Question

Two light waves of intensities \( I \) and \( 2I \) superimpose on each other. If the path difference between the light waves reaching a point is 12.5% of the wavelength of the light, then the resultant intensity at the point is (Both the light waves have the same wavelength)

• A. \( I \)
• B. \( 9I \)
• C. \( 3I \)
• D. \( 5I \)

Hint:

Remember the formula for resultant intensity in terms of individual intensities and phase difference.

Answer 1

The Correct Option is D

We are given that \( I_1 = I \) and \( I_2 = 2I \). The path difference is 12.5% of the wavelength, so we can write: \[ \Delta x = \frac{12.5}{100} \lambda = \frac{1}{8} \lambda. \] The phase difference \( \phi \) is given by: \[ \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{1}{8} \lambda = \frac{\pi}{4}. \] The resultant intensity \( I_R \) is given by the formula: \[ I_R = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi. \] Substitute the values of \( I_1 \), \( I_2 \), and \( \phi \): \[ I_R = I + 2I + 2 \sqrt{I \cdot 2I} \cos \frac{\pi}{4}. \] Simplify: \[ I_R = 3I + 2 \sqrt{2I^2} \cdot \frac{1}{\sqrt{2}}. \] \[ I_R = 3I + 2I = 5I. \] Thus, the resultant intensity at the point is \( \boxed{5I} \).