Light of wavelength `5000 A^(0)` is incident on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 3 m from the slit. Then the width of the slit is

Options

A. `3 cm`
B. `0.3 cm`
C. `0.03 cm`
D. `0.06 cm`

Answer

To solve the problem, we need to find the width of the slit (denoted as 'a') using the given parameters. The first minimum of the diffraction pattern occurs at a specific distance from the central maximum, and we can use the formula for minima in single-slit diffraction to find the slit width.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Distance from the slit to the screen, \( d = 3 \, \text{m} \)
- Distance from the central maximum to the first minimum, \( x_1 = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)

2. **Use the Formula for Minima**:
The position of the first minimum in a single-slit diffraction pattern is given by the formula:
\[ a \cdot \sin(\theta) = n \lambda \]
For small angles, \( \sin(\theta) \approx \tan(\theta) \approx \frac{x}{d} \). Therefore, we can rewrite the formula as:
\[ a \cdot \frac{x_1}{d} = n \lambda \]
where \( n = 1 \) for the first minimum.

3. **Rearranging the Formula**:
Rearranging the above equation to solve for \( a \):
\[ a = \frac{n \lambda d}{x_1} \]

4. **Substituting the Values**:
Substitute \( n = 1 \), \( \lambda = 5000 \times 10^{-10} \, \text{m} \), \( d = 3 \, \text{m} \), and \( x_1 = 5 \times 10^{-3} \, \text{m} \):
\[ a = \frac{1 \cdot (5000 \times 10^{-10}) \cdot 3}{5 \times 10^{-3}} \]

5. **Calculating the Width of the Slit**:
\[ a = \frac{5000 \times 10^{-10} \times 3}{5 \times 10^{-3}} = \frac{15000 \times 10^{-10}}{5 \times 10^{-3}} = \frac{15000}{5} \times 10^{-7} = 3000 \times 10^{-7} \, \text{m} = 3 \times 10^{-4} \, \text{m} \]

6. **Convert to Centimeters**:
To convert meters to centimeters:
\[ a = 3 \times 10^{-4} \, \text{m} = 0.03 \, \text{cm} \]

7. **Final Answer**:
The width of the slit is \( 0.03 \, \text{cm} \).

### Conclusion:
The width of the slit is \( 0.03 \, \text{cm} \).