Interference

Remarks

The formula follows from $\abs{a + b}^2 = \abs{a}^2 + \abs{b}^2 + 2 \operatorname{Re}(a \bar b)$, using the modulus and argument of complex numbers. Over a region where $\Delta\varphi$ varies, the intensity oscillates between $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. The visibility of the fringes is $V = (I_{\max} - I_{\min}) / (I_{\max} + I_{\min}) = 2\sqrt{I_1 I_2} / (I_1 + I_2)$.

Example: two waves of equal intensity $I_0$ give $I = 2 I_0 (1 + \cos \Delta\varphi) = 4 I_0 \cos^2(\Delta\varphi / 2)$. This is $4 I_0$ at constructive points and $0$ at destructive points, with visibility $1$.

Non-example: for vector fields $E_1, E_2$ the interference term is $2 \operatorname{Re}(E_1 \cdot \overline{E_2})$. Two waves with orthogonal linear polarisations have $E_1 \cdot \overline{E_2} = 0$, so their intensities simply add and there are no fringes.