The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

In Young's double slit experiment,if the two slits are illuminated with separate sources,no interference pattern is observed because

In Young's experiment,the ratio of maximum to minimum intensities of the fringe system is $4:1$. The amplitudes of the coherent sources are in the ratio: (in $:1$)

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

Two slits are $1 \, mm$ apart from each other and the distance of the screen is $1 \, m$. If illuminated with light of wavelength $5 \times 10^{-7} \, m$,the distance between the $3^{rd}$ dark fringe and the $5^{th}$ bright fringe is......$mm$.

In Young's double-slit experiment,the distance between the two slits is $2 \times 10^{-3} \, m$ and the distance between the slits and the screen is $2.5 \, m$. The wavelength of the light used ranges from $2000 \, \mathring{A}$ to $9000 \, \mathring{A}$. What wavelength (in $\mathring{A}$) will form a bright fringe at a distance of $10^{-3} \, m$ from the central maximum?