Question

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi / 2$ at point A and $\pi$ at point B. Then the difference between the resultant intensities at A and B is

• A. $4I$
• B. $5I$
• C. $2I$
• D. $3I$

Hint:

When phase difference is $\pi/2$, the waves are orthogonal and the interference term disappears, leaving just the simple sum of individual intensities ($I + 4I = 5I$). When phase difference is $\pi$, the waves interfere destructively to form a minimum, which can be found quickly using $(\sqrt{I_2} - \sqrt{I_1})^2 = (2\sqrt{I} - \sqrt{I})^2 = I$. The difference is simply $5I - I = 4I$.

Answer 1

The Correct Option is A

Step 1: Write the intensity rule.
Two interfering beams of intensities $I_1$ and $I_2$ give a resultant \[ I_{\text{res}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi. \] Here $I_1 = I$ and $I_2 = 4I$.
Step 2: Evaluate the cross term.
$2\sqrt{I\cdot4I} = 2\sqrt{4I^2} = 4I$, so $I_{\text{res}} = 5I + 4I\cos\phi$.
Step 3: Intensity at point A.
With $\phi_A = \dfrac{\pi}{2}$, $\cos\phi_A = 0$, so \[ I_A = 5I + 0 = 5I. \]
Step 4: Intensity at point B.
With $\phi_B = \pi$, $\cos\phi_B = -1$, so \[ I_B = 5I - 4I = I. \]
Step 5: Take the difference.
\[ I_A - I_B = 5I - I = 4I. \]
Step 6: Conclude.
The difference between the resultant intensities is $4I$, option (A). \[ \boxed{I_A - I_B = 4I} \]