Two coherent sources of intensities $I$ and $4I$ interfere. The maximum and minimum intensities of the resultant wave are:

Interference fringes are produced on the screen by using two light sources of intensities $I$ and $9I$. The phase difference between the beams is $\pi / 2$ at point $P$ and $\pi$ at point $Q$ on the screen. The difference between the resultant intensities at points $P$ and $Q$ is $(\cos 90^{\circ}=0, \cos 180^{\circ}=-1)$. (in $I$)

Interference cannot be observed in two sodium lamps of equal power because the two waves do not have a .....

If the wavelength of light is $4000 \text{ Å}$, then the number of waves in $1 \text{ mm}$ length will be . . . . . . -

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

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?