In a YOUNG slit experiment, we use a radiation of wavelength 0.65µm. The distance between the slit and the screen is 1.5m. The distance between the middle fringe and the 2nd bright fringe on the negati side and the 5th dark fringe on the positive side is 51.33mm. Find the distance between the two fringes

In a Young's double-slit experiment, the position of bright and dark fringes on the screen can be determined by using the following formulas:

1. **Bright fringes (constructive interference):**
   \[
   y_m = \frac{m \lambda D}{d}
   \]
   where:
   - \( y_m \) is the distance of the \( m \)-th bright fringe from the central maximum,
   - \( m \) is the fringe order (an integer),
   - \( \lambda \) is the wavelength of light,
   - \( D \) is the distance between the slit and the screen,
   - \( d \) is the distance between the slits.

2. **Dark fringes (destructive interference):**
   \[
   y_n = \frac{(n + \frac{1}{2}) \lambda D}{d}
   \]
   where:
   - \( y_n \) is the distance of the \( n \)-th dark fringe from the central maximum,
   - \( n \) is the fringe order for the dark fringes (an integer).

### Given Information:
- Wavelength \( \lambda = 0.65 \, \mu m = 6.5 \times 10^{-7} \, m \),
- Distance between the slit and the screen \( D = 1.5 \, m \),
- The distance between the middle fringe (central maximum) and the 2nd bright fringe on the negative side (\( m = -2 \)) and the 5th dark fringe on the positive side (\( n = 5 \)) is \( 51.33 \, mm = 51.33 \times 10^{-3} \, m \).

Let’s break the problem into two steps:

### 1. Calculate the position of the 2nd bright fringe:
For the 2nd bright fringe, \( m = -2 \):
\[
y_{\text{bright}} = \frac{-2 \times 6.5 \times 10^{-7} \times 1.5}{d}
\]

### 2. Calculate the position of the 5th dark fringe:
For the 5th dark fringe, \( n = 5 \):
\[
y_{\text{dark}} = \frac{(5 + \frac{1}{2}) \times 6.5 \times 10^{-7} \times 1.5}{d}
\]

### 3. The total distance between the fringes is:
\[
\Delta y = y_{\text{dark}} - y_{\text{bright}} = 51.33 \times 10^{-3} \, m
\]

We can solve these two equations simultaneously to find the slit separation \( d \), and hence, the exact positions of the fringes.

Let me calculate it for you.The distance between the slits, \( d \), is approximately \( 0.142 \, mm \).

Would you like to explore more details, or have any other questions? Here are some related questions:

1. How does the wavelength affect the fringe spacing in a Young's experiment?
2. What happens to the fringe pattern if the slit separation decreases?
3. How do we calculate the position of dark fringes in a single-slit diffraction experiment?
4. Can Young’s double-slit experiment be performed with sound waves?
5. What role does coherence play in the double-slit experiment?

**Tip**: The fringe spacing increases as the wavelength increases or the slit separation decreases.

In a YOUNG slit experiment, we use a radiation of wavelength 0.65µm. The distance between the slit and the screen is 1.5m. The distance between the middle fringe and the 2nd bright fringe on the negative side and the 5th dark fringe on the positive side is 51.33mm. Find the distance between the two fringes.

Learn how to solve a Young's double-slit experiment problem using radiation of wavelength 0.65µm. Understand how to calculate the distance between the 2nd bright fringe on the negative side and the 5th dark fringe on the positive side, using detailed wave interference formulas. Suitable for Grade 12 or college-level physics students.

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