JEE Physics 2025 Question with Solution

Given: The plane wave fronts are given by $x + y + z = C$, where $C$ is a constant.

Find: The angle made by the direction of wave propagation with the $x$-axis.

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

$$ax + by + cz = d$$

the normal vector is

$$\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$$

So for

$$x + y + z = C$$

the propagation direction is along

$$\vec{k} = \hat{i} + \hat{j} + \hat{k}$$

Let the required angle with the positive $x$-axis be $\theta$. Using the dot product with $\hat{i}$,

$$\cos\theta = \frac{\vec{k} \cdot \hat{i}}{|\vec{k}|\,|\hat{i}|}$$

Now,

$$\vec{k} \cdot \hat{i} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = 1$$

and

$$|\vec{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}, \qquad |\hat{i}| = 1$$

Therefore,

$$\cos\theta = \frac{1}{\sqrt{3}}$$

Hence,

$$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$$

Therefore, the correct option is A.

Given: The plane wave fronts are of the form $x + y + z = \text{constant}$.

Find: The angle between the direction of propagation and the $x$-axis.

The direction of propagation is normal to the wave front and is symmetric with respect to the $x$, $y$, and $z$ axes. Hence the angles made with these three axes are equal.

Let these angles be $\alpha$, $\beta$, and $\gamma$. Then

$$\cos\alpha = \cos\beta = \cos\gamma$$

Also, the direction cosines satisfy

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

Since all three are equal,

$$3\cos^2\alpha = 1$$

So,

$$\cos\alpha = \frac{1}{\sqrt{3}}$$

Therefore,

$$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$$

Thus, the angle made with the $x$-axis is $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$, so the correct option is A.