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.

In Young's double slit experiment, two slits $S_1$ and $S_2$ are $d$ distance apart and the separation from slits to screen is $D$ (as shown in figure). Now, if two transparent slabs of equal thickness $0.1 \, mm$ but refractive indices $1.51$ and $1.55$ are introduced in the path of the beam $(\lambda = 4000 \, \mathring{A})$ from $S_1$ and $S_2$ respectively, the central bright fringe spot will shift by $..........$ number of fringes.

$A$ thin plastic sheet of refractive index $1.6$ is used to cover one of the slits of a double slit arrangement. The central point on the screen is now occupied by what would have been the $7^{th}$ bright fringe before the plastic was used. If the wavelength of light is $600 \ nm$, what is the thickness (in $\mu m$) of the plastic sheet?

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 $...... \, \mu m$.

$A$ double slit experiment is performed with light of wavelength $500\, nm$. $A$ thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is: