In a YDSE, $D = 1m \; d= 1mm$ and $\lambda =\large\frac{1}{2}$$mm$ Then the distance between the first and central maxima on screen

Question

$(a)\;\frac{1}{2 \sqrt 3}\;m \\ (b)\;\frac{1}{\sqrt 3} m \\ (c)\;\frac{1}{\sqrt 2}m \\ (d)\;\frac{2}{\sqrt 3}\;m$