Physics Q40 – JEE-MAIN 2025

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 :

Clue 1 of 3: Wavefronts and Propagation Direction

The direction of wave propagation is always perpendicular to the plane of the wavefront.

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Ninja Strategy•Check Domain of Inverse Cosine

Quickly eliminate options where the argument of $\cos^{- 1}$ falls outside the valid range of $[- 1, 1]$ .

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● Video Walkthrough

Step 1: Identify the direction of wave propagation✦ Active

The equation of the plane wavefront is $x + y + z = constant$ . The direction of wave propagation is given by the normal vector to this plane. Comparing with the general plane equation $A x + B y + C z = D$ , we have $A = 1$ , $B = 1$ , $C = 1$ . Thus, the direction vector of wave propagation is $\vec{k} = \hat{i} + \hat{j} + \hat{k}$ .

Step 2: Determine the magnitude of the propagation vector and the x-axis vector○ Expand

The magnitude of the propagation vector is $| \vec{k} | = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}$ . The direction vector for the x-axis is $\hat{i}$ , and its magnitude is $| \hat{i} | = 1$ .

Step 3: Calculate the angle using the dot product○ Expand

The angle $θ$ between the direction of wave propagation $\vec{k}$ and the x-axis $\hat{i}$ is given by $\cos θ = \frac{\vec{k} \cdot \hat{i}}{| \vec{k} | | \hat{i} |}$ . $\vec{k} \cdot \hat{i} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = 1$ . So, $\cos θ = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$ . Therefore, $θ = \cos^{- 1} (\frac{1}{\sqrt{3}})$ .

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Physics Question 40 – JEE-MAIN 2025

Clue 1 of 3: Wavefronts and Propagation Direction

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