In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ 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 $I$)

Obtain the formula for the intensity of light if the phase difference at a point from two coherent sources is $\phi$.

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.

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?