Light of wavelength `lambda = 5890 Å` fall on a double-slit arrangement having separation `d = 0.2 mm`. A thin lens of focal length `f = 1 m` is placed near the slits. The linear separation of fringes on a screen placed in the focal plane of the lens is

Options

A. 3 mm
B. 4 mm
C. 2 mm
D. 1 mm

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given values
- Wavelength of light, \( \lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \)
- Separation of the slits, \( d = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)
- Focal length of the lens, \( f = 1 \, \text{m} \)

### Step 2: Write the formula for fringe width
The formula for the fringe width \( \beta \) in a double-slit experiment with a lens is given by:
\[ \beta = \frac{f \lambda}{d} \]

### Step 3: Substitute the values into the formula
Now, we will substitute the values into the formula:
\[ \beta = \frac{1 \, \text{m} \times 5890 \times 10^{-10} \, \text{m}}{0.2 \times 10^{-3} \, \text{m}} \]

### Step 4: Calculate the numerator and denominator
Calculating the numerator:
\[ 1 \times 5890 \times 10^{-10} = 5890 \times 10^{-10} \, \text{m} \]

Calculating the denominator:
\[ 0.2 \times 10^{-3} = 2 \times 10^{-4} \, \text{m} \]

### Step 5: Perform the division
Now, we can perform the division:
\[ \beta = \frac{5890 \times 10^{-10}}{2 \times 10^{-4}} = \frac{5890}{2} \times 10^{-10 + 4} = 2945 \times 10^{-6} \, \text{m} \]

### Step 6: Convert to millimeters
To convert from meters to millimeters:
\[ \beta = 2945 \times 10^{-6} \, \text{m} = 2.945 \, \text{mm} \approx 3 \, \text{mm} \]

### Conclusion
The linear separation of fringes on the screen placed in the focal plane of the lens is approximately \( 3 \, \text{mm} \).

### Final Answer
The correct option is \( 3 \, \text{mm} \).

---