In a double slit experiment with monochromatic light, fringes are...

We know that the fringe width in YDSE is $\beta = \dfrac{\lambdaD}{d}$

$⇒ \Delta\beta = \dfrac{\lambda}{d}\DeltaD$

$⇒ \lambda = \dfrac{\left(\Delta\beta\right)d}{\DeltaD} = \dfrac{3 \times {10}^{- 3} \times {10}^{- 2} \times {10}^{- 3}}{5 \times {10}^{- 2}} = 6 \times {10}^{- 7}m = 600nm$