In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be

In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

$A$ Young's double slit experiment apparatus has slits separated by $0.2 \ mm$ and a screen $60 \ cm$ away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $\mu = \frac{11}{9}$ and the slits are illuminated with green light $(\lambda = 550 \ nm$ in vacuum$)$. Find the fringe width of the pattern formed on the screen. (in $mm$)

Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

In the 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