Two coherent sources are 0.3 mm apart. They are 0.9m away from the screen. The second dark fringe is at a distance of 0.3cm from the centre. Find the distance of fourth bright fringe from the centre. Also, find the wavelength of light used.

Explanation

To solve the problem step by step, we will first find the wavelength of the light used and then calculate the distance of the fourth bright fringe from the center.

### Step 1: Understand the given data
- Distance between the coherent sources (d) = 0.3 mm = 0.3 × 10^-3 m = 0.0003 m
- Distance from the sources to the screen (D) = 0.9 m
- Distance of the second dark fringe from the center (y) = 0.3 cm = 0.003 m

### Step 2: Use the formula for dark fringes
The position of the nth dark fringe in a double-slit interference pattern is given by the formula:
\[ y_n = \frac{(n + \frac{1}{2}) \lambda D}{d} \]
For the second dark fringe (n = 2):
\[ y_2 = \frac{(2 + \frac{1}{2}) \lambda D}{d} \]
\[ y_2 = \frac{(2.5) \lambda D}{d} \]

### Step 3: Substitute the known values
We know \( y_2 = 0.003 \) m, \( D = 0.9 \) m, and \( d = 0.0003 \) m:
\[ 0.003 = \frac{(2.5) \lambda (0.9)}{0.0003} \]

### Step 4: Solve for the wavelength (λ)
Rearranging the equation to solve for λ:
\[ \lambda = \frac{0.003 \times 0.0003}{2.5 \times 0.9} \]
\[ \lambda = \frac{0.0000009}{2.25} \]
\[ \lambda = 4 \times 10^{-7} \text{ m} = 400 \text{ nm} \]

### Step 5: Find the distance of the fourth bright fringe
The position of the nth bright fringe is given by:
\[ y_n = \frac{n \lambda D}{d} \]
For the fourth bright fringe (n = 4):
\[ y_4 = \frac{4 \lambda D}{d} \]

### Step 6: Substitute the values of λ, D, and d
Substituting the values:
\[ y_4 = \frac{4 \times (4 \times 10^{-7}) \times 0.9}{0.0003} \]
\[ y_4 = \frac{1.44 \times 10^{-6}}{0.0003} \]
\[ y_4 = 0.0048 \text{ m} = 0.48 \text{ cm} \]

### Final Answers:
- Wavelength of light used (λ) = 400 nm
- Distance of the fourth bright fringe from the center = 0.48 cm