In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width

$A$ fringe width of $6 \ mm$ was produced for two slits separated by $1 \ mm$ apart. The screen is placed $10 \ m$ away. The wavelength of light used is $'x' \ nm$. The value of $'x'$ to the nearest integer is

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}$)

In a Young's double-slit experiment,the slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 7500 \, \mathring{A}$ and $\lambda_2 = 9000 \, \mathring{A}$. At what distance from the central maximum on a screen $2 \, m$ away will a bright fringe from one interference pattern coincide with a bright fringe from the other?

In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

In Young's double slit experiment, the fringe width is $0.4 \text{ mm}$. What is the distance between $4^{th}$ dark band and $6^{th}$ bright band on the same side of the interference pattern (in $\text{ mm}$)?