Question

Interference fringes are produced on a 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 \(\left(\cos \frac{\pi }{2}=0,\,\cos \pi =-1\right)\)

• A. \(2I\)
• B. \(4I\)
• C. \(6I\)
• D. \(8I\)

Hint:

Use the resultant intensity formula with the given phase differences and subtract the two results.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The resultant intensity of two coherent beams is \(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi\).

Step 2: Substitute the Intensities
With \(I_1 = I\) and \(I_2 = 9I\), \(\sqrt{I_1I_2} = 3I\).
So \(I_R = 10I + 6I\cos\phi\).

Step 3: Point P
At \(\phi = \dfrac{\pi}{2}\), \(\cos\phi = 0\), so \(I_P = 10I\).

Step 4: Point Q
At \(\phi = \pi\), \(\cos\phi = -1\), so \(I_Q = 10I - 6I = 4I\).

Step 5: Difference
\(I_P - I_Q = 10I - 4I = 6I\)

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