In Young's double-slit experiment,if the separation between the slit and the screen increases,the fringe width:

$A$ beam of light consisting of two wavelengths $7000 \; \mathring{A}$ and $5500 \; \mathring{A}$ is used to obtain an interference pattern in Young's double-slit experiment. The distance between the slits is $2.5 \; mm$ and the distance between the plane of the slits and the screen is $150 \; cm$. The least distance from the central fringe,where the bright fringes due to both the wavelengths coincide,is $n \times 10^{-5} \; m$. The value of $n$ is .............

In an interference experiment, the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5^{th}$ bright fringe at the same point (in $nm$)?

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made