In young`s double slit experiment the fringes are formed at a distance of 1m from double slits of sepration 0.12mm.calculate the distance of 6th bright band from the centre of screen ,given wavelength is 12000A.

Answer

To solve the problem of finding the distance of the 6th bright band from the center of the screen in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Identify and Convert Given Data:**
- Wavelength (λ) = 12000 Å (angstroms)
- Distance from slits to screen (D) = 1 m
- Slit separation (d) = 0.12 mm

**Conversions:**
- Convert wavelength from angstroms to centimeters:
\[ 1 \text{ Å} = 10^{-10} \text{ m} \Rightarrow 12000 \text{ Å} = 12000 \times 10^{-10} \text{ m} = 12 \times 10^{-6} \text{ m} = 12 \times 10^{-5} \text{ cm} \]
- Convert distance from slits to screen from meters to centimeters:
\[ 1 \text{ m} = 100 \text{ cm} \]
- Convert slit separation from mm to cm:
\[ 0.12 \text{ mm} = 0.012 \text{ cm} \]

2. **Use the Formula for Bright Fringe Position:**
The position of the nth bright fringe (Xn) in Young's double slit experiment is given by:
\[ X_n = \frac{n \lambda D}{d} \]
Where:
- \( n \) = order of the fringe (for the 6th bright band, \( n = 6 \))
- \( \lambda \) = wavelength
- \( D \) = distance from the slits to the screen
- \( d \) = slit separation

3. **Substitute the Values into the Formula:**
For the 6th bright band:
\[ X_6 = \frac{6 \times (12 \times 10^{-5} \text{ cm}) \times (100 \text{ cm})}{0.012 \text{ cm}} \]

4. **Calculate the Value:**
\[ X_6 = \frac{6 \times 12 \times 100 \times 10^{-5}}{0.012} \]
\[ = \frac{7200 \times 10^{-5}}{0.012} \]
\[ = \frac{7200}{0.012} \times 10^{-5} \]
\[ = 600000 \times 10^{-5} \text{ cm} \]
\[ = 6 \text{ cm} \]

5. **Final Result:**
The distance of the 6th bright band from the center of the screen is **6 cm**.