Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?

$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.

In Young’s double slit experiment, how many maxima can be seen on a screen (including the central maxima) if $d = \frac{5\lambda}{2}$ (where $\lambda$ is the wavelength of light and $d$ is the distance between the two slits)?

Imagine a Young's double-slit interference experiment performed with waves associated with fast-moving electrons produced from an electron gun. The distance between successive maxima (fringe width) will decrease if:

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

State two conditions for obtaining sustained interference of light. In Young's double-slit experiment,using light of wavelength $400 \, nm$,interference fringes of width $'X'$ are obtained. If the wavelength of light is increased to $600 \, nm$ and the separation between the slits is halved,find the ratio of the distances between the screen and the slits in the two arrangements if the fringe width remains the same.