In a Young's double-slit experiment , the slits are separated by `0.28` mm and screen is placed `1.4` m away . The distance between the central bright fringe and the fourth bright fringe is measured to be `1.2` cm . Determine the wavelength of light used in the experiment .

Options

A. `6xx10^(-7)m`
B. `3xx10^(-7)m`
C. `1.5xx10^(-7)m`
D. `5xx10^(-6)m`

Explanation

To determine the wavelength of light used in a Young's double-slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Slit separation (d) = 0.28 mm = \(0.28 \times 10^{-3}\) m
- Distance to the screen (D) = 1.4 m
- Distance between the central bright fringe and the fourth bright fringe (Xn) = 1.2 cm = \(1.2 \times 10^{-2}\) m
- Order of the fringe (n) = 4

2. **Use the formula for fringe position:**
The position of the nth bright fringe in a Young's double-slit experiment is given by the formula:
\[ X_n = \frac{n \lambda D}{d} \]
where \(X_n\) is the distance from the central maximum to the nth bright fringe, \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits.

3. **Rearranging the formula to solve for wavelength (\(\lambda\)):**
\[ \lambda = \frac{X_n \cdot d}{n \cdot D} \]

4. **Substituting the values into the formula:**
\[ \lambda = \frac{(1.2 \times 10^{-2} \text{ m}) \cdot (0.28 \times 10^{-3} \text{ m})}{4 \cdot (1.4 \text{ m})} \]

5. **Calculating the numerator:**
\[ 1.2 \times 10^{-2} \cdot 0.28 \times 10^{-3} = 3.36 \times 10^{-5} \text{ m}^2 \]

6. **Calculating the denominator:**
\[ 4 \cdot 1.4 = 5.6 \text{ m} \]

7. **Calculating the wavelength (\(\lambda\)):**
\[ \lambda = \frac{3.36 \times 10^{-5}}{5.6} = 6 \times 10^{-7} \text{ m} \]

8. **Final result:**
The wavelength of light used in the experiment is \(6 \times 10^{-7}\) m or \(600 \text{ nm}\).