In an interference experiment,the phase difference for points where the intensity is minimum is $(n=1, 2, 3, \ldots)$

Two point sources $S_1$ and $S_2$ separated by a distance $10 \mu m$ emit light waves of wavelength $4 \mu m$ in phase. $A$ circular wire of radius $40 \mu m$ is placed around the sources as shown in the figure,where $O$ is the centre of the circle and $OS_1 = OS_2$. Then:

If two sources of light emit waves of different amplitudes and interfere, then:

The ratio of the intensities of two waves is $25 : 1$. If interference occurs,then the ratio of the maximum and minimum intensity should be:

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:

Light waves producing interference have their amplitudes in the ratio $3: 2$. The intensity ratio of maximum and minimum of interference fringes is