In YDSE $\large\frac{d}{D}$$=10^4$ (d= distance between slits ; D= distance of screen from slits). at a point P on screen resulting intensity is equal to intensity due to individual slit $I_0$ Then the distance of point P from central maximum is $(\lambda= 6000 A^{\circ})$

Question

$(a)\;2 \;mm \\ (b)\;0.5\;mm \\ (c)\;1\;mm \\ (d)\;4\;mm $