A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m) from a central maximum where the bright fringes due to both the wavelengths coincide ? The distance between the slits is 3 mm and the distance between the plane of the slits and the screen is 150 cm.

Answer

To solve the problem, we need to find the least distance from the central maximum where the bright fringes due to both wavelengths coincide in Young's double-slit experiment.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength 1, \( \lambda_1 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \)
- Wavelength 2, \( \lambda_2 = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m} \)
- Distance between the slits, \( d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 150 \, \text{cm} = 1.5 \, \text{m} \)

2. **Set Up the Condition for Coinciding Bright Fringes:**
- The nth bright fringe for wavelength \( \lambda_2 \) coincides with the (n-1)th bright fringe for wavelength \( \lambda_1 \).
- The condition for bright fringes is given by:
\[ n \lambda_2 = (n - 1) \lambda_1 \]

3. **Substitute the Wavelengths:**
- Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ n \cdot 520 = (n - 1) \cdot 650 \]

4. **Rearranging the Equation:**
- Expanding and rearranging gives:
\[ 520n = 650n - 650 \]
\[ 650 = 130n \]
\[ n = \frac{650}{130} = 5 \]

5. **Calculate the Position of the Coinciding Bright Fringes:**
- The position of the nth bright fringe for wavelength \( \lambda_2 \) is given by:
\[ y_n = \frac{n \lambda_2 D}{d} \]
- Substituting \( n = 5 \), \( \lambda_2 = 520 \times 10^{-9} \, \text{m} \), \( D = 1.5 \, \text{m} \), and \( d = 3 \times 10^{-3} \, \text{m} \):
\[ y_5 = \frac{5 \cdot (520 \times 10^{-9}) \cdot 1.5}{3 \times 10^{-3}} \]

6. **Calculate the Result:**
- Performing the calculation:
\[ y_5 = \frac{5 \cdot 520 \cdot 1.5 \times 10^{-9}}{3 \times 10^{-3}} = \frac{3900 \times 10^{-9}}{3 \times 10^{-3}} = 1.3 \times 10^{-3} \, \text{m} \]
- Converting to millimeters:
\[ y_5 = 1.3 \, \text{mm} \]

### Final Answer:
The least distance from the central maximum where the bright fringes due to both wavelengths coincide is **1.3 mm**.