If $\lambda $ is the wavelength of ligth used and the distanse between the slits in Youngs double slit experiment is d then comment on number of fringes and fringe pattern if ,

1. d is 0.99 $\lambda $

2. $\lambda < d<2 \lambda $

Also the intensity of light falling on 1 slit is 4 times that of other

what I thought :

In a Young double-slit experiment why is there only one maxima formed if distance between the slits approaches wavelength of light.

what I understand so far that

$$I= I_1 +I_2 +2\sqrt {I _1I_2}cos\phi ,$$

where $I $ is resultant intensity

$I_1$ and $I_2$ are intensity of both waves $\phi $ is the phase difference

but then if so then,

how would intensity and the number of fringe pattern vary (even mathematically) because $\phi $=$2\pi \delta x/\lambda$ where $ \delta x$ is the path difference.

So how would that affect intensity and number of fringes?

What I know so far is that if the intensities of both the lights used are same then the contrast would be best and we would be able to differentiate fringes properly.

But I am unable to think about the wavelength and intensity part.

And for second case : atleast one more maximum (besides the central maximum ) will be observed on the screen