In Young's double slit experiment,if the distance between the two slits is halved,then the fringe width will become

Describe Young's double slit experiment.

In a double-slit experiment,the angular width of the fringes is $0.20^o$ for sodium light $(\lambda = 5890 \ \mathring{A})$. In order to increase the angular width of the fringes by $10\%$,the necessary change in the wavelength is:

Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

$A$ double slit experiment is performed by using light of wavelength $6000 \, Å$. If the distance of the screen is $1 \, m$ and the slits are $0.1 \, cm$ apart, calculate the angular position of the $10^{th}$ bright fringe.

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is