The contrast in the fringes in an interference pattern depends on

Four light sources produce the following four waves:
$(i)$ $y_1 = a \sin(\omega t + \phi_1)$
(ii) $y_2 = a \sin(2\omega t)$
(iii) $y_3 = d' \sin(\omega t + \phi_2)$
(iv) $y_4 = d' \sin(3\omega t + \phi)$
Superposition of which two waves gives rise to interference?

Two waves having same intensity $I_{0}$ and originated from two non-coherent sources superpose at a point. The average intensity at that point is . . . . . . .

Two coherent point sources $S_1$ and $S_2$ are separated by a small distance '$d$' as shown. The fringes obtained on the screen will be

Two sources $S_1$ and $S_2$ emit waves of wavelength $\lambda$. For destructive interference at point $P$,the path difference $(S_1P - S_2P)$ must be:

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.....$^\circ$.