In a Young's double-slit experment, the fringe width is `beta`. If the entire arrangement is now placed inside a liquid of refractive index `mu`, the fringe width will become

Options

A. nb
B. `beta/(n+1)`
C. `beta/(n-1)`
D. `beta/n`

Answer

To solve the problem, we need to understand how the fringe width in a Young's double-slit experiment changes when the entire setup is placed in a medium with a refractive index \( \mu \).

### Step-by-Step Solution:

1. **Understanding Fringe Width**:
The fringe width \( \beta \) 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. **Effect of Refractive Index**:
When the entire arrangement is placed in a medium with refractive index \( \mu \), the wavelength of light in the medium changes. The new wavelength \( \lambda' \) in the medium is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
This is because the speed of light in the medium is reduced by the factor of \( \mu \).

3. **Calculating New Fringe Width**:
The new fringe width \( \beta' \) in the medium can be calculated using the modified wavelength:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\left(\frac{\lambda}{\mu}\right) D}{d} \]
Simplifying this gives:
\[ \beta' = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu} \]

4. **Conclusion**:
Therefore, the new fringe width \( \beta' \) when the setup is placed in a medium of refractive index \( \mu \) is:
\[ \beta' = \frac{\beta}{\mu} \]

### Final Answer:
The fringe width will become \( \frac{\beta}{\mu} \).