The first diffraction minimum due to a single slit Fraunhoffer diffraction is at the angle of diffraction `30^(@)` for a light of wavelength `5460A^(@)`. The width of the slit is

Options

A. `1.082 xx 10^(-4) cm`
B. `2.164 xx 10^(-4) cm`
C. `1.082 xx 10^(-3) cm`
D. `0.546 cm`

Answer

To find the width of the slit causing the first diffraction minimum in a single slit Fraunhofer diffraction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Minima**:
The condition for the first diffraction minimum in single slit diffraction is given by the formula:
\[ d \sin \theta = n \lambda \]
where:
- \( d \) is the width of the slit,
- \( \theta \) is the angle of diffraction,
- \( n \) is the order of the minimum (for the first minimum, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

2. **Identify Given Values**:
From the problem statement, we have:
- \( \theta = 30^\circ \)
- \( \lambda = 5460 \, \text{Å} = 5460 \times 10^{-10} \, \text{m} \)

3. **Substitute Known Values**:
For the first minimum, we set \( n = 1 \). Therefore, the equation becomes:
\[ d \sin(30^\circ) = 1 \times 5460 \times 10^{-10} \]

4. **Calculate \( \sin(30^\circ) \)**:
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

5. **Rearranging the Equation**:
Substitute \( \sin(30^\circ) \) into the equation:
\[ d \cdot \frac{1}{2} = 5460 \times 10^{-10} \]
To find \( d \), multiply both sides by 2:
\[ d = 2 \cdot 5460 \times 10^{-10} \]

6. **Calculate \( d \)**:
Now perform the multiplication:
\[ d = 10920 \times 10^{-10} \, \text{m} \]

7. **Convert to Centimeters**:
Since \( 1 \, \text{m} = 100 \, \text{cm} \), we convert \( d \) to centimeters:
\[ d = 10920 \times 10^{-10} \, \text{m} = 1.092 \times 10^{-6} \, \text{m} = 1.092 \times 10^{-4} \, \text{cm} \]

### Final Answer:
The width of the slit \( d \) is:
\[ d = 1.092 \times 10^{-4} \, \text{cm} \]