In young's double slit experiment the fringe width is 4mm. If the experiment is shifted to water of refractive index 4/3 the fringe width becomes (in mm)

Options

A. 3
B. 4
C. 6
D. 8

Answer

To solve the problem regarding the fringe width in Young's double slit experiment when the setup is shifted to water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Fringe Width**:
The fringe width (β) in 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. **Initial Fringe Width**:
We are given that the initial fringe width in air is:
\[ \beta = 4 \text{ mm} \]

3. **Effect of Medium on Wavelength**:
When the experiment is shifted to a medium with a refractive index (μ), the wavelength of light changes. The new wavelength (λ') in the medium is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
where \( \mu = \frac{4}{3} \) for water.

4. **Calculate the New Wavelength**:
Substituting the value of μ:
\[ \lambda' = \frac{\lambda}{\frac{4}{3}} = \frac{3\lambda}{4} \]

5. **New Fringe Width in Water**:
The new fringe width (β') in water can be calculated using the modified wavelength:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\left(\frac{3\lambda}{4}\right) D}{d} \]
Since \( \beta = \frac{\lambda D}{d} \), we can express β' in terms of β:
\[ \beta' = \frac{3}{4} \beta \]

6. **Substituting the Initial Fringe Width**:
Now substituting the initial fringe width:
\[ \beta' = \frac{3}{4} \times 4 \text{ mm} = 3 \text{ mm} \]

7. **Conclusion**:
Therefore, the fringe width when the experiment is shifted to water becomes:
\[ \beta' = 3 \text{ mm} \]

### Final Answer:
The fringe width in water is **3 mm**.