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

Explanation

To solve the problem, we need to understand the relationship between the fringe width in Young's double-slit experiment and the refractive index of the medium in which the experiment is conducted.

### Step-by-Step Solution:

1. **Understand the formula for fringe width in air:**
The fringe width (β) in air 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 two slits.

2. **Determine the wavelength in the new medium:**
When the entire apparatus is placed in a medium with a refractive index \( \mu \), the wavelength of light in that medium (λ') is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]

3. **Write the new fringe width in the liquid:**
The fringe width in the liquid (β') can be expressed as:
\[ \beta' = \frac{\lambda' D}{d} \]
Substituting the expression for \( \lambda' \):
\[ \beta' = \frac{\left(\frac{\lambda}{\mu}\right) D}{d} \]

4. **Simplify the expression for fringe width in the liquid:**
This simplifies to:
\[ \beta' = \frac{\lambda D}{\mu d} \]
Now, we can relate this to the original fringe width (β):
\[ \beta' = \frac{\beta}{\mu} \]

5. **Final result:**
Therefore, the new fringe width when the apparatus is placed in a liquid of refractive index \( \mu \) is:
\[ \beta' = \frac{\beta}{\mu} \]

### Conclusion:
The fringe width in the liquid will be \( \frac{\beta}{\mu} \).

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