Question

Intensity of two sources is same and path difference at point \(A\) and \(B\) are \( \frac{\lambda}{6} \) and \( \frac{\lambda}{3} \) respectively. Ratio of intensity at \(A\) and \(B\) will be.

• A. \(3\)
• B. \(4\)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{4} \)

Answer 1

The Correct Option is A

Concept: Resultant intensity due to interference of two waves is \[ I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi \] where phase difference \[ \phi = \frac{2\pi}{\lambda}\Delta x \] Since both sources have equal intensity \[ I_1 = I_2 = I \] Step 1: Intensity at point \(A\).} Path difference \[ \Delta x_A = \frac{\lambda}{6} \] Phase difference \[ \phi_A = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} \] \[ \phi_A = \frac{\pi}{3} \] Now resultant intensity \[ I_A = I + I + 2\sqrt{I\cdot I}\cos\left(\frac{\pi}{3}\right) \] \[ I_A = 2I + 2I\left(\frac{1}{2}\right) \] \[ I_A = 3I \]
Step 2: Intensity at point \(B\).} Path difference \[ \Delta x_B = \frac{\lambda}{3} \] Phase difference \[ \phi_B = \frac{2\pi}{\lambda}\times\frac{\lambda}{3} \] \[ \phi_B = \frac{2\pi}{3} \] \[ I_B = I + I + 2\sqrt{I\cdot I}\cos\left(\frac{2\pi}{3}\right) \] \[ I_B = 2I + 2I\left(-\frac{1}{2}\right) \] \[ I_B = I \]
Step 3: Find ratio.} \[ \frac{I_A}{I_B}=\frac{3I}{I} \] \[ \frac{I_A}{I_B}=3 \] Final Result \[ \frac{I_A}{I_B}=3 \]