In Young's double slit experiment, the distance between the slits is 1 mm and that between slit and screen is 1 meter and 10th fringe is 5 mm away from the central bright fringe, then wavelength of light used will be

Options

A. `5000 Å`
B. `6000 Å`
C. `7000 Å`
D. `8000 Å `

Answer

To solve the problem, we will use the formula for the position of the bright fringes in Young's double slit experiment. The formula is given by:

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

Where:
- \( y_n \) = position of the nth fringe from the central maximum
- \( n \) = order of the fringe (10 for the 10th fringe)
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance between the slits \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen \( D = 1 \, \text{m} \)
- Position of the 10th fringe from the central maximum \( y_{10} = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Order of the fringe \( n = 10 \)

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

3. **Plug in the Values:**
\[ \lambda = \frac{(5 \times 10^{-3}) \cdot (1 \times 10^{-3})}{10 \cdot 1} \]

4. **Calculate \( \lambda \):**
\[ \lambda = \frac{5 \times 10^{-6}}{10} = 5 \times 10^{-7} \, \text{m} \]

5. **Convert Wavelength to Angstroms:**
Since \( 1 \, \text{m} = 10^{10} \, \text{angstroms} \):
\[ \lambda = 5 \times 10^{-7} \, \text{m} = 5 \times 10^{-7} \times 10^{10} \, \text{angstroms} = 5000 \, \text{angstroms} \]

### Final Answer:
The wavelength of the light used is \( 5000 \, \text{angstroms} \).

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