In a single - slit diffraction pattern , the position of first secondary maximum is at `30^@` , then what will be the angular position of second minima ?

Options

A. `sin ^(-1) (2//3)`
B. `sin^(-1) (1)`
C. `sin^(-1) (1//2)`
D. None

Answer

To solve the problem, we need to find the angular position of the second minima in a single-slit diffraction pattern, given that the first secondary maximum occurs at an angle of \(30^\circ\).

### Step-by-Step Solution:

1. **Understanding the Position of Secondary Maxima**:
The position of the secondary maxima in a single-slit diffraction pattern is given by the formula:
\[ D \sin \theta = \left(n + \frac{1}{2}\right) \lambda \]
where \(D\) is the slit width, \(\theta\) is the angle, \(\lambda\) is the wavelength, and \(n\) is the order of the maximum (0, 1, 2, ...).

2. **Applying the Given Information**:
For the first secondary maximum, \(n = 1\) and \(\theta_1 = 30^\circ\):
\[ D \sin(30^\circ) = \left(1 + \frac{1}{2}\right) \lambda \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we can substitute this into the equation:
\[ D \cdot \frac{1}{2} = \frac{3}{2} \lambda \]

3. **Rearranging to Find \(\frac{\lambda}{D}\)**:
Rearranging the equation gives:
\[ \frac{\lambda}{D} = \frac{1}{3} \]

4. **Finding the Angular Position of the Second Minima**:
The position of the minima is given by:
\[ D \sin \theta = n \lambda \]
For the second minima, \(n = 2\):
\[ D \sin \theta_2 = 2 \lambda \]

5. **Substituting \(\frac{\lambda}{D}\)**:
We can substitute \(\frac{\lambda}{D} = \frac{1}{3}\) into the equation:
\[ D \sin \theta_2 = 2 \left(\frac{1}{3} D\right) \]
Simplifying gives:
\[ \sin \theta_2 = \frac{2}{3} \]

6. **Calculating \(\theta_2\)**:
Finally, we find the angle \(\theta_2\):
\[ \theta_2 = \sin^{-1}\left(\frac{2}{3}\right) \]

### Final Answer:
The angular position of the second minima is:
\[ \theta_2 = \sin^{-1}\left(\frac{2}{3}\right) \]