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 Young's double slit experiment,when two light waves form the third minimum,they have:

In Young's double slit experiment, two slits are illuminated with light of wavelength $\lambda$. The line joining $A_1P$ is perpendicular to $A_1A_2$. If the first minimum is detected at $P$, the value of slit separation '$a$' will be (where $D$ is the distance between the source and the screen):

In Young's double-slit experiment,the distance between the two slits is $0.2 \, mm$ and the screen is $200 \, cm$ away from the slits. If the wavelength of light used is $5000 \, \mathring{A}$,find the distance of the third bright fringe from the central bright fringe in $cm$.

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ 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-a-half times the previous value, then the fringe width is: (in $\text{mm}$)