In a Young's double slit experiment,green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced?

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ In Young's double slit experiment,the fringes produced by red light are closer as compared to those produced by blue light.
Reason $(R) :$ The fringe width is directly proportional to the wavelength of light.
In the light of above statements,choose the correct answer from the options given below $:$

In $Y.D.S.E.$ using red and blue lights of wavelengths $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$,the $n^{th}$ red fringe coincides with the $(n + 1)^{th}$ blue fringe. The value of $n$ is:

In Young's double-slit experiment,which of the following statements is correct?

The maximum intensity in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10d$?

$A$ Young's double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is