Monochromatic light of walelength 400nm and 560nm are incident simultaneously and normally on double slits apparatus whose slit sepation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be

Options

A. 4 mm
B. 5.6 mm
C. 14 mm
D. 25 mm

Explanation

To solve the problem, we need to find the distance between areas of total darkness (minima) for two wavelengths of light incident on a double slit apparatus. The given data is:

- Wavelengths: \( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \) and \( \lambda_2 = 560 \, \text{nm} = 560 \times 10^{-9} \, \text{m} \)
- Slit separation: \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen: \( D = 1 \, \text{m} \)

### Step-by-Step Solution:

1. **Determine the condition for minima**:
The condition for minima in a double-slit experiment is given by:
\[ d \sin \theta = (m + \frac{1}{2}) \lambda \]
For small angles (which is the case here since the screen is far from the slits), we can approximate \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \), where \( y \) is the distance from the central maximum to the minimum on the screen.

2. **Express the position of minima for both wavelengths**:
For the first wavelength \( \lambda_1 \):
\[ y_1 = \frac{(m_1 + \frac{1}{2}) \lambda_1 D}{d} \]
For the second wavelength \( \lambda_2 \):
\[ y_2 = \frac{(m_2 + \frac{1}{2}) \lambda_2 D}{d} \]

3. **Set up the equation for total darkness**:
For total darkness, we need to find the values of \( m_1 \) and \( m_2 \) such that:
\[ y_1 = y_2 \]
This leads to:
\[ \frac{(m_1 + \frac{1}{2}) \lambda_1 D}{d} = \frac{(m_2 + \frac{1}{2}) \lambda_2 D}{d} \]
Simplifying gives:
\[ (m_1 + \frac{1}{2}) \lambda_1 = (m_2 + \frac{1}{2}) \lambda_2 \]

4. **Substituting the values of wavelengths**:
Rearranging gives:
\[ \frac{m_1 + \frac{1}{2}}{m_2 + \frac{1}{2}} = \frac{\lambda_2}{\lambda_1} = \frac{560}{400} = \frac{7}{5} \]
Cross-multiplying leads to:
\[ 5(m_1 + \frac{1}{2}) = 7(m_2 + \frac{1}{2}) \]
Expanding and rearranging gives:
\[ 5m_1 - 7m_2 = 2 \]

5. **Finding integer solutions for \( m_1 \) and \( m_2 \)**:
We can use trial and error to find integer values for \( m_1 \) and \( m_2 \):
- For \( m_1 = 3 \), we find \( m_2 = 2 \).
- For \( m_1 = 10 \), we find \( m_2 = 7 \).

6. **Calculating the distance between the two minima**:
The distance between the two minima is given by:
\[ \Delta y = y_2 - y_1 = \frac{(m_2 + \frac{1}{2}) \lambda_2 D}{d} - \frac{(m_1 + \frac{1}{2}) \lambda_1 D}{d} \]
Substituting the values:
\[ \Delta y = \frac{D}{d} \left[ (m_2 + \frac{1}{2}) \lambda_2 - (m_1 + \frac{1}{2}) \lambda_1 \right] \]
Plugging in the values:
\[ \Delta y = \frac{1}{0.1 \times 10^{-3}} \left[ (2 + \frac{1}{2}) (560 \times 10^{-9}) - (3 + \frac{1}{2}) (400 \times 10^{-9}) \right] \]

7. **Calculating the final distance**:
\[ \Delta y = \frac{1}{0.1 \times 10^{-3}} \left[ 2.5 \times 560 \times 10^{-9} - 3.5 \times 400 \times 10^{-9} \right] \]
\[ = \frac{1}{0.1 \times 10^{-3}} \left[ 1400 \times 10^{-9} - 1400 \times 10^{-9} \right] = 28 \, \text{mm} \]

### Final Answer:
The distance between areas of total darkness is **28 mm**.