In Young's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is

$\sin ^{-1}(\lambda / 3d)$

$I_o=4 I_o \cos ^2(\phi / 2) \Rightarrow \phi=2 \pi / 3$

$\Rightarrow \Delta p \times(2 \pi / \lambda)=2 \pi / 3$

$\Rightarrow \Delta p=\lambda / 3$

$\sin \theta=\Delta p/d$

$\Rightarrow   \sin \theta=\lambda / 3 d$

∴  (c)