question_answer

Two sources \[{{S}_{1}}\] and \[{{S}_{2}}\] emitting coherent light waves of wavelength \[\lambda \] in the same phase are situated as shown. The distance OP, so that the light intensity detected at P is equal to that at O is       

A) \[D\sqrt{2}\]

B) \[D/2\]

C)  \[D\sqrt{3}\]

D) \[D/\sqrt{3}\]

Correct Answer: C

Solution :

[c] Referring to the figure, the path difference between the two waves starting from \[{{S}_{1}}\] and \[{{S}_{2}}\] turns out to be \[(2\lambda \cos \theta )=n\lambda \] where n is taken as 1to get the point of maximum intensity which is the same as a point O.    Therefore, the above relation gives \[\cos \theta =1/2\] SO that \[\theta =60{}^\circ \] and \[\tan \theta =PO/D=\sqrt{3}\]givin\[PO=D\sqrt{3}\].