$A$ 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?

Interference fringes were produced in Young's double slit experiment using light of wavelength $5000 \, Å$. When a film of material $2.5 \times 10^{-3} \, cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe widths. The refractive index of the material of the film is

On replacing a thin film of mica of thickness $12 \times 10^{-5} \ cm$ in the path of one of the interfering beams in Young's double slit experiment using monochromatic light,the fringe pattern shifts through a distance equal to the width of bright fringe. If $\lambda = 6 \times 10^{-5} \ cm$,the refractive index of mica is:

$A$ thin glass plate of thickness $t = \frac{2500}{3} \lambda$ (where $\lambda$ is the wavelength of light used) and refractive index $\mu = 1.5$ is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits,the ratio of the intensities before and after the introduction of the glass plate is

$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.

$A$ double-slit experiment is performed with light of wavelength $500 \ nm$. If a thin plate of thickness $2 \ \mu m$ and refractive index $1.5$ is placed in front of one of the slits,the central fringe will shift by: