In Young's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is
Answer & explanation
Correct answer: option 3
$I_o=4 I_o \cos ^2(\phi / 2) \Rightarrow \phi=2 \pi / 3$
$\Rightarrow \Delta p \times(2 \pi / \lambda)=2 \pi / 3$
$\Rightarrow \Delta p=\lambda / 3$
$\sin \theta=\Delta p/d$
$\Rightarrow \sin \theta=\lambda / 3 d$
∴ (c)