Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will see

Options

A. no interference pattern will be obtained
B. exactly same interference pattern will be obtained with better contrast
C. the fringe width is slightly decreased
D. the fringe width is slightly increased

Explanation

To solve the question, we need to analyze the effects of evacuating the air from an interference chamber on the interference pattern observed.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Initially, interference was observed in the chamber filled with air. The refractive index (μ) of air is slightly greater than 1, which affects the speed of light in that medium.

2. **Effect of Evacuating Air**:
- When the air is evacuated, the refractive index (μ) of the medium changes. In a vacuum, the refractive index is 1 (since there is no medium), which means the speed of light will increase.

3. **Relationship Between Wavelength and Refractive Index**:
- The wavelength of light in a medium is given by the formula:
\[ \lambda = \frac{\lambda_0}{\mu} \]
where \(\lambda_0\) is the wavelength in a vacuum. Since μ decreases when air is evacuated, the wavelength in the chamber will increase.

4. **Fringe Width Calculation**:
- The fringe width (β) in an interference pattern is given by:
\[ \beta = \frac{\lambda d}{D} \]
where \(d\) is the distance between the slits and \(D\) is the distance from the slits to the screen. As the wavelength (λ) increases due to the decrease in μ, the fringe width (β) will also increase.

5. **Conclusion**:
- Therefore, after evacuating the air, the careful observer will see that the fringe width has slightly increased.

### Final Answer:
The correct option is: **D) Fringe width slightly increased.**