In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band. The wavelength of light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

$A$ Young's double slit experiment apparatus has slits separated by $0.2 \ mm$ and a screen $60 \ cm$ away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $\mu = \frac{11}{9}$ and the slits are illuminated with green light $(\lambda = 550 \ nm$ in vacuum$)$. Find the fringe width of the pattern formed on the screen. (in $mm$)

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:

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 ............. $\omega$.