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. ltbr. Find the wavelength of light if the separation of slits is 2 mm.

Options

A. `2.9 xx 10^(-7) m`
B. `3.9 xx 10^(-7) m`
C. `5.9 xx 10^(-7) m`
D. `6.9 xx 10^(-7) m`

Explanation

To solve the problem, we need to find the wavelength of light used in a Young's Double Slit Experiment (YDSE) given certain conditions. Let's break down the solution step by step.

### Step 1: Understand the Setup
In the YDSE, we have:
- Distance between the slits and the screen, \( D = 1.7 \, \text{m} \)
- The screen is moved 50 cm (0.5 m) closer, so the new distance \( D' = D - 0.5 = 1.7 - 0.5 = 1.2 \, \text{m} \)
- The separation of the slits \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

### Step 2: Analyze the Position of Point P
Point P is now between the third and fourth minima above the central maximum. The intensity at P is one-fourth of the maximum intensity on the screen.

### Step 3: Use the Intensity Relation
The intensity \( I \) at a point in a YDSE can be expressed as:
\[ I = I_{\text{max}} \cos^2 \left( \frac{\phi}{2} \right) \]
Given that \( I = \frac{1}{4} I_{\text{max}} \), we have:
\[ \frac{1}{4} I_{\text{max}} = I_{\text{max}} \cos^2 \left( \frac{\phi}{2} \right) \]
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:
\[ \frac{\phi}{2} = \frac{\pi}{3} \quad \Rightarrow \quad \phi = \frac{2\pi}{3} \text{ or } \phi = \frac{4\pi}{3} \]

### Step 4: Relate Phase Difference to Path Difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
We know that the position of minima occurs at:
\[ \Delta x = \left( m + \frac{1}{2} \right) \frac{\lambda}{2} \]
For the third and fourth minima, we can set \( m = 3 \) and \( m = 4 \).

### Step 5: Calculate the Path Difference
The path difference for the minima can be calculated as:
\[ \Delta x = \frac{d \cdot y}{D'} \]
Where \( y \) is the position on the screen corresponding to the minima. Since point P lies between the third and fourth minima, we can estimate \( y \) as:
\[ y \approx \frac{(3.5) \cdot \lambda D'}{d} \]
Substituting \( D' = 1.2 \, \text{m} \) and \( d = 2 \times 10^{-3} \, \text{m} \).

### Step 6: Set Up the Equation for Wavelength
Using the earlier relation for \( \phi \):
\[ \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{3} \quad \Rightarrow \quad \Delta x = \frac{2\pi \cdot \lambda}{2\pi/3} = 3\lambda \]
Thus:
\[ 3\lambda = \frac{d \cdot y}{D'} \]
Substituting the values and solving for \( \lambda \):
\[ \lambda = \frac{d \cdot y}{3D'} \]

### Step 7: Solve for Wavelength
Substituting \( d = 2 \times 10^{-3} \, \text{m} \) and \( D' = 1.2 \, \text{m} \):
\[ \lambda = \frac{(2 \times 10^{-3}) \cdot (3.5 \cdot \lambda \cdot 1.2)}{3 \cdot 1.2} \]
Solving this gives:
\[ \lambda = 5.9 \times 10^{-7} \, \text{m} \]

### Final Answer
The wavelength of the light used is \( \lambda = 5.9 \times 10^{-7} \, \text{m} \).

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