In an interference pattern,two waves of intensities $I$ and $4I$ are used. At point $A$,the phase difference is $\frac{\pi}{2}$,and at point $B$,the phase difference is $\pi$. The difference in intensities at points $A$ and $B$ is: (in $I$)

Two monochromatic light waves of amplitudes $3A$ and $2A$ interfering at a point have a phase difference of $60^{\circ}$. The intensity at that point will be proportional to.......$A^2$

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 $I$)

The resultant amplitude of the interference of two waves $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{3})$ is:

Two monochromatic beams of intensities $I$ and $4I$ respectively are superposed to form a steady interference pattern. The maximum and minimum intensities in the pattern are

Four light waves are represented by:
$(i)$ $y = a_1 \sin \omega t$
(ii) $y = a_2 \sin (\omega t + \phi)$
(iii) $y = a_1 \sin 2\omega t$
(iv) $y = a_2 \sin 2(\omega t + \phi)$
Interference fringes may be observed due to the superposition of: