What is the time-averaged value of the intensity of a light wave,and at what phase difference is the intensity equal to $\frac{1}{2}$ of its maximum value?

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringe width from the centre is

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

The ratio of maximum to minimum intensity due to the superposition of two waves is $\frac{49}{9}$. Then the ratio of the intensities of the component waves is: