In a standard YDSE slit width is 1mm and distance of screen from the slit is 1m. If wavelength of light is 632nm and bright fringe formed at y = 1.270 mm. Find path difference for the point

Options

A. 1.27 microm
B. 2.45 microm
C. 0.27 microm
D. 2.27 microm

Answer

To solve the problem of finding the path difference for the bright fringe formed at \( y = 1.270 \, \text{mm} \) in a Young's Double Slit Experiment (YDSE), we can follow these steps:

### Step 1: Identify the given values
- **Slit width (d)**: \( 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- **Distance from the slit to the screen (L)**: \( 1 \, \text{m} \)
- **Wavelength of light (λ)**: \( 632 \, \text{nm} = 632 \times 10^{-9} \, \text{m} \)
- **Position of the bright fringe (y)**: \( 1.270 \, \text{mm} = 1.270 \times 10^{-3} \, \text{m} \)

### Step 2: Use the small angle approximation
In YDSE, for small angles, we can use the approximation:
\[ \sin \theta \approx \tan \theta \approx \frac{y}{L} \]
where \( y \) is the distance from the central maximum to the fringe and \( L \) is the distance from the slits to the screen.

### Step 3: Calculate \( \tan \theta \)
Using the values given:
\[ \tan \theta = \frac{y}{L} = \frac{1.270 \times 10^{-3} \, \text{m}}{1 \, \text{m}} = 1.270 \times 10^{-3} \]

### Step 4: Calculate the path difference
The path difference \( \Delta x \) for the bright fringe is given by:
\[ \Delta x = d \sin \theta \approx d \tan \theta \]
Substituting the values:
\[ \Delta x = (1 \times 10^{-3} \, \text{m}) \times (1.270 \times 10^{-3}) = 1.270 \times 10^{-6} \, \text{m} \]

### Step 5: Convert to micrometers
To express the path difference in micrometers:
\[ \Delta x = 1.270 \, \mu m \]

### Final Answer
The path difference for the point where the bright fringe is formed at \( y = 1.270 \, \text{mm} \) is:
\[ \Delta x = 1.270 \, \mu m \]

---