In a $YDSE$,if the slits are of unequal width:

In a $YDSE$ apparatus,two identical slits are separated by $1\, mm$ and the distance between the slits and the screen is $1\, m$. The wavelength of light used is $6000\, \mathring{A}$. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is ......$mm$.

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

In a certain region of the screen, the number of fringes formed is observed to be $8$ in Young's double slit experiment when light of wavelength $6600 \text{ Å}$ is used. If light of wavelength $4400 \text{ Å}$ is used, the number of fringes observed in the same region of the screen will be:

Using Young's double slit experiment,a monochromatic light of wavelength $5000 \,\mathring A$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000 \,\mathring A$ is used and the separation between the slits is doubled,then the new fringe width will be ............... $mm$.

If yellow light in the Young's double slit experiment is replaced by red light,the fringe width will