Question

A light wave is propagating with plane wave fronts of the type \( x + y + z = \text{constant} \). The angle made by the direction of wave propagation with the \( x \)-axis is:

• A. \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
• B. \( \cos^{-1} \left( \frac{\sqrt{3}}{3} \right) \)
• C. \( \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \)

Hint:

The symmetry of the wave propagation allows us to use the property that the angle made with the \( x \), \( y \), and \( z \) axes is the same.

Answer 1

The Correct Option is A

The direction of propagation of light is perpendicular to the wave front and is symmetric about the \( x \), \( y \), and \( z \) axes. 
The angle made by the direction of wave propagation with the \( x \)-axis is the same as that with the \( y \)-axis and the \( z \)-axis. 
Thus, the equation can be written as: \[ \cos \theta = \cos \beta = \cos \gamma \quad (\text{where } \alpha, \beta, \gamma \text{ are the angles made by light with the } x, y, z \text{ axes respectively}) \] Also, we know that \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \). Since the angles are equal, we have: \[ \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \quad \Rightarrow \quad 3 \cos^2 \alpha = 1 \quad \Rightarrow \quad \cos \alpha = \frac{1}{\sqrt{3}} \] Thus, the angle is \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \).