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 solve the problem, we will follow these steps:

### Step 1: Understand the problem
We are given:
- Wavelength, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Distance between the slits, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 4 \, \text{m} \)

We need to find the minimum distance \( x \) from the central maximum where the intensity is half of the maximum intensity.

### Step 2: Use the intensity formula
In Young's double-slit experiment, the intensity \( I \) at a point on the screen is given by:
\[ I = I_{\text{max}} \cos^2 \left( \frac{\phi}{2} \right) \]
where \( \phi \) is the phase difference between the two waves arriving at that point.

### Step 3: Set up the condition for half intensity
For the intensity to be half of the maximum intensity:
\[ I = \frac{I_{\text{max}}}{2} \]
This gives us:
\[ \frac{I_{\text{max}}}{2} = I_{\text{max}} \cos^2 \left( \frac{\phi}{2} \right) \]
Dividing both sides by \( I_{\text{max}} \):
\[ \frac{1}{2} = \cos^2 \left( \frac{\phi}{2} \right) \]
Taking the square root:
\[ \cos \left( \frac{\phi}{2} \right) = \frac{1}{\sqrt{2}} \implies \frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2} \]

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

### Step 5: Express path difference in terms of \( x \)
The path difference \( \Delta x \) for a point at distance \( x \) from the central maximum is given by:
\[ \Delta x = \frac{xD}{d} \]
Setting this equal to \( \frac{\lambda}{4} \):
\[ \frac{xD}{d} = \frac{\lambda}{4} \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = \frac{\lambda d}{4D} \]
Substituting the known values:
\[ x = \frac{(500 \times 10^{-9} \, \text{m})(1 \times 10^{-3} \, \text{m})}{4(4 \, \text{m})} \]
Calculating:
\[ x = \frac{500 \times 10^{-12}}{16} = 31.25 \times 10^{-6} \, \text{m} = 3.125 \times 10^{-5} \, \text{m} \]

### Step 7: Convert to required format
To express \( x \) in the form \( ** \times 10^{-4} \, \text{m} \):
\[ 3.125 \times 10^{-5} \, \text{m} = 0.3125 \times 10^{-4} \, \text{m} \]
Thus, the value of \( ** \) is approximately \( 3.125 \).

### Final Answer
The value of \( ** \) is \( 3 \) (if rounded to the nearest whole number).