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. (a)1.0 m
B. (b)1.5 m
C. (c)2.0 m
D. (d)2.3 m

Answer

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

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

### Formula:
For a single slit diffraction, the position of the first minimum is given by the formula:
\[ y_n = \frac{n \lambda D}{a} \]
where:
- \( y_n \) = position of the nth minimum
- \( n \) = order of the minimum (for first minimum, \( n = 1 \))
- \( \lambda \) = wavelength
- \( D \) = distance between the slit and the screen
- \( a \) = width of the slit

### Step 1: Rearranging the Formula
We need to find \( D \), so we rearrange the formula:
\[ D = \frac{y_n \cdot a}{n \lambda} \]

### Step 2: Substitute the Values
Substituting the known values into the rearranged formula:
\[ D = \frac{(4 \times 10^{-3} \, \text{m}) \cdot (3 \times 10^{-4} \, \text{m})}{1 \cdot (6 \times 10^{-7} \, \text{m})} \]

### Step 3: Calculate the Numerator
Calculating the numerator:
\[ 4 \times 10^{-3} \cdot 3 \times 10^{-4} = 12 \times 10^{-7} \, \text{m}^2 \]

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

### Conclusion
Thus, the distance between the slit and the screen is \( D = 2 \, \text{m} \).