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

Answer

To find the angular position of the first minima of diffracted light from a single slit, we can use the formula for the minima in single slit diffraction. The formula is given by:

\[ b \sin \theta = n \lambda \]

where:
- \( b \) is the width of the slit,
- \( \theta \) is the angle of the minima,
- \( n \) is the order of the minima (for the first minima, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

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

2. **Set up the equation for the first minima:**
- For the first minima (\( n = 1 \)):
\[ b \sin \theta = \lambda \]
Substituting the values:
\[ 3 \times 10^{-4} \sin \theta = 6 \times 10^{-7} \]

3. **Solve for \( \sin \theta \):**
\[ \sin \theta = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3} \]

4. **Approximate \( \theta \):**
- Since \( \sin \theta \) is small, we can use the small angle approximation where \( \sin \theta \approx \theta \) (in radians).
\[ \theta \approx 2 \times 10^{-3} \, \text{radians} \]

5. **Conclusion:**
- The angular position of the first minima of diffracted light is approximately \( 2 \times 10^{-3} \, \text{radians} \).

### Final Answer:
The angular position of the first minima of diffracted light is \( 2 \times 10^{-3} \, \text{radians} \).