When a thin plastic film of refractive index $\mu = 1.45$ is placed in the path of one of the interfering waves,the central fringe shifts by a distance equal to $5$ fringes. If the wavelength of light used is $5890 \, \mathring{A}$,find the thickness of the film.

The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,is:

$6300 \mathring A$ wavelength light shines on two narrow slits separated by a distance of $1.0 \ mm$ and illuminates a screen at a distance of $1.5 \ m$ away. When one slit is covered by a thin glass plate of refractive index $1.8$ and the other slit by a thin glass plate of refractive index $\mu$,the central maxima shifts by $6^o$. Both plates have the same thickness of $0.5 \ mm$. The value of the refractive index $\mu$ of the plate is:

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 glass slab of thickness $1.2 \, \mu m$ and refractive index $1.5$ is placed in front of one slit. Another slab of thickness $t$ and refractive index $2.5$ is placed in front of the other slit. If the position of the central fringe remains unchanged,then the thickness $t$ is equal to $...... \, \mu m$.

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