In Young's double-slit experiment performed using a monochromatic light of wavelength $\lambda$,when a glass plate $(\mu=1.5)$ of thickness $t = x \lambda$ is introduced in the path of one of the interfering beams,the intensity at the position where the central maximum occurred previously remains unchanged. The value of $x$ will be..........

$A$ 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 $D = 100 \, cm$.

In Young's double slit experiment,a glass plate is placed before one slit which absorbs half the intensity of light. Under this case:

$A$ 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 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:

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 $d' = 0.8 \, cm$.