Question

Two light sources of intensities \( I \) and \( 9I \) produce interference fringes on a screen. The phase difference between the beams is \( \frac{\pi}{2} \) at point A and at point B on the screen. The difference between resultant intensities at point A and B is

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

Hint:

Interference fringes depend on the phase difference, and the intensity varies accordingly.

Answer 1

The Correct Option is C

The resultant intensity $I_R$ at a point in an interference pattern is given by \[I_R = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi,\]where $I_1$ and $I_2$ are the intensities of the two sources, and $\phi$ is the phase difference between them. At point A, the phase difference is $\phi = \frac{\pi}{2}$, so \[I_A = I + 9I + 2 \sqrt{I \cdot 9I} \cos \frac{\pi}{2} = 10I.\]At point B, the phase difference is $\phi = \pi$, so \[I_B = I + 9I + 2 \sqrt{I \cdot 9I} \cos \pi = 10I - 6I = 4I.\]The difference in intensities is $I_A - I_B = 10I - 4I = \boxed{6I}$. 
Final Answer: \( 6I \).