The correct formula for fringe visibility is

Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$. The intensity at that point will be proportional to (in $A^{2}$)

Two coherent sources produce interference fringes with an intensity ratio of $81:1$. Find the ratio of the maximum to the minimum intensity in the interference pattern.

The figure shows two point sources which emit light of wavelength $\lambda$ in phase with each other and are at a distance $d = 5.5 \lambda$ apart along a line which is perpendicular to a large screen at a distance $L$ from the center of the sources. Assume that $d$ is much less than $L$. Which of the following statement$(s)$ is (are) correct?

Among the two interfering monochromatic sources $A$ and $B$; $A$ is ahead of $B$ in phase by $66^\circ$. If the observation is taken from point $P$,such that $PB - PA = \lambda / 4$. Then the phase difference between the waves from $A$ and $B$ reaching $P$ is.....$^\circ$.

Which characteristic of light determines its color?