A parallel beam of light of wavelength `6000Å` gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :

Options

A. `2 xx 10^(-3)`rad
B. `3 xx 10^(-3)`rad
C. `1.8 xx 10^(-3)`rad
D. `6 xx 10^(-3)`rad

Explanation

To find the angular position of the first minima in a single slit diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Width of the slit, \( a = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} = 3 \times 10^{-4} \, \text{m} \)

2. **Use the Condition for First Minima**:
The condition for the first minima in single slit diffraction is given by:
\[ a \sin \theta = \lambda \]
For the first minima, \( n = 1 \), so we can write:
\[ a \sin \theta = \lambda \]

3. **Rearranging the Equation**:
To find \( \sin \theta \):
\[ \sin \theta = \frac{\lambda}{a} \]

4. **Substituting the Values**:
Substitute the values of \( \lambda \) and \( a \) into the equation:
\[ \sin \theta = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} \]

5. **Calculating the Value**:
Perform the division:
\[ \sin \theta = \frac{6}{3} \times 10^{-7 + 4} = 2 \times 10^{-3} \]

6. **Approximation for Small Angles**:
Since \( \sin \theta \) is small, we can use the small angle approximation:
\[ \theta \approx \sin \theta \]
Therefore, \( \theta \approx 2 \times 10^{-3} \, \text{radians} \).

### Final Answer:
The angular position of the first minima is:
\[ \theta \approx 2 \times 10^{-3} \, \text{radians} \]