In a Young's double slit experiment, the fringe width is found to be `0.4mm`. If the whole apparatus is immersed in water of refractive index `4//3` without disturbing the geometrical arrangement, the new fringe width will be

Explanation

To find the new fringe width when the Young's double slit experiment apparatus is immersed in water, we can follow these steps:

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

### Step 2: Identify the initial conditions
From the problem, we know:
- The initial fringe width \(\beta_1 = 0.4 \, \text{mm}\).
- The refractive index of water \( \mu = \frac{4}{3} \).

### Step 3: Determine the change in wavelength
When the apparatus is immersed in water, the wavelength of light changes. The new wavelength \(\lambda'\) in water is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the value of \(\mu\):
\[ \lambda' = \frac{\lambda}{\frac{4}{3}} = \frac{3\lambda}{4} \]

### Step 4: Relate the fringe widths before and after immersion
The new fringe width \(\beta_2\) can be related to the old fringe width \(\beta_1\) using the ratio of the wavelengths:
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda}{\lambda'} \]
Substituting \(\lambda' = \frac{3\lambda}{4}\):
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda}{\frac{3\lambda}{4}} = \frac{4}{3} \]
Thus, we can express \(\beta_2\) as:
\[ \beta_2 = \frac{3}{4} \beta_1 \]

### Step 5: Calculate the new fringe width
Now substituting \(\beta_1 = 0.4 \, \text{mm}\):
\[ \beta_2 = \frac{3}{4} \times 0.4 \, \text{mm} = 0.3 \, \text{mm} \]

### Conclusion
The new fringe width when the apparatus is immersed in water is:
\[ \beta_2 = 0.3 \, \text{mm} \]

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