In a Young's double slit experiment,two slits are separated by $2 \, mm$ and the screen is placed $1 \, m$ away. When light of wavelength $500 \, nm$ is used,the fringe separation will be ........ $mm$.

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring{A}$ and $5200 \, \mathring{A}$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. What is the minimum distance from the central maximum where the bright fringes of both wavelengths coincide (in $, cm$)?

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

In Young's double slit experiment,the distance between two sources is $0.1 \, mm$. The distance of the screen from the sources is $20 \, cm$. The wavelength of light used is $5460 \, \mathring{A}$. The angular position of the first dark fringe is: (in $^\circ$)

In a double slit interference experiment,the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement,if the slit separation is increased by one-and-half times the previous value,then the fringe width is (in $ \text{ mm}$)