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, we need to determine the new fringe width when the Young's double-slit experiment apparatus is immersed in water with a refractive index of \( \mu = \frac{4}{3} \).

### Step-by-Step Solution:

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

2. **Initial Fringe Width**:
We are given that the initial fringe width (\( \beta \)) is 0.8 mm.

3. **Effect of Medium on Wavelength**:
When the apparatus is immersed in a medium (in this case, water), the wavelength of light changes. The new wavelength (\( \lambda' \)) in the medium can be calculated using the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
where \( \mu \) is the refractive index of the medium.

4. **Calculate New Fringe Width**:
The new fringe width (\( \beta' \)) in the medium can be expressed as:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda}{\mu} \cdot \frac{D}{d} = \frac{\beta}{\mu} \]
Thus, we can substitute the known values:
\[ \beta' = \frac{0.8 \text{ mm}}{\frac{4}{3}} \]

5. **Perform the Calculation**:
To calculate \( \beta' \):
\[ \beta' = 0.8 \text{ mm} \times \frac{3}{4} = 0.8 \times 0.75 = 0.6 \text{ mm} \]

6. **Conclusion**:
The new fringe width when the apparatus is immersed in water is:
\[ \beta' = 0.6 \text{ mm} \]

### Final Answer:
The new fringe width is **0.6 mm**.

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