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 $: 1$)

The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):

In a $YDSE$,bi-chromatic light of wavelengths $400 \, nm$ and $560 \, nm$ are used. The distance between the slits is $0.1 \, mm$ and the distance between the plane of the slits and the screen is $1 \, m$. The minimum distance between two successive regions of complete darkness is......$mm$.

Light of wavelength $6000 \,\mathring A$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. What is the distance between two consecutive minima in $mm$?

Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason : The fringe width is inversely proportional to the distance between the two slits.

If the separation between slits in Young's double slit experiment is reduced to $\frac{1}{3}rd,$ the fringe width becomes $n$ times. The value of $n$ is