Question

The wavefronts of a light wave travelling in vacuum are given by x + y + z = c. The angle made by the direction of propagation of light with the X-axis is
(a) 0°
(b) 45°
(c) 90°
(d) ${\cos }^{-1}\left(1/\sqrt{3}\right)$

Answer 1

(d) ${\cos }^{-1}\left(1/\sqrt{3}\right)$
On writing the given equation in the plane equation form lx + my + nz = p,
where l² + m² + n² = 1 and p>0, we get:

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{c}{\sqrt{3}}$
If $\theta$ is the angle between the normal and +X axis, then
$$\begin{gathered} \cos \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$