What is the width of a single slit if the first minimum is observed at an angle `2^@` with a light of wavelength 9680 Å ?

Options

A. 0.2 mm
B. `2 xx 10^(-2) mm`
C. `2 xx 10^5 mm`
D. `2 mm`

Answer

To find the width of a single slit when the first minimum is observed at an angle of \(2^\circ\) with light of wavelength \(9680 \, \text{Å}\), we can use the formula for single-slit diffraction:

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

where:
- \(a\) is the width of the slit,
- \(\theta\) is the angle at which the first minimum occurs,
- \(n\) is the order of the minimum (for the first minimum, \(n = 1\)),
- \(\lambda\) is the wavelength of the light.

### Step-by-step Solution:

1. **Identify the given values:**
- Wavelength, \(\lambda = 9680 \, \text{Å} = 9680 \times 10^{-10} \, \text{m}\)
- Angle, \(\theta = 2^\circ\)
- Order of minimum, \(n = 1\)

2. **Convert the angle to radians (if necessary):**
- Since we can use \(\sin\) directly in degrees, we will keep \(\theta\) as \(2^\circ\).

3. **Use the formula for the first minimum:**
\[ a \sin(2^\circ) = 1 \times \lambda \]

4. **Rearranging the formula to solve for \(a\):**
\[ a = \frac{\lambda}{\sin(2^\circ)} \]

5. **Substituting the values:**
\[ a = \frac{9680 \times 10^{-10} \, \text{m}}{\sin(2^\circ)} \]

6. **Calculate \(\sin(2^\circ)\):**
- Using a calculator, \(\sin(2^\circ) \approx 0.0349\).

7. **Substituting \(\sin(2^\circ)\) into the equation:**
\[ a = \frac{9680 \times 10^{-10}}{0.0349} \]

8. **Perform the calculation:**
\[ a \approx \frac{9680 \times 10^{-10}}{0.0349} \approx 2.77 \times 10^{-5} \, \text{m} \]

9. **Convert \(a\) from meters to millimeters:**
- Since \(1 \, \text{m} = 1000 \, \text{mm}\):
\[ a \approx 2.77 \times 10^{-5} \, \text{m} \times 1000 \, \text{mm/m} = 2.77 \times 10^{-2} \, \text{mm} \]

10. **Final result:**
- The width of the single slit is approximately \(2.77 \times 10^{-2} \, \text{mm}\).

### Conclusion:
The correct answer is approximately \(2.76 \times 10^{-2} \, \text{mm}\), which corresponds to option B.