Separation batween the slits in Young's double-slit experiment is 0.1 mm and distance of the screen from plane of slits is 1 m. Two wavelengths 400 nm and 560 nm are used in the experiment simultaneously. It is found that `m^(th)` dark from centre corresponding to 400 nm coincides with some dark of 560 nm. What is the minimum value of m?
`{:(0,1,2,3,4,5,6,7,8,9):}`

Explanation

To solve the problem, we need to find the minimum value of \( m \) such that the \( m^{th} \) dark fringe for the wavelength \( \lambda_1 = 400 \, \text{nm} \) coincides with the \( n^{th} \) dark fringe for the wavelength \( \lambda_2 = 560 \, \text{nm} \) in Young's double-slit experiment.

### Step-by-Step Solution:

1. **Identify the formula for dark fringes in Young's double-slit experiment**:
The position of the \( m^{th} \) dark fringe is given by:
\[ Y_{m} = \frac{(2m + 1) \lambda D}{2d} \]
where:
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the separation between the slits.

2. **Set up the equations for both wavelengths**:
For \( \lambda_1 = 400 \, \text{nm} \):
\[ Y_{m} = \frac{(2m + 1) \cdot 400 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} = \frac{(2m + 1) \cdot 400 \times 10^{-9}}{0.0002} \]

For \( \lambda_2 = 560 \, \text{nm} \):
\[ Y_{n} = \frac{(2n + 1) \cdot 560 \times 10^{-9} \cdot 1}{2 \cdot 0.1 \times 10^{-3}} = \frac{(2n + 1) \cdot 560 \times 10^{-9}}{0.0002} \]

3. **Equate the two expressions**:
Since the \( m^{th} \) dark fringe for \( 400 \, \text{nm} \) coincides with the \( n^{th} \) dark fringe for \( 560 \, \text{nm} \), we set the two equations equal:
\[ \frac{(2m + 1) \cdot 400 \times 10^{-9}}{0.0002} = \frac{(2n + 1) \cdot 560 \times 10^{-9}}{0.0002} \]
The \( 0.0002 \) and \( 10^{-9} \) terms cancel out, leading to:
\[ (2m + 1) \cdot 400 = (2n + 1) \cdot 560 \]

4. **Rearranging the equation**:
Rearranging gives:
\[ 400(2m + 1) = 560(2n + 1) \]
Simplifying further:
\[ 2m + 1 = \frac{560}{400}(2n + 1) = \frac{7}{5}(2n + 1) \]

5. **Cross-multiplying**:
Cross-multiplying gives:
\[ 5(2m + 1) = 7(2n + 1) \]
Expanding both sides:
\[ 10m + 5 = 14n + 7 \]
Rearranging gives:
\[ 10m - 14n = 2 \]

6. **Finding integer solutions**:
Rearranging gives:
\[ 5m - 7n = 1 \]
We need to find the minimum positive integer value of \( m \) such that \( n \) is also an integer.

7. **Testing values for \( m \)**:
- For \( m = 3 \):
\[ 5(3) - 7n = 1 \implies 15 - 7n = 1 \implies 7n = 14 \implies n = 2 \]
- Therefore, \( m = 3 \) and \( n = 2 \) is a valid solution.

### Conclusion:
The minimum value of \( m \) such that the \( m^{th} \) dark fringe for \( 400 \, \text{nm} \) coincides with some dark fringe for \( 560 \, \text{nm} \) is:
\[ \boxed{3} \]