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.
So with the artificial delay, now the b_1 's signal arrives at the correlator is:
M_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 simply M_1(t-\tau_{\sin}) to:
M_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} :
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 R_s with the added artificial delay \tau_{\sin} :
R_s &=E[M_0(t)M_1(t-\tau_{\sin})] \\ &=E[M_1(t-\tau_g)M_1(t-\tau_{\sin})] \\ &=E[ M\cos(\omega(t-\tau_g)) M\cos(\omega(t-\tau_{\sin})) ] \\ &=M^2 E[ \cos(\omega(t-\tau_g)+\phi_1{\scriptstyle (t)}) \; \cos(\omega(t-\tau_{\sin})+\phi_1{\scriptstyle (t)}) ] \\ &=M^2 E[ \cos(\omega(t-\tau_g)+\phi_1{\scriptstyle (t)}) \cos(\omega(t-\frac{1}{4\nu})+\phi_1{\scriptstyle (t)}) ] \\ &=M^2 E[ \cos(\omega t -\omega \tau_g+\phi_1{\scriptstyle (t)}) \cos(\omega t-\frac{\omega}{4\nu}+\phi_1{\scriptstyle (t)}) ] \\ &=M^2 E[ \cos(\omega t -\omega \tau_g+\phi_1{\scriptstyle (t)}) \cos(\omega t-\frac{2\pi \nu}{4\nu}+\phi_1{\scriptstyle (t)}) ] \\ &=M^2 E[ \cos(\omega t -\omega \tau_g+\phi_1{\scriptstyle (t)}) \cos(\omega t-\frac{\pi}{2}+\phi_1{\scriptstyle (t)})] \\ &=\frac{1}{2} M^2 E[ \cos(2\omega t -\omega \tau_g+\phi_1{\scriptstyle (t)} -\frac{\pi}{2}+\phi_1{\scriptstyle (t)} ) + \cos( -\omega \tau_g+\phi_1{\scriptstyle (t)} +\frac{\pi}{2}-\phi_1{\scriptstyle (t)} )] \\ &= \frac{1}{2} M^2 E \left[ \underbrace{ \bcancel{\cos (2\omega t -\omega \tau_g+\phi_1{\scriptstyle (t)} -\frac{\pi}{2}+\phi_1{\scriptstyle (t)} )}}_{\text{average}=0} + \cos( \underbrace{\frac{\pi}{2} -\omega \tau_g}_{\text{constant} }+\phi_1{\scriptstyle (t)} -\phi_1{\scriptstyle (t)} ) \right] \\ &=\frac{1}{2}M^2 \cos( \frac{\pi}{2} -\omega \tau_g) \\ &=\frac{1}{2}M^2 \sin( \omega \tau_g ) \\ &(\text{with } I=\frac{1}{2} M^2)\\ &= I \sin(\omega\tau_g) \\ &=I \sin\left( (2\pi\nu)\frac{\vec{b} \cdot \hat{s}}{\nu \lambda} \right) \\ &=I \sin\left(2\pi\frac{\vec{b} \cdot \hat{s}}{ \lambda} \right) \\ &=I \sin(2\pi ul )So the Sine Interferometer output R_s is:
R_s = I \sin(2\pi ul )1.2.3: Sine 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: