Question

Two light waves of intensities \(I_1 = 4I\) and \(I_2 = I\) interfere. If the path difference between the waves is 25 % of the wavelength \(\lambda\), find the resultant intensity at that point.

Answer 1

Step 1: Given intensities: \[ I_1 = 4I, \quad I_2 = I. \] Step 2: Amplitudes of the waves are proportional to the square root of intensities: \[ A_1 = \sqrt{I_1} = \sqrt{4I} = 2\sqrt{I}, \quad A_2 = \sqrt{I_2} = \sqrt{I} = \sqrt{I}. \] Step 3: The phase difference \(\phi\) corresponding to path difference \(\Delta x = 0.25 \lambda\) is: \[ \phi = \frac{2\pi}{\lambda} \times \Delta x = 2\pi \times 0.25 = \frac{\pi}{2}. \] Step 4: The resultant amplitude is: \[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}. \] Step 5: Substitute the values: \[ A = \sqrt{(2\sqrt{I})^2 + (\sqrt{I})^2 + 2 \times 2\sqrt{I} \times \sqrt{I} \times \cos \frac{\pi}{2}} = \sqrt{4I + I + 0} = \sqrt{5I}. \] Step 6: The resultant intensity \(I_r\) is proportional to the square of amplitude: \[ I_r = A^2 = 5I. \] Answer: \[ \boxed{I_r = 5I}. \]