The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

In Young's experiment,obtain the distance between two consecutive bright fringes and two consecutive dark fringes.

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

In Young's double slit arrangement, slits are separated by a gap of $0.5 \, mm$, and the screen is placed at a distance of $0.5 \, m$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890 \, \text{Å}$ is