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 `alpha`

Options

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

Explanation

To solve the problem, we need to calculate the minimum separation between two points on the screen in Young's double slit experiment where the intensities are \(2I_0\) and \(I_0\). Given that the fringe width is equal to \(\alpha\), we will follow these steps:

### Step 1: Understand the relationship between intensity and phase difference
The intensity \(I\) at a point on the screen in Young's double slit experiment is given by:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
where \(I_{\text{max}} = 4I_0\) for two slits of equal intensity \(I_0\).

### Step 2: Set up the equations for the given intensities
1. For intensity \(2I_0\):
\[ 2I_0 = 4I_0 \cos^2\left(\frac{\phi_1}{2}\right) \]
Simplifying this gives:
\[ \cos^2\left(\frac{\phi_1}{2}\right) = \frac{1}{2} \]
Therefore,
\[ \frac{\phi_1}{2} = \frac{\pi}{4} \implies \phi_1 = \frac{\pi}{2} \]

2. For intensity \(I_0\):
\[ I_0 = 4I_0 \cos^2\left(\frac{\phi_2}{2}\right) \]
Simplifying this gives:
\[ \cos^2\left(\frac{\phi_2}{2}\right) = \frac{1}{4} \]
Therefore,
\[ \frac{\phi_2}{2} = \frac{\pi}{3} \implies \phi_2 = \frac{2\pi}{3} \]

### Step 3: Relate phase difference to path difference
The phase difference \(\phi\) is related to the path difference \(\Delta x\) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Thus, we can express the path differences for both intensities:
1. For \(2I_0\):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x_1 \implies \Delta x_1 = \frac{\lambda}{4} \]

2. For \(I_0\):
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x_2 \implies \Delta x_2 = \frac{\lambda}{3} \]

### Step 4: Calculate the minimum separation on the screen
The path difference corresponds to the vertical distance \(y\) on the screen:
\[ \Delta x = y \frac{D}{d} \]
where \(D\) is the distance from the slits to the screen and \(d\) is the distance between the slits. Thus:
1. For \(y_1\):
\[ y_1 = \frac{\Delta x_1 d}{D} = \frac{\lambda d}{4D} \]

2. For \(y_2\):
\[ y_2 = \frac{\Delta x_2 d}{D} = \frac{\lambda d}{3D} \]

### Step 5: Find the minimum separation
The minimum separation \(\Delta y\) between the two points on the screen is:
\[ \Delta y = y_2 - y_1 = \frac{\lambda d}{3D} - \frac{\lambda d}{4D} \]
Finding a common denominator:
\[ \Delta y = \frac{4\lambda d - 3\lambda d}{12D} = \frac{\lambda d}{12D} \]

### Step 6: Relate fringe width to the separation
The fringe width \(\beta\) is given by:
\[ \beta = \frac{\lambda D}{d} \]
Thus, we can express \(\Delta y\) in terms of \(\alpha\):
\[ \Delta y = \frac{\beta}{12} = \frac{\alpha}{12} \]

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