A plane wavefront `(lambda=6xx10^-7m)` falls on a slit `0.4m` wide. A convex lens of focal length `0.8m` placed behind the slit focuses the light on a screen. What is the linear diameter of second maximum?

Options

A. (a) `6mm`
B. (b) `12mm`
C. (c) `3mm`
D. (d) `9mm`

Explanation

To solve the problem of finding the linear diameter of the second maximum in the diffraction pattern created by a slit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength, \( \lambda = 6 \times 10^{-7} \, \text{m} \)
- Width of the slit, \( d = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- Focal length of the lens, \( f = 0.8 \, \text{m} \)

2. **Determine the Position of the Second Maximum:**
- For the second maximum in a single-slit diffraction pattern, the position can be approximated using the formula:
\[ d \sin \theta = (m + 0.5) \lambda \]
where \( m = 2 \) for the second maximum.
- Therefore, we have:
\[ d \sin \theta = (2 + 0.5) \lambda = 2.5 \lambda \]

3. **Substituting the Values:**
- Substitute \( \lambda \) into the equation:
\[ d \sin \theta = 2.5 \times (6 \times 10^{-7} \, \text{m}) = 15 \times 10^{-7} \, \text{m} \]

4. **Using Small Angle Approximation:**
- For small angles, \( \sin \theta \approx \tan \theta \approx \frac{x}{f} \), where \( x \) is the distance from the center to the maximum on the screen.
- Therefore, we can rewrite the equation as:
\[ d \frac{x}{f} = 15 \times 10^{-7} \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ x = \frac{15 \times 10^{-7} \cdot f}{d} \]

6. **Substituting the Known Values:**
- Substitute \( f = 0.8 \, \text{m} \) and \( d = 0.4 \times 10^{-3} \, \text{m} \):
\[ x = \frac{15 \times 10^{-7} \cdot 0.8}{0.4 \times 10^{-3}} \]

7. **Calculating \( x \):**
- Calculate the value:
\[ x = \frac{15 \times 0.8}{0.4} \times 10^{-4} = 30 \times 10^{-4} \, \text{m} = 3 \times 10^{-3} \, \text{m} = 3 \, \text{mm} \]

8. **Finding the Diameter:**
- The diameter of the second maximum is twice the radius:
\[ \text{Diameter} = 2x = 2 \times 3 \, \text{mm} = 6 \, \text{mm} \]

### Final Answer:
The linear diameter of the second maximum is **6 mm**.

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