When two light waves of equal intensity superimpose, the maximum intensity obtained is $I$. If the intensity of one of the waves is quadrupled, then the maximum intensity obtained is

The light beams of intensities in the ratio of $9: 1$ are allowed to interfere. What will be the ratio of the intensities of maxima and minima?

If two sources emit light waves of different amplitudes, then in the interference pattern:

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?

Two coherent sources separated by a distance $d$ are radiating in phase with a wavelength $\lambda$. $A$ detector moves in a large circle around the two sources in the plane of the two sources. The angular position of the $n = 4$ interference maxima is given as

Two light waves of intensities $I$ and $2I$ superimpose on each other. If the path difference between the light waves reaching a point is $12.5 \%$ of the wavelength of the light,then the resultant intensity at the point is (Both the light waves have same wavelength).