$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

In a Young's double slit experiment,the central point on the screen is

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

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:

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

Young's double-slit experiment is performed using microwaves of wavelength $\lambda = 3 \, cm$. The distance between the plane of the slits and the screen is $D = 100 \, cm$,and the distance between the slits is $d = 5 \, cm$. Find $(a)$ the number of maxima and $(b)$ their positions on the screen.