In a Young's double slit experiment,the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength of the light used is $6280 \; \mathring{A}$. The distance between the two coherent sources is $...........\,mm$.

In a $YDSE$ apparatus,if we use white light,then:

Two thin parallel slits are made in an opaque screen. When a monochromatic beam of light passes through them at normal incidence,the first bright fringe in the transmitted light occurs at $\pm 45^{\circ}$ with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid,the same bright fringe now occurs at $\pm 30^{\circ}$. The refractive index of the liquid is:

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to: