a) [5] Calculate the smallest angle of incidence for light incident on the face of a $60^{\circ}$ prism of refractive index $1.5$, such that all of the light on the second face is totally reflected.
Draw a large diagram to show the path of the light and the angles.
b) [12] Consider the step index optical fibre shown in Figure $\boldsymbol{4}$.

Figure 4
The core, cladding and external medium have refractive indices $n_{1}, n_{2}$, and $n_{3}$, respectively. The acceptance angle $\theta$ for light entering the core of the fibre will allow transmission without loss providing the angle $\theta$ between the incident ray and the axis is smaller than the maximum acceptance angle $\theta_{m}$.
(i) Obtain an expression for $\sin \theta_{m}$ in terms of $n_{1}, n_{2}$, and $n_{3}$.
Hint: $\cos ^{2} \theta=1-\sin ^{2} \theta$
(ii) Light enters the fibre optic at angles between $\theta=0^{\circ}$ and $\theta_{m}$. Given a step index fibre with a core refractive index of $1.48585$, and a cladding refractive index of $1.45641$, determine the time broadening of the leading edge of a square pulse over a distance of $100 \mathrm{~km}$.
(iii) If multiple, very brief, square pulses of light are sent down the cable at frequency $f_{\mathrm{p}}$, calculate the maximum possible pulse frequency over this $100 \mathrm{~km}$ distance, such that the pulses do not overlap.
(iv) Show that in general the time spread of a pulse $\Delta t$, in a fibre of length $L$, is given by $\Delta t=\frac{L}{c^{\prime} n_{2}} \Delta n$, with $\Delta n$ the difference in refractive indices between core and cladding, and $c^{\prime}$ the speed of light in the optical fibre.
c) [2] A ray of light from a point source $\mathbf{O}$ on the principal axis of a thin converging lens, passing through the lens, is brought to a focus at the image point $\mathbf{P}$ also on the principal axis, and shown in Figure 5 .

Figure 5
One form of an equation for a thin lens, relating object and image distances, $u, v$ to the focal length $f$ is
$$ \frac{1}{u}+\frac{1}{v}=\frac{1}{f} $$
By taking the limit $u \rightarrow \infty$, explain what the length $f$ represents for a lens.
d) [6] Rays of light from a point source on the principal axis, passing through a converging lens by different paths and reaching a focus point on the principal axis, can be considered to do so by following paths which all have an equal time of travel. (This is a simplified form of a more general principle - Fermat's Principle of Least Time).
A flat, thin, circular disc of transparent material is made so that its refractive index $n$ varies with the distance $r$ from the principal axis normal to the centre of the disc, according to the equation $n=n_{\mathrm{o}}\left(1-k r^{2}\right)$, where $n_{\mathrm{o}}$ and $k$ are constant.
(i) By considering the times taken by light travelling by different paths (take a path not on the principal axis), show for rays travelling very close to the principal axis that the thin disc can behave as a lens. i.e. obtain an expression for $\frac{1}{u}+\frac{1}{v}$, if the thickness of the disc is $d$, with $d \ll u, v$.
(ii) Derive an expression for its focal length in terms of $d, k$ and $n_{\mathrm{o}}$.
Show answer · 显示答案
- (a)$\displaystyle 27.92^{\circ}$
- (b)(i)$\displaystyle \sin \theta_m = \frac{\sqrt{n_1^2 - n_2^2}}{n_3}$
- (b)(ii)$\displaystyle 10 \, \mu \mathrm{s}$
- (b)(iii)$\displaystyle 100 \, \mathrm{kHz}$
- (b)(iv)$\displaystyle \Delta t = \frac{L}{c'} \frac{\Delta n}{n_2}$
- (d)(i)$\displaystyle \frac{1}{u} + \frac{1}{v} = 2 d n_0 k$
- (d)(ii)$\displaystyle f = \frac{1}{2 d n_0 k}$
Worked solutions are in the book.
















