QJEEAdv08Ph800a: JEE Advanced 2008 Physics - Class 12 JEEAdv08Ph800a Solution: In a Young's double slit experiment, the separation between the …

If $$d=\lambda$$, the maximum path difference (path difference is given by $$d\sin \theta$$) will be less than $$\lambda$$. So there will be only central maximum on the screen, because in the equation $$d\sin \theta =n\lambda ,n$$ can take only one value.

If $$\lambda <d<2\lambda$$, then the maximum path difference will be less than 2$$\lambda$$. So there will be two more maximum on the screen in addition to the central maximum.

Intensity of dark fringes becomes zero when intensities at the two slits are equal. Initial intensity at both the slits is unequal so there will some brightness at dark fringes. Hence, when intensity of both slits is made equal, the intensity at dark fringes on screen will reduces to zero.