Question 73: Wave Optics

Question

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 :

Accepted Answer

The light wave is propagating with plane wave fronts described by the equation $x + y + z = 	ext{constant}$. 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}}ight) $

Answer

$\cos ^{-1}(2 / 3)$

Answer

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

Answer

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

Answer

$\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Wave Optics

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 :

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