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

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

In a Young's experiment,when $n$ coherent sources of equal intensity $I_0$ are used,the resultant intensity is $I_1$. When $n$ incoherent sources of equal intensity $I_0$ are used,the resultant intensity is $I_2$. What are the values of $I_1$ and $I_2$?

Two sources of light are $0.6 \, mm$ apart and the screen is placed at a distance of $1.2 \, m$ from them. $A$ light of wavelength $6000 \, \mathring{A}$ is used. The phase difference between the two light waves interfering on the screen at a point at a distance of $3 \, mm$ from the central bright band is:

In Young's double-slit experiment,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from one of the slits,what happens to the interference pattern?