In Young's experiment , the fringe width of the fringes with light of wavelength 6000 Å is `2.0` mm . What will be the fringe width if the entire apparatus is immersed in a liquid of refractive index `1.33` ?

Options

A. 0.5 mm
B. 1 mm
C. 1.5 mm
D. 2 mm

Answer

To solve the problem, we need to find the new fringe width when the entire Young's experiment apparatus is immersed in a liquid with a refractive index of 1.33.

### Step-by-Step Solution:

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

2. **Determine the Wavelength in the Liquid**:
When the apparatus is immersed in a medium with refractive index \( \mu \), the effective wavelength \( \lambda' \) in the medium becomes:
\[ \lambda' = \frac{\lambda}{\mu} \]
Given:
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- \( \mu = 1.33 \)

Therefore,
\[ \lambda' = \frac{6000 \times 10^{-10}}{1.33} \approx 4.51 \times 10^{-10} \, \text{m} \]

3. **Calculate the New Fringe Width**:
The new fringe width \( \beta' \) in the liquid can be expressed as:
\[ \beta' = \frac{\lambda' D}{d} \]
Since \( D \) and \( d \) remain unchanged, we can relate the new fringe width to the original fringe width:
\[ \beta' = \frac{\lambda'}{\lambda} \cdot \beta \]
Substituting the values we have:
\[ \beta' = \frac{\lambda}{\mu} \cdot \beta \]
\[ \beta' = \frac{2.0 \, \text{mm}}{1.33} \approx 1.50 \, \text{mm} \]

4. **Final Result**:
The new fringe width when the apparatus is immersed in a liquid of refractive index 1.33 is approximately \( 1.50 \, \text{mm} \).