Because the difference between \cos and \sin is the phase shift of \frac{\pi}{2} . , so Sine Interferometer is just a Cosine Interferometer with phase shift of \frac{\pi}{2} .
To achieve this phase shift, we add an artificial time delay \tau_{\sin} at station b_1 as shown in the figure below.
Figure
So with the artificial delay, now the b_1 's signal arrives at the correlator is:
x_1(t-\tau_{\sin}) = M\cos(\omega (t-\tau_{\sin})) = M\cos(\omega (t-\tau_{\sin}) + \phi_1{\scriptstyle(t-\tau_{\sin})})We assume that \tau_{\sin} \ll \tau_c , the coherence time, thus:
\phi_1(t-\tau_{\sin}) \approx \phi_1(t)so we can simplify x_1(t-\tau_{\sin}) to:
x_1(t-\tau_{\sin})=M\cos\big( \omega (t-\tau_{\sin}) + \phi_1{\scriptstyle(t)}\big)1.2.1: Solving for
Now what value do we use for \tau_{\sin} ?
Since we we need a \frac{\pi}{2} phase shift, by looking at the figure below, phase = \frac{\pi}{2} corresponds to length = \frac{\lambda}{4} , from which we then can calculate the delay \tau_{\sin} :
Figure
So given the period of the wave is T , the delay \tau_{\sin} is calculated as:
\tau_{\sin} &=\frac{T}{4} \\ &(\text{given }T=\frac{1}{\nu})\\ &=\frac{1/\nu}{4} \\ &=\frac{1}{4\nu} \\1.2.2: Putting It All Together
Now let's see the correlator output v_{R_s} with the added artificial delay \tau_{\sin} :
v_{R_s}(t,\tau_g,\tau_{\sin})|_T =& \langle x_2(t) , x_1(t-\tau_{\sin}) \rangle \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_2(\psi)x_1(\psi-\tau_{\sin})} \, \color{green}d\psi \\ =& {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_1(\psi-\tau_g)x_1(\psi-\tau_{\sin})} \, \color{green}d\psiLet's break v_{R_s}(t,\tau_g,\tau_{\sin})|_T down:
Step 1: Multiplication:
\color{blue}x_1(\psi-\tau_g)x_1(\psi-\tau_{\sin}) &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega (\psi-\tau_{\sin}) + \phi_1{\scriptstyle(\psi)}) \\ &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega (\psi-\frac{1}{4\nu}) + \phi_1{\scriptstyle(\psi)}) \\ &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega \psi-\frac{\omega}{4\nu} + \phi_1{\scriptstyle(\psi)}) \\ &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega \psi-\frac{2\pi \nu}{4\nu} + \phi_1{\scriptstyle(\psi)}) \\ &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega \psi-\frac{\pi}{2} + \phi_1{\scriptstyle(\psi)}) \\ &= \frac{1}{2}M^2\left[\cos\big(2 \omega \psi- \omega \tau_g + 2\phi_1{\scriptstyle(\psi)} -\frac{\pi}{2} \big)+\cos(-\omega \tau_g +\frac{\pi}{2} )\right] \\ \\ &(\text{let } I=\frac{1}{2} M^2 , \text{ where $I$ is referred as Intensity and } I \propto M^2)\\ \\ &= I \cos\big(2 \omega \psi- \omega \tau_g + 2\phi_1{\scriptstyle(\psi)} -\frac{\pi}{2} \big)+I \cos(\omega \tau_g -\frac{\pi}{2} ) \\ &= I \cos\big(2 \omega \psi- \omega \tau_g + 2\phi_1{\scriptstyle(\psi)} -\frac{\pi}{2} \big)+I \sin(\omega \tau_g ) \\
Step 2: Average over integration time T :
(The first integral term = 0 is from Time average of cosine for fast varying random phase :\text{1-0-2} )
So the Sine Interferometer output v_{R_s} is:
v_{R_s}(\tau_g)= I \sin(\omega\tau_g ) = I \sin(2\pi ul ) =v_{R_s}(u,l)1.2.3: Fringe
Below figure you see the Fringe Pattern for Sine Interferometer. The difference in Fringe Pattern compare with Cosine Interferometer a \frac{\pi}{2} phase shift.
Feel free to drag around the blue triangle source \color{blue} s and the brown circle \color{brown} u , and observe that:
Figure :
Interactive \sin Fringe demonstration: draggable {\color{blue}\text{source } s}, {\color{brown}\text{baseline }u}