To solve the problem, we will analyze the given information about the Young's double slit experiment with two different wavelengths of light.
### Step 1: Understanding Fringe Width
The fringe width (β) in a Young's double slit experiment is given by the formula:
\[
\beta = \frac{n \lambda D}{d}
\]
where:
- \( n \) is the order of the fringe,
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.
### Step 2: Comparing Fringe Widths
Given:
- \( \lambda_1 = 400 \, \text{nm} \)
- \( \lambda_2 = 600 \, \text{nm} \)
Since fringe width is directly proportional to the wavelength, we can conclude:
\[
\beta_1 < \beta_2
\]
This means that the fringe width for \( \lambda_1 \) is less than that for \( \lambda_2 \).
### Step 3: Number of Fringes
The number of fringes (m) that can fit within a certain distance \( y \) on one side of the central maximum is given by:
\[
m = \frac{y}{\beta}
\]
Since \( \beta_2 > \beta_1 \), it follows that:
\[
m_1 > m_2
\]
This means that more fringes can fit for the shorter wavelength \( \lambda_1 \) compared to the longer wavelength \( \lambda_2 \).
### Step 4: Overlapping of Maxima and Minima
To check if the third maximum of \( \lambda_2 \) overlaps with the fifth minimum of \( \lambda_1 \), we need to calculate their positions.
1. **Position of Third Maximum for \( \lambda_2 \)**:
\[
y_{3, \lambda_2} = 3 \beta_2 = 3 \left(\frac{600 \, \text{nm} \cdot D}{d}\right) = \frac{1800 \, \text{nm} \cdot D}{d}
\]
2. **Position of Fifth Minimum for \( \lambda_1 \)**:
The position of the nth minimum is given by:
\[
y_{n, \lambda_1} = \left(n - \frac{1}{2}\right) \beta_1
\]
For the fifth minimum:
\[
y_{5, \lambda_1} = \left(5 - \frac{1}{2}\right) \beta_1 = \frac{9}{2} \beta_1 = \frac{9}{2} \left(\frac{400 \, \text{nm} \cdot D}{d}\right) = \frac{1800 \, \text{nm} \cdot D}{d}
\]
Since both positions are equal:
\[
y_{3, \lambda_2} = y_{5, \lambda_1}
\]
This confirms that the third maximum of \( \lambda_2 \) overlaps with the fifth minimum of \( \lambda_1 \).
### Step 5: Angular Separation
The angular separation \( \theta \) is given by:
\[
\theta = \frac{\lambda}{d}
\]
Since \( \lambda_2 > \lambda_1 \), it follows that:
\[
\theta_2 > \theta_1
\]
This means that the angular separation for \( \lambda_2 \) is greater than that for \( \lambda_1 \).
### Conclusion
Based on the analysis:
- \(\beta_2 > \beta_1\) (Correct)
- \(m_1 > m_2\) (Correct)
- The third maximum of \( \lambda_2 \) overlaps with the fifth minimum of \( \lambda_1 \) (Correct)
- \(\theta_1 < \theta_2\) (Incorrect)
Thus, options A, B, and C are correct, while option D is incorrect.
