Question

The condition for maxima in the interference of two waves \( A e^{i(\frac{k}{\sqrt{2}}(\sqrt{3}x + y) - \omega t)} \) and \( A e^{i(\frac{k}{\sqrt{2}}(x + \sqrt{3}y) - \omega t)} \) is given in terms of the wavelength \( \lambda \) and \( m \), an integer, by:

• A. \( (\sqrt{3} - \sqrt{2})x + (1 - \sqrt{2})y = 2m\lambda \)
• B. \( (\sqrt{3} + \sqrt{2})x + (1 - \sqrt{2})y = 2m\lambda \)
• C. \( (\sqrt{3} - \sqrt{2})x - (1 - \sqrt{2})y = m\lambda \)
• D. \( (\sqrt{3} - \sqrt{2})x + (1 - \sqrt{2})y = (2m + 1)\lambda \)

Hint:

For interference maxima, the phase difference must be an integer multiple of \( 2\pi \). Express phase difference geometrically in terms of \( x \) and \( y \).

Answer 1

The Correct Option is A

Step 1: Phase difference for two waves.
For constructive interference, \[ \Delta \phi = \phi_1 - \phi_2 = 2m\pi. \] Step 2: Write down the phases.
\[ \phi_1 = \frac{k}{\sqrt{2}}(\sqrt{3}x + y), \quad \phi_2 = \frac{k}{\sqrt{2}}(x + \sqrt{3}y). \] Then, \[ \Delta \phi = \frac{k}{\sqrt{2}}[(\sqrt{3}x + y) - (x + \sqrt{3}y)] = \frac{k}{\sqrt{2}}[(\sqrt{3} - 1)x + (1 - \sqrt{3})y]. \] Step 3: Simplify for maxima.
\[ \Delta \phi = 2m\pi \Rightarrow \frac{2\pi}{\lambda \sqrt{2}}[(\sqrt{3} - 1)x + (1 - \sqrt{3})y] = 2m\pi. \] \[ (\sqrt{3} - \sqrt{2})x + (1 - \sqrt{2})y = 2m\lambda. \] Step 4: Final Answer.
Hence, the condition for maxima is \( (\sqrt{3} - \sqrt{2})x + (1 - \sqrt{2})y = 2m\lambda. \)