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.

In a double slit experiment,when light of wavelength $400 \; nm$ was used,the angular width of the first minima formed on a screen placed $1 \; m$ away was found to be $0.2^{\circ}$. What will be the angular width of the first minima if the entire experimental apparatus is immersed in water (in $^{\circ}$)? $(\mu_{water} = 4/3)$

In $YDSE$, a thin film $(\mu=1.6)$ of thickness $0.01 \,mm$ is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the $10^{\text{th}}$ bright fringe earlier. The wavelength of the wave is ......... $\mathring{A}$.

In an ideal Young's double-slit experiment,a glass plate of thickness $t$ and refractive index $\mu = 1.5$ is placed in the path of one of the interfering beams. If the central bright fringe shifts to the position originally occupied by the first bright fringe (corresponding to wavelength $\lambda$),then the minimum thickness $t$ of the glass plate is:

In a double-slit experiment,green light $(5303\,\mathring{A})$ falls on a double slit having a separation of $19.44\,\mu m$ and a width of $4.05\,\mu m$. The number of bright fringes between the first and the second diffraction minima is

The figure shows a double-slit experiment. Each slit has a width $W$. $A$ thin glass slab of thickness $\delta$ and refractive index $\mu$ is placed in front of one of the slits. The intensity at the central point $C$ is measured as a function of the thickness $\delta$. At what thickness $\delta$ will the intensity at $C$ be minimum?