If the intensity of each wave in the observed interference pattern in Young's double slit experiment is $I_0$,then for some point $P$ where the phase difference is $\phi$,the resultant intensity $I$ will be:

The intensity ratio of the two interfering beams of light is $\beta$. What is the value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$?

Two light beams of intensities in the ratio of $9: 4$ are allowed to interfere. The ratio of the intensity of maxima and minima will be

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

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} = $ $(\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2})$