Show that for a material with refractive index $\mu \geqslant \sqrt{2}$,light incident at any angle shall be guided along a length perpendicular to the incident face.

(N/A) As shown in the figure,consider a light ray $\overrightarrow{PQ}$ incident on the surface $AB$ of a denser transparent medium with angle of incidence $i$. After refraction,the ray $\overrightarrow{QR}$ is incident at point $R$ on the surface $AC$ of the rarer medium with angle of incidence $\phi$. For the light to be guided without escaping,it must undergo total internal reflection $(TIR)$ at every interface. Thus,we require $\phi \geq C$,where $C$ is the critical angle.
From the geometry,$\phi + r = 90^{\circ}$,so $\phi = 90^{\circ} - r$.
The condition for $TIR$ is $\phi \geq C$,which implies $\sin \phi \geq \sin C$.
Substituting $\phi = 90^{\circ} - r$,we get $\sin(90^{\circ} - r) \geq \sin C$,which simplifies to $\cos r \geq \frac{1}{\mu}$ (since $\sin C = \frac{1}{\mu}$).
Squaring both sides,$\cos^2 r \geq \frac{1}{\mu^2}$,or $1 - \sin^2 r \geq \frac{1}{\mu^2}$.
From Snell's law at point $Q$,$\sin i = \mu \sin r$,so $\sin r = \frac{\sin i}{\mu}$.
Substituting this,$1 - \frac{\sin^2 i}{\mu^2} \geq \frac{1}{\mu^2}$.
Multiplying by $\mu^2$,we get $\mu^2 - \sin^2 i \geq 1$,or $\mu^2 \geq 1 + \sin^2 i$.
Since the maximum value of $\sin^2 i$ is $1$ (at $i = 90^{\circ}$),the condition must hold for all $i$,so $\mu^2 \geq 1 + 1 = 2$,which means $\mu \geq \sqrt{2}$.
Thus,for $\mu \geq \sqrt{2}$,light will always undergo $TIR$ and be guided along the medium.