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 find the least distance from the central maximum where the bright fringes due to both wavelengths coincide in Young's double-slit experiment, we can follow these steps:

### Step 1: Understand the problem
We have two wavelengths of light:
- \( \lambda_1 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m} \)

The distance between the slits \( d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \) and the distance from the slits to the screen \( D = 150 \, \text{cm} = 1.5 \, \text{m} \).

### Step 2: Calculate the fringe width for both wavelengths
The fringe width \( \beta \) is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]

For \( \lambda_1 = 650 \, \text{nm} \):
\[ \beta_1 = \frac{650 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} = \frac{975 \times 10^{-9}}{3 \times 10^{-3}} = 3.25 \times 10^{-4} \, \text{m} \]

For \( \lambda_2 = 520 \, \text{nm} \):
\[ \beta_2 = \frac{520 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} = \frac{780 \times 10^{-9}}{3 \times 10^{-3}} = 2.6 \times 10^{-4} \, \text{m} \]

### Step 3: Find the condition for coincidence of bright fringes
The bright fringes will coincide when:
\[ n_1 \beta_1 = n_2 \beta_2 \]
Where \( n_1 \) and \( n_2 \) are the fringe order numbers for the two wavelengths.

### Step 4: Substitute the values of fringe widths
Substituting the values of \( \beta_1 \) and \( \beta_2 \):
\[ n_1 (3.25 \times 10^{-4}) = n_2 (2.6 \times 10^{-4}) \]
This simplifies to:
\[ \frac{n_1}{n_2} = \frac{2.6}{3.25} \]
Calculating the ratio:
\[ \frac{n_1}{n_2} = \frac{2.6 \times 100}{3.25 \times 100} = \frac{260}{325} = \frac{52}{65} = \frac{4}{5} \]
This means:
\[ 5n_1 = 4n_2 \]

### Step 5: Find the smallest integer values for \( n_1 \) and \( n_2 \)
The smallest integer values satisfying \( 5n_1 = 4n_2 \) are:
- Let \( n_1 = 4 \) and \( n_2 = 5 \).

### Step 6: Calculate the distance from the central maximum
The distance from the central maximum to the \( n_1 \)th bright fringe is:
\[ y_1 = n_1 \beta_1 = 4 \times 3.25 \times 10^{-4} = 1.3 \times 10^{-3} \, \text{m} \]

### Final Answer
The least distance from the central maximum where the bright fringes due to both wavelengths coincide is:
\[ \boxed{1.3 \times 10^{-3} \, \text{m}} \]