\circ }$ nothing leaves and the whole wave is reflected.

Beyond the critical angle: the evanescent wave

What happens to the transmitted wave when $\sin {\theta }_t>1$? Phase matching Eq. (6.4.9) still holds: the transmitted field must vary along the interface as ${\mathrm{e}}^{-\mathrm{j}{\beta }_{1}y\sin {\theta }_i}$, with ${\beta }_2\sin {\theta }_t={\beta }_1\sin {\theta }_i>{\beta }_2$. Its $x$-dependence is ${\mathrm{e}}^{-\mathrm{j}{\beta }_{2}x\cos {\theta }_t}$ with ${\cos }^2{\theta }_t=1-{\sin }^2{\theta }_t<0$, so $\cos {\theta }_t$ is imaginary. Of the two roots we must take

because then ${\mathrm{e}}^{-\mathrm{j}{\beta }_{2}x\cos {\theta }_t}={\mathrm{e}}^{-{\alpha }_{2}x}$ decays into medium 2. The other root would give a field that grows without bound away from the interface, which no source in medium 1 can produce.

The Fresnel formulas remain valid with the imaginary $\cos {\theta }_t$. Writing $\cos {\theta }_t=-\mathrm{j}|\cos {\theta }_t|$ in Eq. (6.4.13),

Numerator and denominator are complex conjugates, so $|{\Gamma }_{\perp }|=1$: the reflected wave carries the full incident amplitude, and total reflection only shifts its phase. The same holds for ${\Gamma }_{\parallel }$ (Subsection 6.4.8). The transmitted magnetic field has $H_y^t=-(E_0^t/{\eta }_2)\cos {\theta }_t{\mathrm{e}}^{-{\alpha }_{2}x}\cdots$, in phase quadrature with $E_z^t$, so the normal component of the time-average Poynting vector, ${\mathcal{P}}_x=-\frac{1}{2}\mathrm{Re}\{E_z{H}_y^{\ast }\}\propto \mathrm{Re}\{\cos {\theta }_t^{\ast }\}$, is zero: no time-average power crosses the interface. The component ${\mathcal{P}}_y$, built from $E_z$ and the real $H_x\propto \sin {\theta }_t$, is not zero: power flows along the face, inside the thin evanescent layer, but not away from it.

Below, light of ${\lambda }_0=600\mathrm{nm}$ meets a glass-air face from inside the glass ($n_1=1.5$). Drag ${\theta }_i$ through ${\theta }_c=41.8^{\circ }$ and watch $|\Gamma |$ reach $1$, after which only its phase changes and $1/{\alpha }_2$ shrinks, to $115\mathrm{nm}$ at $60^{\circ }$.

⚙Application 6.4.1 · Total internal reflection: prisms and optical fibres

Prisms. For glass ($n=1.5$) the critical angle is $41.8^{\circ }$, so a beam that meets the hypotenuse of a $45^{\circ }$–$90^{\circ }$–$45^{\circ }$ prism at $45^{\circ }$ from inside is reflected completely (Figure 6.4.8a). It is a mirror with no metal, no tarnish and no absorption, which works for any glass with $n>\sqrt{2}=1.414$. Binoculars (the Porro prisms), periscopes and retroreflectors use it. The field outside the hypotenuse is evanescent, as Problem 6.4.2 shows with numbers. A second prism brought within a fraction of a wavelength lets light tunnel across: beam-splitter cubes and fingerprint sensors rely on such frustrated total reflection.

Optical fibres. A glass core of index $n_1$ is surrounded by a cladding of slightly lower index $n_2$. A ray that meets the core-cladding boundary at $\theta \ge {\theta }_c=\arcsin (n_2/n_1)$ is totally reflected at every bounce and is trapped in the core over kilometres (Figure 6.4.8b). A trapped ray makes the angle $\varphi =90^{\circ }-\theta$ with the fibre axis, at most ${\varphi }_{\max }=90^{\circ }-{\theta }_c$. At the end face Snell's law turns this into the largest angle ${\theta }_a$ at which light can be launched from air: $\sin {\theta }_a=n_1\sin {\varphi }_{\max }=n_1\cos {\theta }_c=n_1\sqrt{1-n_2^2/n_1^2}$.

Numerical aperture of a step-index fibre

$$\begin{array}{rl}& \mathrm{NA}=\sin {\theta }_a=\sqrt{n_1^2-n_2^2}\\ & \text{(launched from air)},\end{array}\text{\hspace{1em}(6.4.21)}$$

where ${\theta }_a$, the acceptance angle, is the half-angle of the cone of rays that the fibre guides.

In plain words

The larger the index step between core and cladding, the wider the cone of light that a fibre accepts. An index step of only $0.01$ already gives an acceptance angle of about $10^{\circ }$ (Problem 6.4.3).

• Telecom fibre, $n_1=1.465$, $n_2=1.460$: ${\theta }_c=85.3^{\circ }$, $\mathrm{NA}=\sqrt{1.465^2-1.460^2}=0.121$, ${\theta }_a=6.9^{\circ }$. Guided rays are almost parallel to the axis. Problem 6.4.3 adds the delay spread between the axial and the steepest ray.
• Multimode fibre, $n_1=1.48$, $n_2=1.46$: ${\theta }_c=80.6^{\circ }$, $\mathrm{NA}=0.24$, ${\theta }_a=14.0^{\circ }$. For a ray at $85^{\circ }$ at ${\lambda }_0=850\mathrm{nm}$, ${\alpha }_2={\beta }_0\sqrt{1.48^2{\sin }^{2}85^{\circ }-1.46^2}=1.52\times 10^{6}Np/m$, so the evanescent tail reaches only $1/{\alpha }_2=0.66\mu \mathrm{m}$ into the cladding.
• The evanescent tail runs in the cladding, which is therefore made many wavelengths thick: a standard fibre has a $125\mu \mathrm{m}$ cladding around a $9\mu \mathrm{m}$ core, and touching its outside does not disturb the light.

?Quick check

What is the smallest refractive index for which a $45^{\circ }$ prism reflects totally in air?

Total reflection at $45^{\circ }$ needs ${\theta }_c\le 45^{\circ }$, $\sin {\theta }_c=1/n\le \sin 45^{\circ }$, so $n\ge \sqrt{2}=1.414$. Every ordinary optical glass qualifies.

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