Young's experiment demonstrates that . . . . . .

In Young's double-slit experiment,when two light waves form the third minimum,they have:

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.

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 $:$

The intensity of each source 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. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

In a Young's Double slit experiment,the first maxima is observed at a fixed point $P$ on the screen. Now,the screen is continuously moved away from the plane of the slits. The ratio of the intensity at point $P$ to the intensity at point $O$ (centre of the screen) is: