Monochromatic light of wavelength $500 \, nm$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $= 1.5$), the central maximum is shifted to a position previously occupied by the $4^{th}$ bright fringe. The thickness of the glass plate is ..................... $\mu m$.

In a Young's double slit experimental arrangement shown here,if a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of the slit $S_1$,then the path difference $(S_1P - S_2P)$

Calculate the wavelength of light used in an interference experiment from the following data: Fringe width $\beta = 0.03 \, cm$. The distance between the slits and the eyepiece is $D = 1 \, m$. The distance between the images of the virtual source,when a convex lens of focal length $f = 16 \, cm$ is used at a distance of $v = 80 \, cm$ from the eyepiece,is $d' = 0.8 \, cm$.

When a thin transparent plate of thickness $t$ and refractive index $\mu$ is placed in the path of one of the two interfering waves of light,then the path difference changes by

$A$ monochromatic light source $S$ of wavelength $440 \,nm$ is placed slightly above a plane mirror $M$ as shown below. The image of $S$ in $M$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0 \,cm$ and the distance of the screen from $O$ is $100.0 \,cm$ (figure is not to scale). If the angle $\theta = 0.50 \times 10^{-3} \,radians$, then the width of the interference fringes observed on the screen is ............... $mm$.

In the $YDSE$ shown,the two slits are covered with thin sheets having thickness $t$ and $2t$,and refractive index $2\mu$ and $\mu$ respectively. Find the position $(y)$ of the central maxima.