In Young's double-slit experiment `lambda = 500 nm, d = 1 mm`, and `D = 4 m`. The minimum distance from the central maximum for which the intensity is half of the maximum intensity is `'**' xx 10^(-4) m`. What is the value of `'**'`?

Explanation

To find the minimum distance from the central maximum for which the intensity is half of the maximum intensity in Young's double-slit experiment, we can follow these steps:

### Step 1: Understand the relationship between intensity and phase difference
In Young's double-slit experiment, the intensity \( I \) at a point on the screen is given by:
\[ I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_0 \) is the intensity due to each slit and \( \phi \) is the phase difference.

### Step 2: Set up the equation for half maximum intensity
The maximum intensity \( I_{\text{max}} \) occurs when \( \phi = 0 \), which gives:
\[ I_{\text{max}} = 4 I_0 \]
We want to find the position where the intensity is half of the maximum:
\[ I = \frac{1}{2} I_{\text{max}} = 2 I_0 \]
Substituting this into the intensity equation:
\[ 2 I_0 = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( 2 I_0 \):
\[ 1 = 2 \cos^2\left(\frac{\phi}{2}\right) \]
Thus,
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2} \]

### Step 3: Find the phase difference
Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

### Step 4: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \Delta x = \frac{\lambda}{2\pi} \phi \]
Substituting \( \phi = \frac{\pi}{2} \):
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{2} = \frac{\lambda}{4} \]

### Step 5: Calculate the path difference using geometry
The path difference can also be expressed in terms of the geometry of the setup:
\[ \Delta x = \frac{d y}{D} \]
where \( d \) is the slit separation, \( y \) is the distance from the central maximum, and \( D \) is the distance from the slits to the screen.

### Step 6: Set the equations equal to each other
Equating the two expressions for path difference:
\[ \frac{d y}{D} = \frac{\lambda}{4} \]
Rearranging gives:
\[ y = \frac{\lambda D}{4d} \]

### Step 7: Substitute the known values
Given:
- \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- \( D = 4 \, \text{m} \)

Substituting these values into the equation:
\[ y = \frac{(500 \times 10^{-9}) \times 4}{4 \times 10^{-3}} = \frac{500 \times 10^{-9} \times 4}{4 \times 10^{-3}} = \frac{500 \times 10^{-9}}{10^{-3}} = 500 \times 10^{-6} = 5 \times 10^{-4} \, \text{m} \]

### Step 8: Final answer
The minimum distance from the central maximum for which the intensity is half of the maximum intensity is:
\[ y = 5 \times 10^{-4} \, \text{m} \]
Thus, the value of '**' is 5.