= 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. Hence
$$
\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1},\qquad
\boxed{\theta_2=\arcsin\!\left(\frac{n_1}{n_2}\sin\theta_1\right).}
$$
Here 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 when
$$
\boxed{\theta_1>\theta_c=\arcsin(n_2/n_1).}
$$
This 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> satisfies
$$
\boxed{\sin\theta_{i,\max}=\sqrt{n_1^2-n_2^2}\equiv\mathrm{NA},\qquad
\theta_{i,\max}=\arcsin\sqrt{n_1^2-n_2^2}.}
$$
The <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 $t=n_1L/(c\cos\beta)$. Therefore
$$
\begin{aligned}
t_{\max}&=\frac{n_1^2L}{cn_2}=5.00438\,\mu\mathrm s,\\
t_{\mathrm{half}}&=\frac{n_1L}{c\sqrt{1-\sin^2(\theta_{i,\max}/2)/n_1^2}}
=4.95365\,\mu\mathrm s,
\end{aligned}
\qquad
\boxed{\Delta t\simeq50.7\,\mathrm{ns}.}
$$
Here $L=1000\,\mathrm m$ and $c=299792458\,\mathrm{m\,s^{-1}}$; using $3.00\times10^8\,\mathrm{m\,s^{-1}}$ 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 $\sin\beta=\sin10^\circ/n_1$. 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 gives
$$
\sin i_{\mathrm{out}}=\frac{R\cos\beta}{R+r}.
$$
At the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius gives
$$
\boxed{R_c=\frac{n_2r}{n_1\cos\beta-n_2}
=\frac{n_2r}{\sqrt{n_1^2-\sin^2 10^\circ}-n_2}
=14.93\,\mathrm{mm}.}
$$
For 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.
\b[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 gives
$$
R_{c,\mathrm{inner}}=r\frac{n_1\cos\beta+n_2}{n_1\cos\beta-n_2}=29.96\,\mathrm{mm}.
$$
Thus $14.93\,\mathrm{mm}$ 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.
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