Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

In a certain double-slit experimental arrangement,interference fringes of width $1.0 \ mm$ each are observed when light of wavelength $5000 \ \mathring{A}$ is used. Keeping the setup unaltered,if the source is replaced by another source of wavelength $6000 \ \mathring{A}$,the fringe width will be $...... \ mm$.

In Young's double-slit experiment,which of the following statements is correct?

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

In the $Young's$ double slit experiment,the intensity produced by each of the individual slits is $I_0$. The distance between the two slits is $2 \ mm$. The distance of the screen from the slits is $10 \ m$. The wavelength of light is $6000 \ \mathring{A}$. What is the intensity of light on the screen in front of one of the slits?