A young's double slit apporatus is immersed in a liquid of refractive index 1.33.It has slit separation of 1 mm and interference pattern is observed on the screen at a distance 1.33 m from plane of slits.The wavelength in air is `6300 Å`
Calculate the fringe width.

Options

A. 0.63 mm
B. 1.26mm
C. 1.67mm
D. 2.2mm

Explanation

To calculate the fringe width in Young's double slit experiment when immersed in a liquid, follow these steps:

### Step 1: Understand the Given Data
- Wavelength in air, \( \lambda = 6300 \, \text{Å} = 6300 \times 10^{-10} \, \text{m} \)
- Refractive index of the liquid, \( \mu = 1.33 \)
- Slit separation, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1.33 \, \text{m} \)

### Step 2: Calculate the Wavelength in the Liquid
The wavelength of light in the liquid can be calculated using the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the values:
\[ \lambda' = \frac{6300 \times 10^{-10}}{1.33} \]

### Step 3: Calculate the Fringe Width
The fringe width \( \beta \) is given by the formula:
\[ \beta = \frac{\lambda' D}{d} \]
Substituting \( \lambda' \) from Step 2 into the fringe width formula:
\[ \beta = \frac{\left(\frac{6300 \times 10^{-10}}{1.33}\right) \times 1.33}{1 \times 10^{-3}} \]

### Step 4: Simplify the Expression
Notice that \( 1.33 \) in the numerator and denominator cancels out:
\[ \beta = \frac{6300 \times 10^{-10}}{1 \times 10^{-3}} \]

### Step 5: Calculate the Final Value
Now, calculate \( \beta \):
\[ \beta = 6300 \times 10^{-7} \, \text{m} = 0.63 \times 10^{-3} \, \text{m} = 0.63 \, \text{mm} \]

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

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