In a $YDSE$ apparatus,if we use white light,then:

In Young's double-slit experiment,the two slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 12000 \, \mathring{A}$ and $\lambda_2 = 10000 \, \mathring{A}$. At what minimum distance from the common central bright fringe on the screen will a bright fringe from one interference pattern coincide with a bright fringe from the other? The distance between the slits and the screen is $2 \, m$.

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.