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

Options

A. 0.30 mm
B. 0.40 mm
C. 0.53 mm
D. 450 mm

Answer

To solve the problem of finding 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:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance from the slits to the screen,
- \(d\) is the distance between the slits.

### Step 2: Determine the effect of immersion in water
When the apparatus is immersed in water, the wavelength of light changes due to the refractive index of the medium. The relationship between the wavelength in vacuum (\(\lambda_0\)) and the wavelength in the medium (\(\lambda\)) is given by:
\[ \lambda = \frac{\lambda_0}{\mu} \]
where \(\mu\) is the refractive index of the medium. For water, \(\mu = \frac{4}{3}\).

### Step 3: Substitute the new wavelength into the fringe width formula
The new fringe width (\(\beta'\)) when the apparatus is in water can be expressed as:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\frac{\lambda_0}{\mu} D}{d} = \frac{\lambda_0 D}{\mu d} \]

### Step 4: Relate the new fringe width to the original fringe width
We can relate the new fringe width to the original fringe width:
\[ \beta' = \frac{\beta}{\mu} \]
where \(\beta\) is the original fringe width.

### Step 5: Calculate the new fringe width
Given that the original fringe width \(\beta = 0.4 \, \text{mm}\) and \(\mu = \frac{4}{3}\):
\[ \beta' = \frac{0.4 \, \text{mm}}{\frac{4}{3}} = 0.4 \, \text{mm} \times \frac{3}{4} = 0.3 \, \text{mm} \]

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

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