Young's experiment is performed in air and then performed in water, the fringe width:

Options

A. will remain same
B. will decreases
C. will increases
D. all the above types of waves

Explanation

To solve the problem regarding the effect on fringe width when Young's experiment is performed in air and then in water, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Fringe Width**:
The fringe width (β) in Young's double-slit experiment is given by the formula:
\[ \beta = \frac{\lambda d}{t} \]
where:
- \( \lambda \) is the wavelength of light,
- \( d \) is the distance between the slits,
- \( t \) is the distance from the slits to the screen.

2. **Effect of Medium on Wavelength**:
When light travels from air (a rarer medium) to water (a denser medium), its wavelength changes. The wavelength in a medium is given by:
\[ \lambda' = \frac{\lambda}{n} \]
where \( n \) is the refractive index of the medium. For water, \( n \) is approximately 1.33.

3. **Calculating New Wavelength in Water**:
Since the refractive index of water is greater than that of air, the wavelength of light in water will be:
\[ \lambda_{water} = \frac{\lambda_{air}}{1.33} \]
This shows that the wavelength decreases when light enters water.

4. **Substituting into Fringe Width Formula**:
Now, substituting the new wavelength into the fringe width formula:
\[ \beta_{water} = \frac{\lambda_{water} d}{t} = \frac{\left(\frac{\lambda_{air}}{1.33}\right) d}{t} \]
This can be simplified to:
\[ \beta_{water} = \frac{\beta_{air}}{1.33} \]
where \( \beta_{air} \) is the fringe width in air.

5. **Conclusion**:
Since \( \beta_{water} < \beta_{air} \), we conclude that the fringe width decreases when the experiment is performed in water compared to air.

### Final Answer:
The fringe width will **decrease** when Young's experiment is performed in water compared to when it is performed in air.

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