Two coherent monochromatic light beams of intensities $I$ and $4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are

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 plane light waves of equal amplitude make a small angle $\alpha (\alpha \ll 1)$ with each other. They fall almost normally on a screen. If $\lambda$ is the wavelength of the light waves,the fringe width $\Delta x$ of the interference pattern of the two sets of waves on the screen is:

Two light waves of intensity $I$ and $4I$ superpose at point $A$ with zero phase difference and at point $B$ with a phase difference of $\frac{\pi}{2}$. Calculate the difference of resultant intensities at point $A$ and $B$. (in $I$)

In a double-slit interference experiment,if the path difference at a point on the screen for yellow light is $3\lambda/2$,then the fringe at that point will be . . . . . .

In the interference of two coherent sources of different intensities,the ratio of maximum and minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{144}{81}$. What is the ratio of their amplitudes?