In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: If Young's double slit experiment is performed in an optically denser medium than air,then the consecutive fringes come closer.
Reason $(R)$: The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements,choose the most appropriate answer from the options given below:

In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In Young's double slit experiment,carried out with light of wavelength $5000 \ \mathring{A}$,the distance between the slits is $0.3 \ \text{mm}$ and the screen is at $200 \ \text{cm}$ from the slits. The central maximum is at $x=0 \ \text{cm}$. The value of $x$ for the third maxima is ............. $\text{mm}$.