In Young's experiment,the ratio of maximum to minimum intensities of the fringe system is $4:1$. The amplitudes of the coherent sources are in the ratio: (in $:1$)

In a two-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by $5 \times 10^{-2} \, m$ towards the slits, the change in fringe width is $3 \times 10^{-5} \, m$. If the separation between the slits is $10^{-3} \, m$, the wavelength of light used is: (in $\text{Å}$)

Two slits are $1 \, mm$ apart from each other and the distance of the screen is $1 \, m$. If illuminated with light of wavelength $5 \times 10^{-7} \, m$,the distance between the $3^{rd}$ dark fringe and the $5^{th}$ bright fringe is......$mm$.

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?