Question

A light wave is propagating with plane wave fronts of the type \( x + y + z = \text{constant} \). The angle made by the direction of wave propagation with the \( x \)-axis is:

• A. \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
• B. \( \cos^{-1} \left( \frac{\sqrt{3}}{3} \right) \)
• C. \( \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \)

Hint:

The symmetry of the wave propagation allows us to use the property that the angle made with the \( x \), \( y \), and \( z \) axes is the same.

Answer 1

The Correct Option is A

The direction of propagation of light is perpendicular to the wave front and is symmetric about the \( x \), \( y \), and \( z \) axes. 
The angle made by the direction of wave propagation with the \( x \)-axis is the same as that with the \( y \)-axis and the \( z \)-axis. 
Thus, the equation can be written as: \[ \cos \theta = \cos \beta = \cos \gamma \quad (\text{where } \alpha, \beta, \gamma \text{ are the angles made by light with the } x, y, z \text{ axes respectively}) \] Also, we know that \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \). Since the angles are equal, we have: \[ \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \quad \Rightarrow \quad 3 \cos^2 \alpha = 1 \quad \Rightarrow \quad \cos \alpha = \frac{1}{\sqrt{3}} \] Thus, the angle is \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \).

Answer 2

This problem asks for the angle between the direction of propagation of a light wave and the positive x-axis, given that the wave fronts are planes described by the equation \( x + y + z = \text{constant} \).

Concept Used:

The key concept is that the direction of propagation of a wave is always perpendicular (normal) to its wave fronts. For a plane described by the equation \( ax + by + cz = d \), the vector normal to the plane is given by \( \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} \). This normal vector represents the direction of wave propagation.

The angle \( \theta \) between a vector \( \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} \) and the positive x-axis (represented by the unit vector \( \hat{i} \)) can be found using the dot product formula:

\[ \cos\theta = \frac{\vec{A} \cdot \hat{i}}{|\vec{A}| |\hat{i}|} = \frac{A_x}{\sqrt{A_x^2 + A_y^2 + A_z^2}} \]

Step-by-Step Solution:

Step 1: Determine the vector representing the direction of wave propagation.

The equation of the plane wave fronts is given as:

\[ x + y + z = C \quad (\text{where C is a constant}) \]

This equation is in the form \( ax + by + cz = d \), with \( a=1, b=1, \) and \( c=1 \).

The vector normal to this plane, which represents the direction of wave propagation (\( \vec{k} \)), is:

\[ \vec{k} = 1\hat{i} + 1\hat{j} + 1\hat{k} = \hat{i} + \hat{j} + \hat{k} \]

Step 2: Find the angle between the propagation vector and the x-axis.

Let \( \theta \) be the angle between the direction of propagation \( \vec{k} \) and the positive x-axis, which is represented by the unit vector \( \hat{i} \).

Using the dot product formula:

\[ \cos\theta = \frac{\vec{k} \cdot \hat{i}}{|\vec{k}| |\hat{i}|} \]

Step 3: Calculate the components of the dot product formula.

First, calculate the dot product \( \vec{k} \cdot \hat{i} \):

\[ \vec{k} \cdot \hat{i} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i}) = (1)(1) + (1)(0) + (1)(0) = 1 \]

Next, calculate the magnitude of the propagation vector \( |\vec{k}| \):

\[ |\vec{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

The magnitude of the unit vector \( \hat{i} \) is \( |\hat{i}| = 1 \).

Final Computation & Result:

Substitute the calculated values back into the cosine formula:

\[ \cos\theta = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}} \]

Therefore, the angle \( \theta \) made by the direction of wave propagation with the x-axis is:

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