In a YDSE using monochromatic visible light, the distance between the plate of slits and the screen is 1.7 m. At point P on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved 50 cm closer to the plane of slits. Point P now lies between third and fouth minima above the central maxima and the intensity at P in one-fourth of the maxima intensity on the screen.
Find the value of n.

Options

A. 4
B. 6
C. 2
D. 8

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and break it down into clear steps.

### Step 1: Understanding the Setup
In a Young's Double Slit Experiment (YDSE), we have two slits and a screen. The distance from the slits to the screen is initially given as 1.7 m. When the screen is moved 50 cm closer, the new distance from the slits to the screen becomes:
\[ D' = 1.7 \, \text{m} - 0.5 \, \text{m} = 1.2 \, \text{m} \]

### Step 2: Path Difference for Maximum Intensity
At point P, which is directly in front of the upper slit, the path difference is given by:
\[ \Delta x = \frac{Y \cdot d}{D} \]
Where:
- \( Y \) is the distance from the central maximum to point P,
- \( d \) is the distance between the slits,
- \( D \) is the distance from the slits to the screen.

Since point P is at maximum intensity, we can express this as:
\[ \Delta x = n \lambda \]
Where \( n \) is the order of the maximum and \( \lambda \) is the wavelength of the light.

### Step 3: Setting Up the Equation
Given that point P lies between the third and fourth minima after moving the screen, we know that the intensity at P is one-fourth of the maximum intensity:
\[ I = \frac{I_{\text{max}}}{4} \]

Using the intensity formula:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
We can set up the equation:
\[ \frac{I_{\text{max}}}{4} = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \]
Thus:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{3} \quad \text{or} \quad \frac{\phi}{2} = \frac{5\pi}{3} \]
So:
\[ \phi = \frac{2\pi}{3} \quad \text{or} \quad \phi = \frac{10\pi}{3} \]

### Step 4: Relating Phase to Path Difference
The phase difference \( \phi \) can also be expressed in terms of the path difference:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting for \( \Delta x \):
\[ \phi = \frac{2\pi}{\lambda} \left(\frac{Y \cdot d}{D'}\right) \]
Where \( D' = 1.2 \, \text{m} \).

### Step 5: Finding the Value of n
Now we can set the two expressions for \( \phi \) equal to each other:
1. From the intensity condition:
- For \( \phi = \frac{2\pi}{3} \):
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \left(\frac{Y \cdot d}{1.2}\right) \]
- For \( \phi = \frac{10\pi}{3} \):
\[ \frac{10\pi}{3} = \frac{2\pi}{\lambda} \left(\frac{Y \cdot d}{1.2}\right) \]

### Step 6: Solving for n
From the equations above, we can solve for \( n \):
1. For \( \phi = \frac{2\pi}{3} \):
\[ \frac{Y \cdot d}{1.2} = \frac{\lambda}{3} \]
- Rearranging gives us \( n = 2 \).

2. For \( \phi = \frac{10\pi}{3} \):
\[ \frac{Y \cdot d}{1.2} = \frac{5\lambda}{3} \]
- Rearranging gives us \( n = 6 \).

### Conclusion
The value of \( n \) that satisfies the conditions of the problem is:
\[ n = 2 \]