Two slits are separated by a distance of `0.5mm` and illuminated with light of `lambda=6000Å`. If the screen is placed `2.5m` from the slits. The distance of the third bright image from the centre will be

Options

A. 1.5 mm
B. 3mm
C. 6 mm
D. 9 mm

Explanation

To find the distance of the third bright fringe from the center in a double-slit interference pattern, we can use the formula for the position of the bright fringes:

\[ y_n = \frac{n \lambda D}{d} \]

Where:
- \( y_n \) = distance of the nth bright fringe from the center
- \( n \) = order of the bright fringe (in this case, \( n = 3 \))
- \( \lambda \) = wavelength of the light (in meters)
- \( D \) = distance from the slits to the screen (in meters)
- \( d \) = distance between the slits (in meters)

### Step-by-Step Solution:

1. **Convert the given values to appropriate units:**
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Distance between the slits \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \)
- Distance from the slits to the screen \( D = 2.5 \, \text{m} \)

2. **Substitute the values into the formula for the third bright fringe (n = 3):**
\[ y_3 = \frac{3 \cdot \lambda \cdot D}{d} \]

3. **Plug in the values:**
\[ y_3 = \frac{3 \cdot (6 \times 10^{-7} \, \text{m}) \cdot (2.5 \, \text{m})}{5 \times 10^{-4} \, \text{m}} \]

4. **Calculate the numerator:**
\[ 3 \cdot (6 \times 10^{-7}) \cdot (2.5) = 4.5 \times 10^{-6} \, \text{m} \]

5. **Now divide by \( d \):**
\[ y_3 = \frac{4.5 \times 10^{-6} \, \text{m}}{5 \times 10^{-4} \, \text{m}} = 9 \times 10^{-3} \, \text{m} \]

6. **Convert the result to micrometers:**
\[ 9 \times 10^{-3} \, \text{m} = 9 \, \text{mm} \]

### Final Answer:
The distance of the third bright image from the center is \( 9 \, \text{mm} \).