Answer:
Option B
Explanation:
The linear width of the principal maxima,
$\beta=\frac{2D\lambda}{d}$
It is given that, the linear width of the principal maxima is equal to the width of the slit.
$\therefore$ $d=\frac{2D\lambda}{d}\Rightarrow D=\frac{d^{2}}{2\lambda}$