In Young's double-slit experiment,if the distance between the slits is reduced to half and the distance between the slit and the screen is doubled,then the fringe width

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 fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?

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?

Young's experiment demonstrates that . . . . . .