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APhO 2003, Теория — Задача 2. II. Optical Gyroscope

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4th Asian Physics Olympiad — Theoretical Competition — 23 April 2003 · 23 април 2003 г.

Условие

In 1913 Georges Sagnac (1869-1926) considered the use of a ring resonator to search for the aether drift relative to a rotating frame. However, as often happen, his results turned out to be useful ways that Sagnac himself never dreamt of. One of those applications is the Fibre-Optic Gyroscope (FOG) which is based upon a simple phenomena, first observed by Sagnac. The essential physics associated with the Sagnac effect is due to the phase shift caused by two coherent beams of light being sent around a rotating ring of optical fibre in the opposite directions. This phase shift is also used to determine the angular speed of the ring.

As shown in a schematic diagram in Fig. 1, a light wave enters a circular optical fibre light path of radius $R$ at point P on the rotating platform with a uniform angular speed $\Omega$ , in the clockwise direction. Here the light wave is split into two waves which travels in the opposite directions, clockwise (CW) and counter clockwise (CCW), through the ring. The refractive index of optical fibre material is $\mu$ . Assuming the light traveling inside the fibre-optic cable is a smooth circular path of radius $R$ .

A circular optical fibre ring (double line) of radius R drawn from the centre; light enters from the left at point P and splits into two beams shown by arrows going both ways around the ring; a curved arrow at the centre marks the clockwise rotation Ω; a label points to the ring: 'Optical fibre light path'. — Fig.1

a) Practically, the orbital speed of the ring is much less than the speed of light such that $(R\Omega)^2 \ll c^2$ , find the time difference $\Delta t = t^{+} - t^{-}$ where $t^{+}$ and $t^{-}$ denote the round-trip transit time of the CW and CCW beam respectively. Give your answer in term of area $A$ enclosed by the ring. [2 т.]

b) Find the optical path difference, $\Delta L$ , for the CW and CCW beams to complete one round-trip of the light within the rotating ring. [2 т.]

c) For a circular fibre-optic of radius, $R = 1\ \mathrm{m}$ , what is the maximum value of $\Delta L$ for the rotation of the earth? Given $\mu = 1.5$ . [1 т.]

d) In part b), the measurement could be amplified by increasing number of turns in fibre-optic coil, $N$ , find the phase difference, $\Delta\theta$ , for lights to complete the turns. [1 т.]

e) The second scheme of the Optical Gyroscope is Ring Laser Gyroscope (RLG). This could be accomplished by the inclusion of active laser cavity into an equilateral triangular ring, of total length $L$ , as illustrated in Fig.2. The laser source here will generate two amplified coherent light sources propagating in the opposite directions. In order to sustain the laser oscillation in this triangular ring resonator, the perimeter of the ring must be equal to the integer multiple of wavelength $\lambda$ . Etalon, additional component inserted into the ring, is possible to cause frequency selective losses in the ring resonator, so that the undesired modes can be damped and suppressed.

Equilateral triangular ring with mirrors M1 (top, 50% coated mirror), M2 and M3 at the base corners, each side L/3 and 60° angles; a laser tube (orange) on the bottom side emits beams both ways; an etalon on the left side; a dashed circular arrow Ω at the centre; light leaving through M1 passes a lens to a Detector. — Fig.2: Schematic illustration of the Ring Laser Gyroscope

Rendered illustration of a ring laser gyroscope: a triangular platform on a round rotating base carrying mirrors and a laser tube with a red beam path around the triangle, and electronic instruments including an oscilloscope in the background. — Fig. 3: Illustration of the Ring Laser Gyroscope discussed in this problem

Find the time difference of the transit in clockwise and counterclockwise, $\Delta t$ , for the case of the triangular ring as shown in the figure 2. Give the answer in terms of $\Omega$ and the area $A$ enclosed by the ring. Show that this result is the same as that of the circular ring. [2 т.]

f) If the ring is rotating with an angular frequency $\Omega$ as shown in Fig. 2, there will be frequency difference between CW and CCW measurements. What is the observed beat frequency, $\Delta\nu$ , between the CW and CCW beams in terms of $L, \Omega, \lambda$ . [2 т.]

Отговори

Решение

Solution and Marking Scheme

Theory

II. Optical Gyroscope

The light wave moves with speed $c' = \dfrac{c}{\mu}$ in the medium having refractive index $\mu$ . Wavelength of light in medium $\lambda' = \dfrac{\lambda}{\mu}$ , where $\lambda$ is the wavelength of light in vacuum.

a) (2 points)

transit time for the CW beam: $t^{+} = \dfrac{2\pi R + R\Omega t^{+}}{c'} = \dfrac{2\pi R}{c'}\left(1 - \dfrac{R\Omega}{c'}\right)^{-1}$

transit time for the CCW beam: $t^{-} = \dfrac{2\pi R - R\Omega t^{-}}{c'} = \dfrac{2\pi R}{c'}\left(1 + \dfrac{R\Omega}{c'}\right)^{-1}$

the time difference between $t^{+}$ and $t^{-}$ : $\Delta t = \dfrac{4\pi R^2\Omega}{(c')^2 - R^2\Omega^2}$

since $(R\Omega)^2 \ll (c')^2$ $\quad \Delta t \approx \dfrac{4\pi R^2\Omega}{(c')^2}$

b) (2 points) the round-trip optical path difference, $\Delta L$ , is given by

\Delta L = c'\Delta t = \frac{4\pi R^2\Omega}{c'}

c) (1 point) $\quad \Delta \mathrm{L} \cong 4.5\mathrm{x}10^{-12}\ \mathrm{m}.$

d) (1 point) the corresponding optical phase difference $\Delta\theta$ is,

\Delta\theta = \frac{2\pi\Delta L}{\lambda'} = \frac{8\pi^2 R^2\Omega}{c\lambda'}, \text{ where } \lambda' = \frac{\lambda}{\mu}

for $N$ turns of fiber optic ring,

\Delta\theta = \frac{8\pi^2 R^2 N\Omega}{c\lambda'}

e) (2 points)

Equilateral triangle ABC with base BC of length L/3 marked by a double arrow; dotted medians meet at the centre O; from O a line R goes to a point on side AC, where the velocity RΩ is drawn at angle θ to the side and its component RΩ cosθ along AC is shown; h is the perpendicular distance from O to AC, with angle θ at O.

The figure shows the triangular ring rotating about the centre o with the angular speed $\Omega$ in the clockwise direction. Without loosing generality, let‘s first consider the velocity of light along AC in the CW and CCW direction,

$v_{\pm} = c \pm R\Omega\cos\theta = c \pm \Omega h$ , where h is constant.

\tau_{\pm} = \frac{L/3}{v_{\pm}} = \frac{L/3}{c \pm \Omega h} \approx \frac{L/3}{c}\left(1 \mp \frac{\Omega h}{c}\right)

where $\tau_{\pm}$ is the time taken for light travelling along AC in the CW and CCW.

$t_{\pm} = \dfrac{L}{v_{\pm}} = \dfrac{L}{c \pm \Omega h} \approx \dfrac{L}{c}\left(1 \mp \dfrac{\Omega h}{c}\right)$ , where L is the perimeter of the triangular ring.

Therefore, the time difference of light travelling in one complete cycle.

$\Delta t = \dfrac{2\Omega L h}{c^2} = \dfrac{4\Omega}{c^2}\left(\dfrac{1}{2}Lh\right) = \dfrac{4\Omega A}{c^2}$ , where $A$ is the area of the triangular ring.

f) The resonance frequencies associated with $L_{\pm}$ corresponding to the effective cavity lengths seen by CW and CCW propagating beams respectively is,

L_{+} = ct^{+} \approx L\left(1 - \frac{\Omega h}{c}\right)

L_{-} = ct^{-} \approx L\left(1 + \frac{\Omega h}{c}\right)

where $L_{\pm}$ is the perimeter of the equilateral triangle in the CW (+) and CCW (-) and we also use the fact that $h\Omega \ll c$ . Therefore,

\Delta L = L_{-} - L_{+} = 2L\frac{\Omega h}{c} = \frac{4\Omega A}{c} = \frac{\Omega L^2}{\sqrt{3}\ c}

The condition to sustain the laser oscillation (given in the problem),

$\nu_{\pm} = \dfrac{m}{L_{\pm}}c$ , m = 1, 2, 3, … integers (1 point)

$\Delta\nu = \nu_{-} - \nu_{+} = \dfrac{m}{L_{-}}c - \dfrac{m}{L_{+}}c \approx mc\dfrac{\Delta L}{L^2} = \nu\dfrac{\Delta L}{L}$ (1 point)

the approximation arises from $L_{+}L_{-} \approx L^2$

where L is the perimeter of the triangular ring. Hence,

$\Delta\nu = \dfrac{\Delta L}{L}\nu = \dfrac{4A}{Lc}\nu\Omega = \dfrac{1}{\sqrt{3}}\dfrac{L}{\lambda}\Omega$ (1 point)

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf