In a Young's double slit experiment, slits are separated by `0.5 mm` and the screen is placed `150 cm` away. A beam of light consisting of two wavelengths, `650 nm` and `520 nm`, is used to obtain interference fringes on the screen. The least distance from the commom central maximum to the point where the bright fringes fue to both the wavelengths coincide is

Options

A. `9.75 mm`
B. `15.6 mm`
C. `1.56 mm`
D. `7.8 mm`

Explanation

To solve the problem of finding the least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide in a Young's double slit experiment, we can follow these steps:

### Step 1: Understand the setup
In a Young's double slit experiment, we have two slits separated by a distance \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \) and a screen placed at a distance \( D = 150 \, \text{cm} = 1.5 \, \text{m} \) from the slits. The wavelengths of light used are \( \lambda_1 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \) and \( \lambda_2 = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m} \).

### Step 2: Write the condition for maxima
The position of the bright fringes (maxima) on the screen is given by the formula:
\[ x = \frac{n \lambda D}{d} \]
where \( n \) is the order of the maximum, \( \lambda \) is the wavelength, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

### Step 3: Set up equations for both wavelengths
For the first wavelength \( \lambda_1 \):
\[ x_1 = \frac{n_1 \lambda_1 D}{d} \]
For the second wavelength \( \lambda_2 \):
\[ x_2 = \frac{n_2 \lambda_2 D}{d} \]

### Step 4: Equate the two positions
Since we want to find the position where the bright fringes coincide, we set \( x_1 = x_2 \):
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
Cancelling \( D \) and \( d \) from both sides gives:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

### Step 5: Substitute the values of wavelengths
Substituting the values of the wavelengths:
\[ n_1 \cdot 650 \times 10^{-9} = n_2 \cdot 520 \times 10^{-9} \]
This simplifies to:
\[ n_1 \cdot 650 = n_2 \cdot 520 \]

### Step 6: Rearrange to find the ratio of \( n_1 \) and \( n_2 \)
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{520}{650} = \frac{52}{65} = \frac{4}{5} \]
This means \( n_1 = 4k \) and \( n_2 = 5k \) for some integer \( k \).

### Step 7: Find the least common multiple
The least common multiple of \( n_1 \) and \( n_2 \) is when \( k = 1 \):
- \( n_1 = 4 \)
- \( n_2 = 5 \)

### Step 8: Calculate the position of the coinciding fringe
Now, we can calculate the position \( x \) using either wavelength. Using \( n_1 = 4 \) and \( \lambda_1 = 650 \, \text{nm} \):
\[ x = \frac{n_1 \lambda_1 D}{d} = \frac{4 \cdot (650 \times 10^{-9}) \cdot (1.5)}{0.5 \times 10^{-3}} \]
Calculating this gives:
\[ x = \frac{4 \cdot 650 \cdot 1.5 \times 10^{-9}}{0.5 \times 10^{-3}} = \frac{3900 \times 10^{-9}}{0.5 \times 10^{-3}} = \frac{3900 \times 10^{-9}}{0.5 \times 10^{-3}} = 7.8 \, \text{mm} \]

### Final Answer
The least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide is \( 7.8 \, \text{mm} \).

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