Fringe shift
An optical-fibre temperature sensor is constructed using a Mach-Zehnder interferometer arrangement. The length of fibre in one arm that is exposed to the measurement is $4 cm$. The refractive index of the fibre is $1.49$ and its temperature coefficients of linear expansion and refractive index are $6.5*10^{-7} (1/deg C)$ and $9.5*10^{-6} (1/ deg C)$ respectively. Calculate the fringe shift per degree for the sensor when the light source is an $820$ nm laser diode.
The total path traveled by the light in an arm of length $L_0 =4 cm$ is
$z =n_0*L_0$
By differentiating this path length z we get:
$dz =n_0*d L +L_0*d n$
But from the thermal expansion law we know that ($\alpha_1$ and $\alpha_2$ are the thermal expansion coefficients)
$d L =\alpha_1*L_0*d T$ and $d n=\alpha_2*n_0*d T$
$dz =n_0*L_0*(\alpha_1+\alpha2)*d T$
For $d T = 1 deg C$ we have a difference of path for the light of
$dz =1.49*0.04*(6.5*10^{-7} +9.5*10^{-6})= 604.94 nm$
Making the observation that for a complete shift of 1 fringe the wave phase difference of light is $2\pi$ and writing the phase difference as
$\Phi =(2\pi*dz)/\lambda$
one can say that 1 fringe shift is equivalent to a path difference $dz = \lambda$
Therefore the fringe shift is ($D_f$ is the distance between two consecutive fringes)
$d(D_f)/D_f =dz/\lambda =604.94/820=0.738$
