JEE PHYSICS MCQ: WAVE OPTICS

In Young's double slit experiment $\frac{\mathrm{d}}{\mathrm{D}}=10^{-4}$ ( $\mathrm{d}=$ distance between slits, $\mathrm{D}=$ distance of screen from the slits). At a point $\mathrm{P}$ on the screen the resulting intensity is equal to the intensity due to individual slit $\mathrm{I}_0$. Then the minimum distance of point $\mathrm{P}$ from the central maximum is $(\lambda=6000 \mathop {\text{A}}\limits^ \circ)$ :-