On replacing a thin film of mica of thickness $12 \times 10^{-5} \ cm$ in the path of one of the interfering beams in Young's double slit experiment using monochromatic light,the fringe pattern shifts through a distance equal to the width of bright fringe. If $\lambda = 6 \times 10^{-5} \ cm$,the refractive index of mica is:

In a double-slit experiment,the wavelength $\lambda$ of the light source is $400 \ nm$,the slit separation $d$ is $20 \ \mu m$,and the individual slit width $a$ is $4 \ \mu m$. Considering the interference of light from the two slits and the diffraction of light through each slit,find the number of bright interference fringes that lie within the central peak of the diffraction envelope.

$A$ 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:

In $YDSE$,the source $S$ placed symmetrically with respect to the slits $S_1$ and $S_2$ is now moved parallel to the plane of the slits so that it is closer to the upper slit $S_1$,as shown. Then,

The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,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: