In a Young's double-slit experiment , the slits are separated by $0.28$ mm and screen is placed $1.4$ m away . The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2$ cm . Determine the wavelength of light used in the experiment .

Solution: Distance between the silits d=0.28 ,mm $=0.28\times {10}^{-3}$ m
Distance between the slits and the screen $,D=1.4m$
Distance between the central fringe and the fouth (n=4) Ftinge ,
$u=1.2cm=1.2\times {10}^{-2}m$
In case of a sonstrctive interence , we have the relation for the the distance between the two fringes as :
$u=n\lambda \frac{D}{d}$
where
n= order of fringes =4
$\lambda$= wavelength of light used
$\therefore \lambda =\frac{ud}{nD}$
$=\frac{1.2\times {10}^{-2}\times 0.28\times {10}^{-3}}{4\times 1.4}$
$=6\times {10}^{-7}$
=600 nm
Hence , the wavelength of the light is 600 nm .