If a transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is inserted in front of one of the slits of Young's Double Slit experiment,how much will be the shift in the interference pattern? The distance between the slits is $0.5 \, mm$ and that between slits and screen is $100 \, cm$. (Answer in $cm$)

In Young's double slit experiment, the aperture screen distance is $2 \, m$. The fringe width is $1 \, mm$. Light of $600 \, nm$ is used. If a thin plate of glass $(\mu = 1.5)$ of thickness $0.06 \, mm$ is placed over one of the slits, then there will be a lateral displacement of the fringes by $... \, cm$.

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

Monochromatic light of wavelength $500 \, nm$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $= 1.5$), the central maximum is shifted to a position previously occupied by the $4^{th}$ bright fringe. The thickness of the glass plate is ..................... $\mu m$.

In Young's double slit arrangement,water is filled in the space between the screen and the slits. Then:

$A$ thin mica sheet of thickness $2 \times 10^{-6} \ m$ and refractive index $\mu = 1.5$ is introduced in the path of the first wave. The wavelength of the wave used is $5000 \ \mathring{A}$. The central bright maximum will shift: