A light wave is propagating with plane wave fronts of the type $x+y+z=$ constant. Th angle made by the direction of wave propagation with the $x$-axis is :

<p>The light wave is propagating with plane wave fronts described by the equation $x + y + z = \text{constant}$. The wave propagates in a direction perpendicular to these wave fronts. This direction is symmetric with respect to the $x$, $y$, and $z$ axes.</p> <p>Since the wave is symmetric about these axes, the angle it makes with each axis is the same. Let these angles be $\alpha$, $\beta$, and $\gamma$, which are the angles made by the light with the $x$, $y$, and $z$ axes, respectively.</p> <p>This implies:</p> <p>$ \cos \alpha = \cos \beta = \cos \gamma $</p> <p>According to the sum of the squares of the direction cosines:</p> <p>$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $</p> <p>Substituting the equality of direction cosines, we get:</p> <p>$ 3 \cos^2 \alpha = 1 $</p> <p>$ \cos^2 \alpha = \frac{1}{3} $</p> <p>Thus, the angle $\alpha$ is given by:</p> <p>$ \alpha = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right) $</p>