In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

In Young's double slit experiment,the central bright fringe can be identified

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:

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

In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is