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,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from slit $A$,then:

In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are sources of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is:

For Young's double slit experiment,two statements are given below :
Statement $I:$ If the screen is moved away from the plane of the slits,the angular separation of the fringes remains constant.
Statement $II:$ If the monochromatic source is replaced by another monochromatic source of larger wavelength,the angular separation of the fringes decreases.
In the light of the above statements,choose the correct answer from the options given below :

In a Young's double-slit experiment, the angular width of a fringe is $1^\circ$ for light of wavelength $6000 \, \mathring{A}$. What is the distance between the slits in $mm$?

In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.