The intensity of each of the two slits in Young's double slit experiment is `I_0`. Calculate the minimum separation between the two points on the screen where intensities are `2I_0` and `I_0`. Given, the fringe width equal to `beta.`

Options

A. `beta/4`
B. `beta/3`
C. `beta/12`
D. None of these

Explanation

To solve the problem of finding the minimum separation between two points on the screen in Young's double slit experiment where the intensities are \(2I_0\) and \(I_0\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Intensities**:
- The intensity at a point on the screen due to two coherent sources can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\delta) \]
- Here, \(I_1 = I_0\) and \(I_2 = I_0\) (intensities from both slits). Thus, the maximum intensity \(I_{max} = 4I_0\) when the waves are in phase.

2. **Finding the Condition for \(I = 2I_0\)**:
- For \(I = 2I_0\):
\[ 2I_0 = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\delta) \]
\[ 2I_0 = 2I_0 + 2I_0 \cos(\delta) \]
\[ 0 = 2I_0 \cos(\delta) \]
- This implies \(\cos(\delta) = \frac{1}{\sqrt{2}}\), leading to a phase difference of:
\[ \delta_1 = \frac{\pi}{4} \times 2 = \frac{\pi}{2} \]

3. **Finding the Condition for \(I = I_0\)**:
- For \(I = I_0\):
\[ I_0 = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\delta) \]
\[ I_0 = 2I_0 + 2I_0 \cos(\delta) \]
\[ -I_0 = 2I_0 \cos(\delta) \]
- Thus, \(\cos(\delta) = -\frac{1}{2}\), leading to a phase difference of:
\[ \delta_2 = \frac{2\pi}{3} \]

4. **Calculating the Phase Difference**:
- The phase difference between the two points is:
\[ \Delta \delta = \delta_1 - \delta_2 = \frac{\pi}{2} - \frac{2\pi}{3} \]
- Converting to a common denominator:
\[ \Delta \delta = \frac{3\pi}{6} - \frac{4\pi}{6} = -\frac{\pi}{6} \]
- Taking the absolute value:
\[ |\Delta \delta| = \frac{\pi}{6} \]

5. **Relating Phase Difference to Position on the Screen**:
- The position difference \(y\) on the screen is related to the phase difference by:
\[ \Delta y = \frac{\beta}{2\pi} |\Delta \delta| \]
- Substituting the values:
\[ \Delta y = \frac{\beta}{2\pi} \cdot \frac{\pi}{6} = \frac{\beta}{12} \]

### Final Answer:
The minimum separation between the two points on the screen where the intensities are \(2I_0\) and \(I_0\) is:
\[ \Delta y = \frac{\beta}{12} \]