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: Understand the given data
- Refractive index of the liquid, \( \mu = 1.33 \)
- Separation between the slits, \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Distance from the slits to the screen, \( D = 1.33 \text{ m} \)
- Wavelength of light, \( \lambda = 6300 \text{ Å} = 6300 \times 10^{-10} \text{ m} = 6.3 \times 10^{-7} \text{ m} \)

### Step 2: Calculate the effective wavelength in the liquid
The wavelength of light in a medium is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the values:
\[ \lambda' = \frac{6.3 \times 10^{-7} \text{ m}}{1.33} \]
Calculating this gives:
\[ \lambda' \approx 4.73 \times 10^{-7} \text{ m} \]

### Step 3: Use the formula for fringe width
The fringe width \( \beta \) in a double-slit experiment is given by:
\[ \beta = \frac{\lambda' D}{d} \]
Substituting the values we have:
\[ \beta = \frac{(4.73 \times 10^{-7} \text{ m})(1.33 \text{ m})}{1 \times 10^{-3} \text{ m}} \]

### Step 4: Calculate the fringe width
Calculating the above expression:
\[ \beta = \frac{4.73 \times 10^{-7} \times 1.33}{1 \times 10^{-3}} \]
\[ \beta \approx 6.293 \times 10^{-4} \text{ m} \]
Converting this to mm:
\[ \beta \approx 0.6293 \text{ mm} \approx 0.63 \text{ mm} \]

### Final Answer
The fringe width \( \beta \) is approximately \( 0.63 \text{ mm} \).

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