Q74 / 75 · JEE Main 2025 (2 Apr Shift 1) · Physics
A light wave is propagating with plane wave fronts of the type $x+y+z=$ constant. The angle made by the direction of wave propagation with the $x$-axis is:
- Correct answerA$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
- B$\cos ^{-1}\left(\frac{2}{3}\right)$
- C$\cos ^{-1}\left(\frac{1}{3}\right)$
- D$\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
Solution
The direction of propagation of light is perpendicular to the wave front and is symmetric about $\mathrm{x}, \mathrm{y}$ and z axis.
$\therefore$ Angle made by the light with $\mathrm{x}, \mathrm{y} \& \mathrm{z}$ axis is same.
$\therefore \quad \cos \alpha=\cos \beta=\cos \gamma(\alpha, \beta \& \gamma$ are angle made by light with $\mathrm{x}, \mathrm{y} \& \mathrm{z}$ axis respectively)
Also $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$ [Sum of direction cosines]
$\therefore \alpha=\cos ^{-1} \frac{1}{\sqrt{3}}$
$\therefore$ Angle made by the light with $\mathrm{x}, \mathrm{y} \& \mathrm{z}$ axis is same.
$\therefore \quad \cos \alpha=\cos \beta=\cos \gamma(\alpha, \beta \& \gamma$ are angle made by light with $\mathrm{x}, \mathrm{y} \& \mathrm{z}$ axis respectively)
Also $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$ [Sum of direction cosines]
$\therefore \alpha=\cos ^{-1} \frac{1}{\sqrt{3}}$
Hence, the correct answer is option: 'A'.