In a $YDSE$ setup,the intensity due to two coherent beams differs from each other by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima 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 $I$)

Two waves have their amplitudes in the ratio $1 : 9$. The maximum and minimum intensities when they interfere are in the ratio

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).

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

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?