In the phenomenon of interference:

The two coherent sources of equal intensity produce a maximum intensity of $100$ units at a point. If the intensity of one of the sources is reduced by $36\%$ by reducing its width,then the intensity of light at the same point will be:

Two waves of intensity $I$ undergo interference. The maximum intensity obtained is

When a light wave undergoes reflection at the interface from air to glass,the change in phase of the reflected wave is equal to

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 intensity of light depends upon: