The wavefront of a light beam is given by the equation `x + 2y + 3x = c` (where c is arbitrary constant), then the angle made by the direction of light with the y-axis is

Options

A. `cos^(1) .(1)/(sqrt(14))`
B. `sin^(1).(2)/(sqrt(14))`
C. `cos^(1).(2)/(sqrt(14))`
D. `sin^(1).(3)/(sqrt(14))`

Explanation

To find the angle made by the direction of light with the y-axis given the wavefront equation \( x + 2y + 3z = c \), we can follow these steps:

### Step 1: Identify the Direction Vector
The wavefront equation can be interpreted as a vector in space. The coefficients of \( x \), \( y \), and \( z \) in the equation represent the components of the direction vector \( \mathbf{n} \).
- From the equation \( x + 2y + 3z = c \), we can identify the direction vector as:
\[ \mathbf{n} = (1, 2, 3) \]

### Step 2: Calculate the Magnitude of the Direction Vector
The magnitude of the vector \( \mathbf{n} \) can be calculated using the formula:
\[ |\mathbf{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} \]
Substituting the components:
\[ |\mathbf{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Find the Cosine of the Angle with the y-axis
To find the angle \( \beta \) between the direction vector \( \mathbf{n} \) and the y-axis, we use the dot product formula:
\[ \cos \beta = \frac{\mathbf{n} \cdot \mathbf{j}}{|\mathbf{n}| \cdot |\mathbf{j}|} \]
Where \( \mathbf{j} \) is the unit vector in the y-direction, which is \( (0, 1, 0) \).

Calculating the dot product:
\[ \mathbf{n} \cdot \mathbf{j} = (1, 2, 3) \cdot (0, 1, 0) = 0 \cdot 1 + 2 \cdot 1 + 0 \cdot 3 = 2 \]

Since the magnitude of \( \mathbf{j} \) is 1, we have:
\[ \cos \beta = \frac{2}{|\mathbf{n}|} = \frac{2}{\sqrt{14}} \]

### Step 4: Calculate the Angle \( \beta \)
To find the angle \( \beta \), we take the inverse cosine:
\[ \beta = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \]

### Final Answer
Thus, the angle made by the direction of light with the y-axis is:
\[ \beta = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \]

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