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, we need to find the path difference (Δx) for the point where a bright fringe is formed in a Young's Double Slit Experiment (YDSE). Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Slit width (d) = 1 mm = \(1 \times 10^{-3}\) m
- Distance from the slit to the screen (D) = 1 m
- Wavelength of light (λ) = 632 nm = \(632 \times 10^{-9}\) m
- Position of the bright fringe on the screen (y) = 1.270 mm = \(1.270 \times 10^{-3}\) m

2. **Understanding Path Difference**:
- In YDSE, the path difference (Δx) between the two waves arriving at a point on the screen is given by:
\[ \Delta x = d \sin \theta \]
- Where \(d\) is the distance between the two slits and \(\theta\) is the angle made by the line connecting the point on the screen to the midpoint between the slits.

3. **Finding \(\sin \theta\)**:
- From the geometry of the setup, we can relate \(\sin \theta\) to the position on the screen (y) and the distance to the screen (D):
\[ \sin \theta \approx \frac{y}{D} \]
- Substituting the values:
\[ \sin \theta = \frac{1.270 \times 10^{-3}}{1} = 1.270 \times 10^{-3} \]

4. **Calculating Path Difference**:
- Now, substituting \(\sin \theta\) back into the path difference equation:
\[ \Delta x = d \sin \theta \]
- Substituting the values:
\[ \Delta x = (1 \times 10^{-3}) \times (1.270 \times 10^{-3}) \]
- Performing the multiplication:
\[ \Delta x = 1.270 \times 10^{-6} \text{ m} \]

5. **Converting to Micrometers**:
- To express the path difference in micrometers:
\[ \Delta x = 1.270 \text{ micrometers} \]

### Final Answer:
The path difference for the point where the bright fringe is formed is \(1.270 \, \mu m\).

---