In a Young's double slit experiment `lamda= 500nm, d=1.0 mm andD=1.0m`. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Explanation

To solve the problem of finding the minimum distance from the central maximum in a Young's double slit experiment where the intensity is half of the maximum intensity, we can follow these steps:

### Step 1: Understand the relationship between intensity and amplitude
In a Young's double slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the amplitude \( A \) of the waves coming from the slits:
\[ I = 4A^2 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the two waves.

### Step 2: Set up the equation for half maximum intensity
The maximum intensity \( I_{max} \) occurs when \( \cos^2\left(\frac{\phi}{2}\right) = 1 \), thus:
\[ I_{max} = 4A^2 \]
According to the problem, we want to find the position where the intensity is half of the maximum intensity:
\[ I = \frac{I_{max}}{2} = 2A^2 \]
Setting the two expressions for intensity equal gives:
\[ 4A^2 \cos^2\left(\frac{\phi}{2}\right) = 2A^2 \]

### Step 3: Simplify the equation
Dividing both sides by \( 2A^2 \) (assuming \( A \neq 0 \)):
\[ 2 \cos^2\left(\frac{\phi}{2}\right) = 1 \]
This simplifies to:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2} \]

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

### Step 5: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
Solving for the path difference \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

### Step 6: Calculate the distance from the central maximum
The distance \( y \) from the central maximum to the point where the intensity is half can be calculated using the formula:
\[ y = \Delta x \cdot \frac{D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

Given:
- \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- \( D = 1 \, \text{m} \)
- \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)

Substituting these values:
\[ y = \left(\frac{500 \times 10^{-9}}{4}\right) \cdot \frac{1}{1 \times 10^{-3}} \]
Calculating \( y \):
\[ y = \left(\frac{500 \times 10^{-9}}{4}\right) \cdot 1000 = \frac{500 \times 10^{-6}}{4} = 125 \times 10^{-6} \, \text{m} = 125 \, \mu\text{m} \]

### Final Answer
The minimum distance from the central maximum for which the intensity is half of the maximum intensity is:
\[ y = 125 \, \mu\text{m} \]