A light of wavelength `6000Å` is incident on a single slit . First minimum is obtained at a distance of 0.4 cm from the centre . If width of the slit is 0.3 mm, the distance between slit and screen will be

Options

A. 1.0 m
B. 1.5 m
C. 2.0 m
D. 2.3 m

Answer

To solve the problem, we need to find the distance \( D \) between the slit and the screen using the information provided about the single slit diffraction pattern.

### Given Data:
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Position of the first minimum \( y_1 = 0.4 \, \text{cm} = 0.4 \times 10^{-2} \, \text{m} = 4 \times 10^{-3} \, \text{m} \)
- Width of the slit \( a = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} = 3 \times 10^{-4} \, \text{m} \)

### Formula:
The position of the \( n^{th} \) minimum in a single slit diffraction pattern is given by the formula:
\[ y_n = \frac{n \lambda D}{a} \]
For the first minimum (\( n = 1 \)):
\[ y_1 = \frac{\lambda D}{a} \]

### Step-by-Step Solution:
1. **Substituting the known values into the formula:**
\[ 4 \times 10^{-3} = \frac{(6 \times 10^{-7}) D}{(3 \times 10^{-4})} \]

2. **Rearranging the equation to solve for \( D \):**
\[ D = \frac{4 \times 10^{-3} \times (3 \times 10^{-4})}{6 \times 10^{-7}} \]

3. **Calculating the numerator:**
\[ 4 \times 10^{-3} \times 3 \times 10^{-4} = 12 \times 10^{-7} \, \text{m}^2 \]

4. **Now substituting back into the equation for \( D \):**
\[ D = \frac{12 \times 10^{-7}}{6 \times 10^{-7}} = 2 \, \text{m} \]

### Final Answer:
The distance between the slit and the screen \( D \) is \( 2 \, \text{m} \).