Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in the figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1$ $(D >> \lambda)$. Find the distance $OP$.

In Young's double slit arrangement,water is filled in the space between the screen and the slits. Then:

Calculate the wavelength of light used in an interference experiment from the following data: Fringe width $\beta = 0.03 \, cm$. The distance between the slits and the eyepiece is $D = 1 \, m$. The distance between the images of the virtual source,when a convex lens of focal length $f = 16 \, cm$ is used at a distance of $v = 80 \, cm$ from the eyepiece,is $d' = 0.8 \, cm$.

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 . . . . . . .

$A$ 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

$A$ transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is placed in front of one of the slits in a Young's double-slit experiment. By what distance (in $cm$) will the interference pattern shift? The distance between the two slits is $d = 0.5 \, mm$ and the distance between the screen and the slits is $D = 100 \, cm$.