Two coherent light beams of intensities I and 4I interfere to form a fringe pattern. The phase difference between the beams at a point P is 3 . If the resultant intensity at point P is greater than the resultant intensity at another point Q by 4I , the phase difference between the beams at point Q is :

Step-by-step solution

The resultant intensity in interference is given by I_ res = I₁ + I₂ + 2 I₁ I₂ At point P , the phase difference is _P = 3 . The resultant intensity at P is: I_P = I + 4I + 2 I 4I ( 3 ) I_P = 5I + 2(2I) ( 1 2 ) = 7I It is given that I_P - I_Q = 4I , so the intensity at Q is: I_Q = 7I - 4I = 3I Let the phase difference at Q be _Q . Then: I_Q = I + 4I + 2 I 4I _Q 3I = 5I + 4I _Q -2I = 4I _Q _Q = - 1 2 Therefore, the phase difference at Q is _Q = 2 3 . Answer: 2 3