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?

In a Young's double-slit experiment,a transparent plate of thickness $2.5 \times 10^{-5} \, m$ and refractive index $1.5$ is placed in the path of one of the beams. What will be the shift in the fringe pattern (in $, cm$)? The distance between the two slits $S_1$ and $S_2$ is $0.5 \, mm$ and the distance between the slits and the screen is $100 \, cm$.

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. $A$ thin film of thickness $t$ and refractive index $\mu$ is placed in front of $A$. Let $\beta =$ fringe width. The central maximum will shift:

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

If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

In a double slit arrangement,fringes are produced using light of wavelength $4800 \ \mathring A$. One slit is covered by a thin plate of glass of refractive index $1.4$ and the other with another glass plate of the same thickness but of refractive index $1.7$. By doing so,the central bright fringe shifts to the original fifth bright fringe from the center. The thickness of the glass plate is ...... $\mu m$.