In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

In a Young's double-slit experiment,the distance between the two sources is $0.1 / \pi \, mm$. The distance between the source and the screen is $25 \, cm$. The wavelength of light is $5000 \, \mathring{A}$. The angular position of the first dark fringe is ........$^o$.

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

In Young's double-slit experiment,if the separation between two narrow slits is doubled,then in order to maintain the same fringe spacing,the distance $D$ of the screen from the slits must be changed to:

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}$)