In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength `lamda`. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

Explanation

To solve the problem step by step, we will analyze the conditions for the intensity of light in a Young's double slit interference experiment.

### Given:
- Distance from slits to screen: \( D \)
- Distance between slits: \( d \)
- Wavelength of light: \( \lambda \)

### (a) Finding the distance from the central point where the intensity falls to half the maximum:

1. **Understanding Maximum Intensity**:
The maximum intensity \( I_{\text{max}} \) in a double slit experiment is given by:
\[ I_{\text{max}} = 4a^2 \]
where \( a \) is the amplitude of the light waves from each slit.

2. **Setting Up the Intensity Equation**:
The intensity at a point on the screen can be expressed as:
\[ I = 4a^2 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the two waves arriving at that point.

3. **Finding the Condition for Half Maximum Intensity**:
To find the position where the intensity is half of the maximum:
\[ I = \frac{I_{\text{max}}}{2} = 2a^2 \]
Setting the two equations equal:
\[ 4a^2 \cos^2\left(\frac{\phi}{2}\right) = 2a^2 \]
Dividing both sides by \( 4a^2 \):
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2} \]

4. **Finding the Phase Difference**:
Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
Therefore:
\[ \frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2} \]

5. **Relating 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 \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

6. **Finding the Position on the Screen**:
The position \( y \) on the screen corresponding to the path difference \( \Delta x \) is given by:
\[ y = \frac{\Delta x \cdot D}{d} \]
Substituting \( \Delta x = \frac{\lambda}{4} \):
\[ y = \frac{\frac{\lambda}{4} \cdot D}{d} = \frac{\lambda D}{4d} \]

### (b) Finding the distance from the central point where the intensity falls to one fourth of the maximum:

1. **Setting Up the Intensity Equation**:
For intensity to be one fourth of the maximum:
\[ I = \frac{I_{\text{max}}}{4} = a^2 \]
Setting the two equations equal:
\[ 4a^2 \cos^2\left(\frac{\phi}{2}\right) = a^2 \]
Dividing both sides by \( 4a^2 \):
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \]

2. **Finding the Phase Difference**:
Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \]
Therefore:
\[ \frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3} \]

3. **Relating Phase Difference to Path Difference**:
Using the same relation:
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{3} \]

4. **Finding the Position on the Screen**:
The position \( y \) on the screen corresponding to the path difference \( \Delta x \) is given by:
\[ y = \frac{\Delta x \cdot D}{d} \]
Substituting \( \Delta x = \frac{\lambda}{3} \):
\[ y = \frac{\frac{\lambda}{3} \cdot D}{d} = \frac{\lambda D}{3d} \]

### Final Answers:
- (a) The distance from the central point where the intensity falls to half the maximum is \( \frac{\lambda D}{4d} \).
- (b) The distance from the central point where the intensity falls to one fourth of the maximum is \( \frac{\lambda D}{3d} \).