If the ratio of the intensities of two waves producing interference is $49: 16$,then the ratio of the resultant maximum intensity to minimum intensity will be

$A$ and $B$ are two interfering sources where $A$ is ahead in phase by $54^{\circ}$ relative to $B$. The observation is taken from point $P$ such that $PB-PA=2.5 \lambda$. Then the phase difference between the waves from $A$ and $B$ reaching point $P$ is (in rad) (in $\pi$)

In an interference experiment,the ratio of amplitudes of coherent waves is $\frac{a_1}{a_2} = \frac{1}{3}$. The ratio of maximum and minimum intensities of fringes will be

Two coherent plane waves of identical frequency, polarization, and intensity $I$ interfere at a point where they differ in phase by $60^{\circ}$. What is the resulting intensity?

Two coherent monochromatic light beams of intensities $I$ and $4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are

Two point sources $X$ and $Y$ emit waves of the same frequency and speed,but $Y$ lags in phase behind $X$ by $2\pi l$ radians. If there is a maximum in direction $D$,the distance $XO$ (where $n$ is an integer) is given by: