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?

In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\frac{I_1 + I_2}{I_0} = $ $(\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2})$

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

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

The path difference between two interfering waves at a point on the screen is $\frac{\lambda}{8}$. The ratio of intensity at this point and that at the central fringe will be

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