In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

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

To make the central fringe at the centre $O$,a mica sheet of refractive index $1.5$ is introduced. Choose the correct statement$(s)$.

As shown in the figure,in Young's double slit experiment,a thin plate of thickness $t = 10\,\mu m$ and refractive index $\mu_1 = 1.2$ is inserted in front of slit $S_1$. The experiment is conducted in air $(\mu = 1)$ and uses a monochromatic light of wavelength $\lambda = 500\,nm$. Due to the insertion of the plate,the central maxima is shifted by a distance of $x\beta_0$,where $\beta_0$ is the fringe-width before the insertion of the plate. The value of $x$ is $.............$

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

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: