If two sources emit light waves of different amplitudes, then:

Two waves having same intensity $I_{0}$ and originated from two non-coherent sources superpose at a point. The average intensity at that point is . . . . . . .

The distance between a point source of light and a screen is $60 \ cm$. If this distance is increased to $180 \ cm$,what will be the intensity on the screen compared to the original intensity?

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,

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

"If you illuminate two pinholes using two lamps, the interference pattern will not be observed" - Explain.