In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

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 $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

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?

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$,the path difference (in terms of an integer $n$) corresponding to any point having half the peak intensity is:

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda / 6$ is $I'$. If $I_0$ denotes the maximum intensity,then $I'/I_0$ is equal to

In the Young's double slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda / 4$ will be: