In $YDSE$ experiment,when two light rays form the third minima,they have:

In a double slit experiment with monochromatic light,fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} \,m$ towards the slits,the change in fringe width is $3 \times 10^{-3} \,cm$. If the distance between the slits is $1 \,mm$,then the wavelength of the light will be . . . . . . $nm$.

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,which of the following statements is correct?

The slits in a Young's double-slit experiment have equal widths, and the source is placed symmetrically relative to the slits. The intensity at the central fringe is $I_0$. If one of the slits is closed, the intensity at this point will be

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)