In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be:

Options

A. 28
B. 24
C. 30
D. 18

Explanation

To solve the problem, we need to determine how many fringes will be observed when the wavelength of light is changed from 700 nm to 400 nm in a Young's double slit experiment, given that 16 fringes were observed with the 700 nm wavelength.

### Step-by-Step Solution:

1. **Understanding the relationship between the number of fringes and wavelength**:
In a Young's double slit experiment, the number of fringes (N) observed in a segment of the screen is inversely proportional to the wavelength (λ) of the light used. This can be expressed as:
\[ N \propto \frac{1}{\lambda} \]
This means that if the wavelength decreases, the number of fringes observed will increase.

2. **Setting up the equation**:
Let \( N_1 \) be the number of fringes observed with wavelength \( \lambda_1 = 700 \, \text{nm} \), and \( N_2 \) be the number of fringes observed with wavelength \( \lambda_2 = 400 \, \text{nm} \).
Given that \( N_1 = 16 \), we can write:
\[ N_1 \lambda_1 = N_2 \lambda_2 \]

3. **Substituting the known values**:
We substitute \( N_1 = 16 \), \( \lambda_1 = 700 \, \text{nm} \), and \( \lambda_2 = 400 \, \text{nm} \) into the equation:
\[ 16 \times 700 = N_2 \times 400 \]

4. **Solving for \( N_2 \)**:
Rearranging the equation to solve for \( N_2 \):
\[ N_2 = \frac{16 \times 700}{400} \]
Simplifying the right-hand side:
\[ N_2 = \frac{11200}{400} = 28 \]

5. **Conclusion**:
Therefore, the number of fringes observed when the wavelength is changed to 400 nm will be \( N_2 = 28 \).

### Final Answer:
The number of fringes observed in the same segment of the screen when the wavelength is changed to 400 nm is **28**.