In the setup shown in the figure,the two slits,$S_1$ and $S_2$,are not equidistant from the slit $S$. The central fringe at $O$ is 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 Young's double-slit experiment,the fringe width is found to be $0.4 \, mm$ when performed in air. If the entire apparatus is immersed in water,the new fringe width will be ........ (Refractive index of water $n = 4/3$).

$A$ 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

In $YDSE$,the source $S$ placed symmetrically with respect to the slits $S_1$ and $S_2$ is now moved parallel to the plane of the slits so that it is closer to the upper slit $S_1$,as shown. Then,

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: