In YDSE, bichromatic light of wavelengths 400 nm and 560 nm
are used. The distance between the slits is 0.1 mm and the distance between the
plane of the slits and the screen is 1m. The minimum distance between two
successive regions of complete darkness is
(a) 4mm (b) 5.6mm (c) 14 mm (d)28 mm .

Options

A. `4mm`
B. `5.6mm`
C. `14 mm`
D. `28 mm`

Explanation

To solve the problem of finding the minimum distance between two successive regions of complete darkness in Young's Double Slit Experiment (YDSE) using bichromatic light of wavelengths 400 nm and 560 nm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelengths: \( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 560 \, \text{nm} = 560 \times 10^{-9} \, \text{m} \)
- Distance between slits: \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen: \( D = 1 \, \text{m} \)

2. **Determine the Condition for Dark Fringes**:
The condition for dark fringes in YDSE is given by:
\[ y_n = \frac{(2n - 1) \lambda D}{2d} \]
where \( n \) is the order of the dark fringe.

3. **Calculate the Positions of Dark Fringes for Each Wavelength**:
For the first wavelength \( \lambda_1 \):
\[ y_{n1} = \frac{(2n_1 - 1) \cdot 400 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} \]
For the second wavelength \( \lambda_2 \):
\[ y_{n2} = \frac{(2n_2 - 1) \cdot 560 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} \]

4. **Set the Two Equations Equal for Complete Darkness**:
To find the minimum distance between two successive regions of complete darkness, we set the two equations equal:
\[ \frac{(2n_1 - 1) \cdot 400}{0.0002} = \frac{(2n_2 - 1) \cdot 560}{0.0002} \]
This simplifies to:
\[ (2n_1 - 1) \cdot 400 = (2n_2 - 1) \cdot 560 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{2n_1 - 1}{2n_2 - 1} = \frac{560}{400} = \frac{7}{5} \]
Let \( 2n_1 - 1 = 7k \) and \( 2n_2 - 1 = 5k \) for some integer \( k \).

6. **Solving for \( n_1 \) and \( n_2 \)**:
From \( 2n_1 - 1 = 7k \):
\[ n_1 = \frac{7k + 1}{2} \]
From \( 2n_2 - 1 = 5k \):
\[ n_2 = \frac{5k + 1}{2} \]
Both \( n_1 \) and \( n_2 \) must be integers, which means \( k \) must be odd.

7. **Finding the Minimum Value of \( k \)**:
Let \( k = 1 \):
\[ n_1 = \frac{7 \cdot 1 + 1}{2} = 4 \quad \text{and} \quad n_2 = \frac{5 \cdot 1 + 1}{2} = 3 \]

8. **Calculating the Positions of Dark Fringes**:
For \( n_1 = 4 \):
\[ y_{4} = \frac{(2 \cdot 4 - 1) \cdot 400 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} = \frac{7 \cdot 400 \times 10^{-9}}{0.0002} = 14 \, \text{mm} \]
For \( n_2 = 3 \):
\[ y_{3} = \frac{(2 \cdot 3 - 1) \cdot 560 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} = \frac{5 \cdot 560 \times 10^{-9}}{0.0002} = 14 \, \text{mm} \]

9. **Finding the Minimum Distance**:
The minimum distance between two successive regions of complete darkness is:
\[ \Delta y = y_{4} - y_{3} = 28 \, \text{mm} \]

### Final Answer:
Thus, the minimum distance between two successive regions of complete darkness is \( \boxed{28 \, \text{mm}} \).