In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

In the $Young's$ double slit experiment, the intensity produced by each of the individual slits is $I_0$. The distance between the two slits is $2 \ mm$. The distance of the screen from the slits is $10 \ m$. The wavelength of light is $6000 \ \mathring{A}$. What is the intensity of light on the screen in front of one of the slits?

In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

Imagine a Young's double-slit interference experiment performed with waves associated with fast-moving electrons produced from an electron gun. The distance between successive maxima (fringe width) will decrease if:

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)