When one of the slits in Young's experiment is covered with a transparent sheet of thickness $3.6 \times 10^{-3} \ cm$,the central fringe shifts to a position originally occupied by the $30^{th}$ bright fringe. If $\lambda = 6000 \ \mathring{A}$,then what is the refractive index of the sheet?

Young's double slit experiment is performed with monochromatic light. $A$ thin film is introduced in front of one of the slits.

$A$ 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 figure shown,if a parallel beam of white light is incident on the plane of the slits,then the distance of the white spot on the screen from $O$ is [Assume $d << D, \lambda << d$].

$A$ thin mica sheet of thickness $2 \times 10^{-6} \ m$ and refractive index $\mu = 1.5$ is introduced in the path of the first wave. The wavelength of the wave used is $5000 \ \mathring{A}$. The central bright maximum will shift:

In $YDSE,$ the source placed symmetrically with respect to the slits is now moved parallel to the plane of the slits so that it is closer to the upper slit,as shown. Then,