The distance between two coherent sources is $1 \ mm$. The screen is placed at a distance of $1 \ m$ from the sources. If the distance of the third bright fringe is $1.2 \ mm$ from the central fringe, the wavelength of light used is:

In Young's double-slit experiment,when two light waves form the third minimum,the path difference between them is:

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

The maximum intensity in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10d$?

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

Write the formula for fringe width.