JEE Main 2025 — Wave Optics Question with Solution

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.

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.

This implies:

$\cos \alpha =\cos \beta =\cos \gamma$

According to the sum of the squares of the direction cosines:

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

Substituting the equality of direction cosines, we get:

$3{\cos }^2\alpha =1$

${\cos }^2\alpha =\frac{1}{3}$

Thus, the angle $\alpha$ is given by:

$\alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$