If the amplitude ratio of two sources producing interference is $3:5$,the ratio of intensities at maxima and minima is

Two monochromatic coherent light waves of amplitudes $A$ and $2A$,interfering at a point,have a phase difference of $60^{\circ}$. The intensity at that point will be proportional to (in $A^2$)

Two monochromatic and coherent point sources of light are placed at a certain distance from each other in the horizontal plane. The locus of all those points in the horizontal plane which have constructive interference will be

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

Four light sources produce the following four waves:
$(i)$ $y_1 = a \sin(\omega t + \phi_1)$
(ii) $y_2 = a \sin(2\omega t)$
(iii) $y_3 = d' \sin(\omega t + \phi_2)$
(iv) $y_4 = d' \sin(3\omega t + \phi)$
Superposition of which two waves gives rise to interference?

The optical path difference between two identical light waves arriving at a point is $31.5 \lambda$,where $\lambda$ is the wavelength of light used. The point is [Given: Two light sources are coherent]