With two slits spaced 0.2 mm apart and a screen at a distance of 1 m, the third bright fringe is found to be at 7.5 mm from the central fringe. The wavelength of light used is

Options

A. 400nm
B. 500nm
C. 550nm
D. 600nm

Explanation

To find the wavelength of light used in the double-slit experiment, we can use the formula for the position of the bright fringes on the screen. The formula for the position of the nth bright fringe is given by:

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

Where:
- \( y_n \) = position of the nth bright fringe from the central maximum
- \( n \) = order of the fringe (for the third bright fringe, \( n = 3 \))
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step-by-Step Solution:

1. **Identify Given Values**:
- Distance between the slits, \( d = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)
- Distance to the screen, \( D = 1 \, \text{m} \)
- Position of the third bright fringe, \( y_3 = 7.5 \, \text{mm} = 7.5 \times 10^{-3} \, \text{m} \)
- Order of the fringe, \( n = 3 \)

2. **Use the Formula**:
We need to rearrange the formula to solve for the wavelength \( \lambda \):
\[ y_n = \frac{n \lambda D}{d} \implies \lambda = \frac{y_n d}{n D} \]

3. **Substitute the Values**:
Substitute the known values into the equation:
\[ \lambda = \frac{(7.5 \times 10^{-3} \, \text{m}) \times (0.2 \times 10^{-3} \, \text{m})}{3 \times (1 \, \text{m})} \]

4. **Calculate**:
- First, calculate the numerator:
\[ 7.5 \times 10^{-3} \times 0.2 \times 10^{-3} = 1.5 \times 10^{-6} \, \text{m}^2 \]
- Now, divide by the denominator:
\[ \lambda = \frac{1.5 \times 10^{-6}}{3} = 0.5 \times 10^{-6} \, \text{m} = 500 \times 10^{-9} \, \text{m} \]

5. **Convert to Nanometers**:
\[ \lambda = 500 \, \text{nm} \]

### Final Answer:
The wavelength of the light used is \( 500 \, \text{nm} \).