In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

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 ........ $mm$.

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

In a Young's double slit experiment,two slits are illuminated with light of wavelength $\lambda = 800 \, nm$. The line joining $A_1 P$ is perpendicular to $A_1 A_2$ as shown in the figure. If the first minimum is detected at $P$,the value of slit separation '$a$' will be $......... \, mm$. The distance of the screen from the slits is $D = 5 \, cm$.

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

In a Young's double slit experiment,green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced?