Young's double slit experiment is first performed in air and then in a medium other than air. It is found than 8th bright fringe in the medium lies where 5th daek fringe lies in air. The re3fractive index of the medium is nearly

Options

A. `1.25`
B. `1.59`
C. `1.69`
D. `1.78`

Answer

To solve the problem, we need to analyze the conditions of Young's double slit experiment in both air and another medium.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two scenarios: one in air and the other in a medium with a refractive index \( \mu \).
- The 8th bright fringe in the medium corresponds to the position of the 5th dark fringe in air.

2. **Path Difference for Bright and Dark Fringes**:
- For the 8th bright fringe in the medium, the path difference is given by:
\[ \text{Path difference} = 8\lambda \quad \text{(where } \lambda \text{ is the wavelength in the medium)} \]
- For the 5th dark fringe in air, the path difference is given by:
\[ \text{Path difference} = \left(2n - 1\right)\frac{\lambda_0}{2} = 9\frac{\lambda_0}{2} \quad \text{(where } \lambda_0 \text{ is the wavelength in air)} \]

3. **Equating the Path Differences**:
- Since both fringes occur at the same position, we can set the two path differences equal to each other:
\[ 8\lambda = 9\frac{\lambda_0}{2} \]

4. **Expressing Wavelengths**:
- Rearranging the equation gives:
\[ 16\lambda = 9\lambda_0 \]
- Thus, we can express the ratio of the wavelengths:
\[ \frac{\lambda_0}{\lambda} = \frac{16}{9} \]

5. **Finding the Refractive Index**:
- The refractive index \( \mu \) of the medium is given by the ratio of the wavelengths:
\[ \mu = \frac{\lambda_0}{\lambda} = \frac{16}{9} \]

6. **Final Calculation**:
- Calculating the numerical value gives:
\[ \mu \approx 1.78 \]

### Conclusion:
The refractive index of the medium is nearly \( 1.78 \).