In Young's double-slit experiment,if the separation between two narrow slits is doubled,then in order to maintain the same fringe spacing,the distance $D$ of the screen from the slits must be changed to:

In an interference experiment,the distance between a maximum and the adjacent minimum is .......

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

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

In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to:

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.