If young's double slit experiment is performed in water

Options

A. the fringe width will decrease
B. the fringe width will increases
C. the fringe width will remain unchanged
D. there will be no fringe.

Answer

To solve the problem regarding Young's double slit experiment performed in water, we will analyze how the fringe width is affected by the change in medium.

### 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. **Identifying Changes in Medium**:
When the experiment is performed in water instead of air, the refractive index (μ) of water comes into play. The refractive index of water is greater than 1 (approximately 1.33).

3. **Effect on Wavelength**:
The wavelength of light in a medium is related to its wavelength in vacuum (or air) by the formula:
\[ \lambda_w = \frac{\lambda}{\mu} \]
where \( \lambda_w \) is the wavelength in water and \( \lambda \) is the wavelength in air.

4. **Calculating Wavelength in Water**:
Since the refractive index of water (μ) is greater than 1, the wavelength of light in water will be:
\[ \lambda_w < \lambda \]
This means that the wavelength decreases when moving from air to water.

5. **Substituting in Fringe Width Formula**:
Since \( d \) and \( t \) remain unchanged, the fringe width in water becomes:
\[ \beta_w = \frac{\lambda_w d}{t} = \frac{\left(\frac{\lambda}{\mu}\right) d}{t} \]
This shows that:
\[ \beta_w = \frac{\beta}{\mu} \]
Since \( \mu > 1 \), it follows that:
\[ \beta_w < \beta \]
Therefore, the fringe width decreases when the experiment is performed in water.

6. **Conclusion**:
The correct option is that the fringe width will decrease when Young's double slit experiment is performed in water.

### Final Answer:
**The fringe width will decrease.**

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