Question

Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is \(\pi / 2\) at point P and \(\pi\) at point Q on the screen. The difference between the resultant intensities at points P and Q is \(\left(\cos 90^{\circ}=0, \cos \pi=-1\right)\)

• A. 2 I
• B. 4 I
• C. 6 I
• D. 8 I

Hint:

Use I = I1 + I2 + 2 sqrt(I1 I2) cos(phi) at each point.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Resultant intensity is \(I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\) with \(I_1=I\) and \(I_2=9I\).

Step 2: At P and Q
\(2\sqrt{I\cdot9I}=6I\).
At P (\(\phi=\pi/2\)): \(I_P=10I+6I\times0=10I\).
At Q (\(\phi=\pi\)): \(I_Q=10I-6I=4I\).

Step 3: Difference
\(I_P-I_Q=10I-4I=6I\), which is option (C).

Final Answer:
The difference in resultant intensities is 6I. \[ \boxed{\text{(C) }6I} \]