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

$A$ thin transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the two interfering waves. The path difference between these waves is changed by half the wavelength of light $\lambda$. The thickness of the plate is

The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.

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 $D$.

Young's double slit experiment is carried out with two thin sheets of thickness $t = 10.4 \, \mu m$ each and refractive indices $\mu_1 = 1.52$ and $\mu_2 = 1.40$ covering the slits $S_1$ and $S_2$,respectively. If white light of range $400 \, nm$ to $780 \, nm$ is used,then which wavelength will form maxima exactly at point $O$,the centre of the screen?