In Young's double slits experiment, light of wavelength `4000Å` is used to produced bright fringes of width 0.6 mm, at a distance of `2m`. If the whole apparatus is dipped in a liquid of refractive index 1.5, then the fringe width will be

Options

A. `0.2mm`
B. `0.3mm`
C. `0.4mm`
D. `1.2mm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between fringe width and wavelength
In Young's double-slit experiment, the fringe width (β) 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

### Step 2: Identify the initial conditions
From the problem, we have:
- Wavelength \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- Initial fringe width \( \beta_1 = 0.6 \, \text{mm} = 0.6 \times 10^{-3} \, \text{m} \)
- Distance \( D = 2 \, \text{m} \)

### Step 3: Determine the new wavelength in the liquid
When the apparatus is dipped in a liquid with a refractive index \( \mu = 1.5 \), the wavelength of light in the liquid (\( \lambda_2 \)) is given by:
\[ \lambda_2 = \frac{\lambda_1}{\mu} \]
Substituting the values:
\[ \lambda_2 = \frac{4000 \times 10^{-10}}{1.5} = \frac{4000}{1.5} \times 10^{-10} \, \text{m} = 2666.67 \times 10^{-10} \, \text{m} \]

### Step 4: Calculate the new fringe width in the liquid
Using the relationship between the fringe widths before and after dipping in the liquid:
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \]
Rearranging gives:
\[ \beta_2 = \beta_1 \cdot \frac{\lambda_2}{\lambda_1} \]
Substituting the values:
\[ \beta_2 = 0.6 \times 10^{-3} \cdot \frac{2666.67 \times 10^{-10}}{4000 \times 10^{-10}} \]
Calculating \( \frac{2666.67}{4000} \):
\[ \frac{2666.67}{4000} = 0.66667 \]
Thus:
\[ \beta_2 = 0.6 \times 10^{-3} \cdot 0.66667 = 0.4 \times 10^{-3} \, \text{m} = 0.4 \, \text{mm} \]

### Final Answer:
The new fringe width when the apparatus is dipped in the liquid is \( 0.4 \, \text{mm} \).

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