The energy in the phenomenon of interference

In the figure below,$P$ and $Q$ are two equally intense coherent sources emitting radiation of wavelength $20\, m$. The separation between $P$ and $Q$ is $5\, m$ and the phase of $P$ is ahead of that of $Q$ by $90^{\circ}$. $A, B$ and $C$ are three distinct points of observation,each equidistant from the midpoint of $PQ$. The intensities of radiation at $A, B, C$ will be in the ratio

Two waves of same intensity $I_0$ emitted from two sources having same phase difference $(\phi)$. Due to superposition of two waves, the intensity of resultant wave is directly proportional to . . . . . . .

Radio waves coming at an angle $\alpha$ to the vertical are received by a radar after reflection from a nearby water surface and directly. What should be the height of the antenna from the water surface so that it records a maximum intensity? (wavelength $= \lambda$)

If the ratio of the amplitudes of two interfering sources is $3 : 5$,then the ratio of the intensity of maxima and minima in the interference pattern will be

Two coherent monochromatic point sources $S_1$ and $S_2$ of wavelength $\lambda = 600 \ nm$ are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance $d = 1.8 \ mm$. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct?
$[A]$ $A$ dark spot will be formed at the point $P_2$
$[B]$ At $P_2$ the order of the fringe will be maximum
$[C]$ The total number of fringes produced between $P_1$ and $P_2$ in the first quadrant is close to $3000$
$[D]$ The angular separation between two consecutive bright spots decreases as we move from $P_1$ to $P_2$ along the first quadrant