In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width

If the screen is moved away from the plane of the slits in a Young's double slit experiment,then the

Two slits separated by a distance of $1\, mm$ are illuminated with red light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed $1\, m$ from the slits. Find the distance between the third dark fringe and the fifth bright fringe on the same side of the central maxima.

In a Young's double slit experiment,the slit separation is $1 \, mm$ and the screen is $1 \, m$ from the slit. For a monochromatic light of wavelength $500 \, nm$,the distance of the $3rd$ minima from the central maxima is.....$mm$

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves: