Q. Among the two interfering monochromatic sources $A$ and $B$; $A$ is ahead of $B$ in phase by $66{^\circ}$. If the observation be taken from point $P$, such that $PB - PA = \lambda/4$. Then the phase difference between the waves from $A$ and $B$ reaching $P$ is
A) $156{^\circ}$
B) $140{^\circ}$
C) $136{^\circ}$
D) $126{^\circ}$
Explanation
(a) Total phase difference
= Initial phase difference + Phase difference due to path
$= 66{^\circ} + \frac{360{^\circ}}{\lambda} \times \mathrm{\Delta}x = 66{^\circ} + \frac{360{^\circ}}{\lambda} \times \frac{\lambda}{4} = 66{^\circ} + 90 = 156{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.