Question-13 Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is \frac{\pi}{2} at point P and \pi at point Q on the screen. The difference between the resultant intensities at points P and Q is (\cos90^\circ=0, \cos\pi=-1)
(A) 2I
(B) 4I
(C) 6I
(D) 8I
Answer : (C)
Explanation:
Resultant intensity is given by,
I_R=I_1+I_2+2\sqrt{I_1 I_2}\cos\phiAt point P, \phi=\frac{\pi}{2}
∴ (I_R)_P=I+9I+0 ….(∵ \cos90^\circ=0)
(I_R)_P=10I ….(i)
At point Q, \phi=\pi
∴ (I_R)_Q=I+9I-2\sqrt{I\times9I} ….(∵ \cos\pi=-1)
=10I-6I(I_R)_Q=4I ….(ii)
∴ Difference between resultant intensities at point P and Q is
∴ (I_R)_P-(I_R)_Q=10I-4I ….[From (i) and (ii)]
∴ (I_R)_P-(I_R)_Q=6I