In Young's experiment, two coherent sources are `1.5 mm` apart and the fringes are obtained at a distance of `2.5 m` from them. If the sources produce light of wavelength `589.3 nm`, find the number of fringes in the interference pattern, which is `4.9 xx 10^(-3) m` long.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Distance between the two coherent sources (d) = 1.5 mm = \(1.5 \times 10^{-3}\) m
- Distance from the sources to the screen (D) = 2.5 m
- Wavelength of light (\(\lambda\)) = 589.3 nm = \(589.3 \times 10^{-9}\) m
- Length of the interference pattern (L) = \(4.9 \times 10^{-3}\) m

### Step 2: Calculate the fringe width (β)
The fringe width (β) can be calculated using the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{(589.3 \times 10^{-9} \text{ m}) \times (2.5 \text{ m})}{(1.5 \times 10^{-3} \text{ m})} \]

### Step 3: Perform the calculation
Calculating the numerator:
\[ 589.3 \times 10^{-9} \times 2.5 = 1.47325 \times 10^{-6} \text{ m} \]
Now, divide by \(1.5 \times 10^{-3}\):
\[ \beta = \frac{1.47325 \times 10^{-6}}{1.5 \times 10^{-3}} = 0.00098217 \text{ m} = 982.17 \times 10^{-6} \text{ m} \]

### Step 4: Calculate the number of fringes (N)
The number of fringes (N) in the given length (L) can be calculated using the formula:
\[ N = \frac{L}{\beta} \]
Substituting the values:
\[ N = \frac{4.9 \times 10^{-3}}{982.17 \times 10^{-6}} \]

### Step 5: Perform the calculation
Calculating:
\[ N = \frac{4.9 \times 10^{-3}}{982.17 \times 10^{-6}} \approx 4.98 \]

### Conclusion
The number of fringes in the interference pattern is approximately 4.98, which can be rounded to 5.

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