For constructive interference to take place between two monochromatic light waves of wavelength $\lambda$, the path difference should be

The frequency of a light ray having a wavelength of $3000 \ \mathring A$ is

In a double-slit interference experiment,if the path difference at a point on the screen for yellow light is $3\lambda/2$,then the fringe at that point will be . . . . . .

Light waves producing interference have their amplitudes in the ratio $3: 2$. The intensity ratio of maximum and minimum of interference fringes is

Four monochromatic and coherent sources of light,emitting waves in phase of wavelength $\lambda$,are placed at the points $x = 0, d, 2d$,and $3d$ on the $x$-axis. The intensity of the waves reaching a point $P$ far away on the $+x$ axis from each of the four sources is almost the same,and equal to $I_0$. Then,

Interference fringes are produced on a screen by using two light sources of intensities $I$ and $9I$. The phase difference between the beams is $\frac{\pi}{2}$ at the point $P$ and $\pi$ at the point $Q$ on the screen. The difference between the resultant intensities at point $P$ and $Q$ is (in $I$)