In a standard $YDSE$ setup,two coherent sources of light of intensity ratio $\beta$ produce an interference pattern. The value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ is equal to (where $I_{\max}$ and $I_{\min}$ are the maximum and minimum intensities of the resultant wave).

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 $: 1$)

The two coherent sources produce interference with intensity ratio $b$. In the interference pattern, the ratio $\frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}}$ will be

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 intensity variation in the interference pattern obtained with the help of two coherent sources is $5\%$ of the average intensity. Find out the ratio of intensities of the two sources.

Four identical monochromatic sources $A, B, C, D$ as shown in the figure produce waves of the same wavelength $\lambda$ and are coherent. Two receivers $R_1$ and $R_2$ are at great but equal distances from $B$.
$(i)$ Which of the two receivers picks up the larger signal?
$(ii)$ Which of the two receivers picks up the larger signal when $B$ is turned off?
$(iii)$ Which of the two receivers picks up the larger signal when $D$ is turned off?
$(iv)$ Which of the two receivers can distinguish which of the sources $B$ or $D$ has been turned off?