A double-slit experiment is immersed in a liquid of refractive index 1.33. The separation between the slits is 1mm and the distance between the slits and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is `6300 Å`, then fringe width will be

Options

A. 6.3mm
B. 63 mm
C. 0.63 mm
D. None of these

Answer

To solve the problem of finding the fringe width in a double-slit experiment immersed in a liquid, we can follow these steps:

### Step 1: Identify the Given Values
- Refractive index (μ) = 1.33
- Separation between the slits (d) = 1 mm = \(1 \times 10^{-3}\) m
- Distance between the slits and the screen (D) = 1.33 m
- Wavelength of light (λ) = 6300 Å = \(6300 \times 10^{-10}\) m = \(6.3 \times 10^{-7}\) m

### Step 2: Calculate the Effective Wavelength in the Liquid
The effective wavelength (λ') in the medium can be calculated using the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the values:
\[ \lambda' = \frac{6.3 \times 10^{-7}}{1.33} \]
Calculating this gives:
\[ \lambda' \approx 4.73 \times 10^{-7} \text{ m} \]

### Step 3: Calculate the Fringe Width (β)
The fringe width (β) is given by the formula:
\[ \beta = \frac{\lambda' D}{d} \]
Substituting the values:
\[ \beta = \frac{(4.73 \times 10^{-7}) \times (1.33)}{(1 \times 10^{-3})} \]

### Step 4: Perform the Calculation
Calculating the numerator:
\[ 4.73 \times 10^{-7} \times 1.33 \approx 6.2939 \times 10^{-7} \]
Now, divide by \(1 \times 10^{-3}\):
\[ \beta = \frac{6.2939 \times 10^{-7}}{1 \times 10^{-3}} = 6.2939 \times 10^{-4} \text{ m} \]

### Step 5: Convert to Millimeters
To convert meters to millimeters:
\[ \beta = 6.2939 \times 10^{-4} \text{ m} = 0.62939 \text{ mm} \approx 0.63 \text{ mm} \]

### Final Answer
Thus, the fringe width is approximately **0.63 mm**.

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