If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

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:

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

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:

In a Young's double slit experiment,the fringe width will remain the same if ($D =$ distance between screen and plane of slits,$d =$ separation between two slits,and $\lambda =$ wavelength of light used).

In Young's double slit experiment,white light is used. The separation between the slits is $b$. The screen is at a distance $d$ $(d >> b)$ from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are