In Y.D.S.E., light of wavelength 600 nm is used. When slits are 1.2 mm apart, interference pattern of span `5 xx 10^(-3)m` is observed on a screen placed 2m away. Calculate the number of fringes observed in the interference pattern.

Answer

To calculate the number of fringes observed in the interference pattern in Young's Double Slit Experiment (Y.D.S.E.), we can follow these steps:

### Step 1: Identify the given values
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance between the slits, \( d = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 2 \, \text{m} \)
- Total width of the interference pattern, \( \beta_t = 5 \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the fringe width (\( \beta \))
The fringe width in Y.D.S.E. is given by the formula:
\[ \beta = \frac{D \lambda}{d} \]
Substituting the values:
\[ \beta = \frac{2 \times 600 \times 10^{-9}}{1.2 \times 10^{-3}} \]
Calculating this:
\[ \beta = \frac{1200 \times 10^{-9}}{1.2 \times 10^{-3}} = \frac{1200}{1.2} \times 10^{-6} = 1000 \times 10^{-6} = 10^{-3} \, \text{m} \]

### Step 3: Calculate the number of fringes (\( n \))
The number of fringes can be calculated by dividing the total width of the interference pattern by the fringe width:
\[ n = \frac{\beta_t}{\beta} \]
Substituting the values:
\[ n = \frac{5 \times 10^{-3}}{10^{-3}} = 5 \]

### Final Answer
The number of fringes observed in the interference pattern is \( n = 5 \).

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