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 a double-slit experiment,instead of taking slits of equal widths,one slit is made twice as wide as the other. Then,in the interference pattern:

When a glass plate of refractive index $1.44$ is introduced in the path of one of the interfering beams, the fringes are displaced by a distance '$y$'. If this plate is replaced by another plate of same thickness but of refractive index $1.66$, the fringes will be displaced by a distance

$A$ plate of refractive index $1.6$ is introduced in the path of light from one of the slits in Young's double slit experiment,then:

In a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.

The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,the central maximum shifts by a distance equal to $n$ fringe widths. If the wavelength of light used is $\lambda$,then $t$ will be: