MCQ
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$0^\circ$
- B$45^\circ$
- C$90^\circ$
- ✓$\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
✓
Answer
Correct option: D.
$\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
On writing the given equation in the plane equation form $lx + my + nz = p,$
Where $l^2 + m^2 + n^2$ and $p > 0$, we get:
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{\text{c}}{\sqrt{3}}$
If $\theta$ is the angle between the normal and $+x$ axis, then
$\cos\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
Where $l^2 + m^2 + n^2$ and $p > 0$, we get:
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{\text{c}}{\sqrt{3}}$
If $\theta$ is the angle between the normal and $+x$ axis, then
$\cos\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
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