In Young's experiment, fringe width was found to be 0.8 mm. If whole apparatus is immersed in water of refractive Index, `mu = 4/3`, new fringe width is

Options

A. 0.6
B. 0.8
C. 0.3
D. 0.4

Answer

To solve the problem of finding the new fringe width in Young's experiment when the apparatus is immersed in water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Fringe Width Formula**:
The fringe width (β) in Young's experiment is given by the formula:
\[ \beta = \frac{D \cdot \lambda}{d} \]
where:
- \(D\) = distance between the slits and the screen
- \(\lambda\) = wavelength of the light used
- \(d\) = separation between the slits

2. **Identify Given Values**:
We are given:
- Initial fringe width, \(\beta = 0.8 \, \text{mm}\)
- Refractive index of water, \(\mu = \frac{4}{3}\)

3. **Determine the New Wavelength**:
When the apparatus is immersed in water, the wavelength of light changes. The new wavelength (\(\lambda'\)) can be calculated using the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the value of \(\mu\):
\[ \lambda' = \frac{\lambda}{\frac{4}{3}} = \frac{3\lambda}{4} \]

4. **Relate the New Fringe Width to the Old Fringe Width**:
The new fringe width (\(\beta'\)) can be expressed in terms of the new wavelength:
\[ \beta' = \frac{D \cdot \lambda'}{d} \]
Substituting \(\lambda' = \frac{3\lambda}{4}\):
\[ \beta' = \frac{D \cdot \frac{3\lambda}{4}}{d} \]

5. **Set Up the Ratio of Fringe Widths**:
We can set up a ratio of the old fringe width to the new fringe width:
\[ \frac{\beta}{\beta'} = \frac{\lambda}{\lambda'} \]
Substituting \(\lambda' = \frac{3\lambda}{4}\):
\[ \frac{\beta}{\beta'} = \frac{\lambda}{\frac{3\lambda}{4}} = \frac{4}{3} \]

6. **Solve for the New Fringe Width**:
Rearranging the equation gives:
\[ \beta' = \frac{3}{4} \cdot \beta \]
Substituting the value of \(\beta = 0.8 \, \text{mm}\):
\[ \beta' = \frac{3}{4} \cdot 0.8 \, \text{mm} = 0.6 \, \text{mm} \]

### Final Answer:
The new fringe width when the apparatus is immersed in water is:
\[ \beta' = 0.6 \, \text{mm} \]