In normal YDSE experiment maximum intensity is `4I_(0)` in Column I, y-coordinate is given corresponding to centre line In Coloum II resultant intensities are given Match the two columns.
`{:(,"Column I",,"Column II"),("A",y=(lambdaD)/d,,p. I=I_(0)),("B",y=(lambdaD)/(2d),,q. I=2I_(0)),("C",y=(lambdaD)/(3d),,r. I=4I_(0)),("D",y=(lambdaD)/(4d),,s. I=zero):}`

Explanation

To solve the problem of matching the y-coordinates with their corresponding resultant intensities in the Young's Double Slit Experiment (YDSE), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Maximum Intensity**:
The maximum intensity in the YDSE is given as \(4I_0\). This occurs at the central maximum where the path difference is zero.

2. **Phase Difference Calculation**:
The phase difference \(\phi\) at any point on the screen can be expressed in terms of the path difference \(\Delta x\) as:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
The path difference \(\Delta x\) can be calculated using the formula:
\[ \Delta x = \frac{y \cdot d}{D} \]
where \(y\) is the distance from the central maximum, \(d\) is the distance between the slits, and \(D\) is the distance from the slits to the screen.

3. **Intensity Formula**:
The resultant intensity \(I\) at any point on the screen is given by:
\[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]

4. **Evaluate Each Case**:
We will evaluate the intensity for each y-coordinate given in Column I.

- **Case A**: \(y = \frac{\lambda D}{d}\)
- Path difference: \(\Delta x = \frac{\lambda D}{d}\)
- Phase difference: \(\phi = 2\pi\)
- Intensity:
\[ I = 4I_0 \cos^2(\pi) = 4I_0 \cdot 1 = 4I_0 \]

- **Case B**: \(y = \frac{\lambda D}{2d}\)
- Path difference: \(\Delta x = \frac{\lambda D}{2d}\)
- Phase difference: \(\phi = \pi\)
- Intensity:
\[ I = 4I_0 \cos^2\left(\frac{\pi}{2}\right) = 4I_0 \cdot 0 = 0 \]

- **Case C**: \(y = \frac{\lambda D}{3d}\)
- Path difference: \(\Delta x = \frac{\lambda D}{3d}\)
- Phase difference: \(\phi = \frac{2\pi}{3}\)
- Intensity:
\[ I = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \cdot \left(\frac{1}{2}\right)^2 = 4I_0 \cdot \frac{1}{4} = I_0 \]

- **Case D**: \(y = \frac{\lambda D}{4d}\)
- Path difference: \(\Delta x = \frac{\lambda D}{4d}\)
- Phase difference: \(\phi = \frac{\pi}{2}\)
- Intensity:
\[ I = 4I_0 \cos^2\left(\frac{\pi}{4}\right) = 4I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \cdot \frac{1}{2} = 2I_0 \]

5. **Matching Columns**:
Now we can match the results from Column I with Column II:
- A: \(y = \frac{\lambda D}{d} \rightarrow I = 4I_0\) (matches with r)
- B: \(y = \frac{\lambda D}{2d} \rightarrow I = 0\) (matches with s)
- C: \(y = \frac{\lambda D}{3d} \rightarrow I = I_0\) (matches with p)
- D: \(y = \frac{\lambda D}{4d} \rightarrow I = 2I_0\) (matches with q)

### Final Matching:
- A → r
- B → s
- C → p
- D → q