Past exam of the physics 2 paper of the University of Oxford / 2019 resit / 10 / Solution

In the Huygens-Fresnel principle construction, after a time $\Delta t$ the incident disturbance advances $v_1\Delta t$ while the secondary disturbance in the second medium advances $v_2\Delta t$. The two right triangles share the segment along the interface. HenceHere both angles are measured from the normal vector, and $v_j=c/n_j$ are phase velocities. This is Snell's law. For $n_1>n_2$, no real transmitted angle exists whenThis is total internal reflection; equality is the limiting grazing geometry. The statement concerns an ideal transparent interface, not material absorption or losses at a nearby second boundary.

Let $\beta$ be the internal angle to the optical axis in the straight step-index optical fiber. At its flat entrance, Snell's law gives $\sin {\theta }_i=n_1\sin \beta$. At the side wall a meridional ray has angle of incidence $\pi /2-\beta$. The limiting total internal reflection condition is $\cos \beta =n_2/n_1$. Thus the optical fiber acceptance angle satisfiesThe numerical aperture expression assumes the square root is at most one for air; otherwise all forward launch angles are admitted by this ideal ray test. Strictly confined meridional rays lie inside the limiting cone.

For $n_1=1.48$ and $n_2=1.46$, $\mathrm{NA}=0.242487$ and ${\theta }_{i,\max }=14.0334^{\circ }$. A light ray at internal angle $\beta$ travels distance $L/\cos \beta$ along a guide of axial length $L$. In the nondispersive approximation its travel time is $t=n_{1}L/(c\cos \beta )$. ThereforeHere $L=1000\mathrm{m}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$; using $3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ gives $5.001\mu \mathrm{s}$ and $4.950\mu \mathrm{s}$. The halving applies to the external launch angle, not to $\beta$. For actual short pulses the group refractive index $n_{\mathrm{g}}$ replaces the prefactor $n_1$ in $t=n_{\mathrm{g}}L/(c\cos \beta )$; the source supplies no optical dispersion data. The difference calculated here is geometrical intermodal dispersion in a nondispersive geometrical optics model.

Bending lowers the outer-wall angle of incidence, allowing a previously guided light ray to escape into the cladding. To calculate a definite meridional ray bend threshold, continue the on-axis launch assumption and take a meridional ray in the plane of the circular bend. Its initial internal angle to the centreline tangent obeys $\sin \beta =\sin 10^{\circ }/n_1$. Its straight path has distance $b_{\mathrm{ray}}=R\cos \beta$ from the bend centre. At the outer wall of radius $R+r$, elementary right-triangle geometry givesAt the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius givesFor this launch, $R<R_c$ violates the total internal reflection condition at the outer wall; $R=R_c$ is the limiting case. A real optical fiber has gradual bend loss governed by its electromagnetic wave modes, rather than a universal abrupt cutoff.

Launch dependence. The final sentence does not separately specify the entry position or orientation of the bend relative to the incident light ray. For a planar launch displaced by $u$ from the centreline toward the outer wall, replace the numerator of $\sin i_{\mathrm{out}}$ by $(R+u)\cos \beta$. For example, a launch at the inner edge, $u=-r$, instead givesThus $14.93\mathrm{mm}$ is the result for the stated on-axis continuation; it is not an entry-independent threshold. No missing figure is needed for these explicitly specified ray models.