The central fringe of an interference pattern produced by light of wavelength $6000 \, \mathring{A}$ is found to shift to the position of the fourth bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit. The thickness of the glass plate would be ...... $\mu m$.

In a $YDSE$,light of wavelength $\lambda = 5000 \; \mathring{A}$ is used,which emerges in phase from two slits separated by a distance $d = 3 \times 10^{-7} \; m$. $A$ transparent sheet of thickness $t = 1.5 \times 10^{-7} \; m$ and refractive index $\mu = 1.17$ is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern,and what is its linear position $y$ from the center of the screen?

To make the central fringe appear at the centre $O$ of the screen,a mica sheet of refractive index $\mu = 1.5$ is introduced in front of one of the slits. Choose the correct statement.

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:

As shown in the figure,two coherent sources $S_1$ and $S_2$ are separated by a very small distance $d$. The fringes produced on the screen will be ........

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index $1.56$, the central fringe shifts to the position of $7^{th}$ bright fringe, obtained with both slits uncovered. If the light source wavelength is $450 \text{ nm}$, the thickness of mica sheet is $\alpha \times 10^{-9} \text{ m}$. The value of $\alpha$ is . . . . . . .