In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength `600nm` is used. If the wavelength of light is changed to `400nm`, number of fringes observed in the same segment of the screen is given by

Options

A. 12
B. 18
C. 24
D. 30

Answer

To solve the problem of how many fringes will be observed when the wavelength of light is changed from `600 nm` to `400 nm`, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between fringe width and wavelength**:
The fringe width (β) in a Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance from the slits to the screen,
- \(d\) is the distance between the slits.

2. **Determine the number of fringes**:
The number of fringes (n) observed in a segment of the screen can be expressed as:
\[ n = \frac{L}{\beta} \]
where \(L\) is the length of the segment of the screen being observed.

3. **Relate the number of fringes for different wavelengths**:
Since the length of the segment \(L\) remains constant, we can set up the relationship between the number of fringes for the two different wavelengths:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
Here, \(n_1\) and \(n_2\) are the number of fringes for wavelengths \(\lambda_1\) and \(\lambda_2\) respectively.

4. **Substitute the known values**:
From the problem, we know:
- \(n_1 = 12\) (for \(\lambda_1 = 600 nm\))
- \(\lambda_1 = 600 nm\)
- \(\lambda_2 = 400 nm\)

Plugging these values into the equation:
\[ 12 \times 600 = n_2 \times 400 \]

5. **Solve for \(n_2\)**:
Rearranging the equation to find \(n_2\):
\[ n_2 = \frac{12 \times 600}{400} \]
Simplifying this gives:
\[ n_2 = \frac{7200}{400} = 18 \]

6. **Conclusion**:
Therefore, the number of fringes observed when the wavelength is changed to `400 nm` is **18**.