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 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 intensities of two interfering waves is $9:1$. The ratio of the maximum amplitude to the minimum amplitude of the resultant wave is ........

In the Young's double slit experiment,the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$,then $\frac{I_1}{I_2}$ is

In Young's double slit experiment,one of the slits is wider than the other,so that the amplitude of the light from one slit is double that of the other. If $I_m$ is the maximum intensity,the resultant intensity $I$ when they interfere at a phase difference $\phi$ is given by

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?