$A$ monochromatic beam of light falls on a $YDSE$ apparatus at an angle $\theta$ as shown in the figure. $A$ thin sheet of glass of thickness $t$ and refractive index $\mu$ is inserted in front of the lower slit $S_2$. The central bright fringe (path difference $= 0$) will be obtained:

$A$ 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 ............... $mm$.

$A$ 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:

$6300 \mathring A$ wavelength light shines on two narrow slits separated by a distance of $1.0 \ mm$ and illuminates a screen at a distance of $1.5 \ m$ away. When one slit is covered by a thin glass plate of refractive index $1.8$ and the other slit by a thin glass plate of refractive index $\mu$,the central maxima shifts by $6^o$. Both plates have the same thickness of $0.5 \ mm$. The value of the refractive index $\mu$ of the plate is:

The slits in a double-slit interference experiment are illuminated by orange light $(\lambda = 600 \ nm)$. $A$ thin transparent plastic sheet of thickness $t$ is placed in front of one of the slits. The number of fringes $(N)$ shifting on the screen is plotted versus the refractive index $\mu$ of the plastic in the graph shown. The value of $t$ is

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: