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?

$S_{1}$ and $S_{2}$ are coherent point sources of light located in the $xy$-plane at points $(0,0)$ and $(0,3\lambda)$ respectively. Here $\lambda$ is the wavelength of light. At which one of the following points (given as coordinates) will the intensity of interference be maximum?

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_{1}$ through a medium of refractive index $n_{1}$ while the second wave travels a path of length $L_{2}$ through a medium of refractive index $n_{2}$. After this the phase difference between the two waves is:

Two coherent sources of intensities,$I_1$ and $I_2$ produce an interference pattern. The maximum intensity in the interference pattern will be

If the ratio of amplitudes of two interfering waves is $4: 3$,then the ratio of maximum and minimum intensity 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 $A^{2}$)