Question

The waveforms of a light wave traveling 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. $0^o$
• B. ${45}^o$
• C. ${90}^o$
• D. $co{s}^{-1}\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $co{s}^{-1}\frac{1}{\sqrt{3}}$

Let any $\overrightarrow{R}$ its components are$\overrightarrow{R}$ = $\overrightarrow{R}$x + $\overrightarrow{R}$y + $\overrightarrow{R}$z
with | $\overrightarrow{R}$ | =$\sqrt{R{x}^2+R{y}^2+R{z}^2}$
& cos${\theta }_x=\frac{Rx}{R},cos{\theta }_y=\frac{Ry}{R},cos{\theta }_z=\frac{Rz}{R}$
there cos${\theta }_x$, cos${\theta }_y$ and cos${\theta }_z$ one called direction cosines.
Hence x + y + z = c $(=\overrightarrow{R})$
So, magnitude of c = $\sqrt{I^2+I^2+I^2}=\sqrt{3}$
and cos${\theta }_x$ = $\frac{1}{\sqrt{3}}$