Write the formula for fringe width.

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

If the source of light used in a Young's double slit experiment is changed from red to violet:

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ 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-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In a Young's double slit experiment with slit separation $0.1\, mm$,one observes a bright fringe at angle $\frac{1}{40}\, rad$ by using light of wavelength $\lambda_1$. When the light of wavelength $\lambda_2$ is used,a bright fringe is seen at the same angle in the same setup. Given that $\lambda_1$ and $\lambda_2$ are in the visible range ($380\, nm$ to $740\, nm$),their values are: