In a young's double slit experiment, two wavelength of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again ? Take `D//d = 10^(3)` . Symbols have their usual meanings.

Explanation

To solve the problem of finding the minimum distance from the central maximum where the maximas of two wavelengths coincide in a Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelengths: \( \lambda_1 = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Wavelengths: \( \lambda_2 = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m} \)
- Ratio \( \frac{D}{d} = 10^3 \)

2. **Set Up the Condition for Coinciding Maxima:**
- Let the \( n_1 \)-th maxima of wavelength \( \lambda_1 \) coincide with the \( n_2 \)-th maxima of wavelength \( \lambda_2 \).
- The condition for maxima is given by:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

3. **Express the Ratio of Maxima:**
- Rearranging the equation gives:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]
- Substituting the values of the wavelengths:
\[ \frac{n_1}{n_2} = \frac{700 \times 10^{-9}}{500 \times 10^{-9}} = \frac{700}{500} = \frac{7}{5} \]

4. **Determine the Corresponding Maxima:**
- From the ratio \( \frac{n_1}{n_2} = \frac{7}{5} \), we can conclude:
- The 7th maxima of \( \lambda_1 \) coincides with the 5th maxima of \( \lambda_2 \).
- Additionally, the 14th maxima of \( \lambda_1 \) coincides with the 10th maxima of \( \lambda_2 \).

5. **Find the Minimum Integral Value for \( n_1 \):**
- The minimum integral value for \( n_1 \) is 7.

6. **Calculate the Minimum Distance from the Central Maximum:**
- The distance \( y \) from the central maximum for the \( n_1 \)-th maxima is given by:
\[ y = \frac{n_1 \lambda_1 D}{d} \]
- Substituting the known values:
\[ y = \frac{7 \times (500 \times 10^{-9}) \times D}{d} \]
- Since \( \frac{D}{d} = 10^3 \):
\[ y = 7 \times (500 \times 10^{-9}) \times 10^3 \]
- Calculating:
\[ y = 7 \times 500 \times 10^{-6} = 3500 \times 10^{-6} = 3.5 \, \text{mm} \]

### Final Answer:
The minimum distance from the central maximum where their maximas coincide again is **3.5 mm**.