Width of slit in a single slit diffraction experiment such that 20 maxima of double slit interference pattern are obtained within central maxima of the diffraction pattern is ( Slit separation for double slit arrangement =2mm)

Options

A. 0.05mm
B. 0.1mm
C. 0.2 mm
D. 0.4 mm

Answer

To solve the problem of finding the width of the slit in a single-slit diffraction experiment such that 20 maxima of a double-slit interference pattern are obtained within the central maximum of the diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a double-slit interference pattern with a slit separation \( d = 2 \, \text{mm} \).
- We need to find the width of the single slit \( a \) such that there are 20 maxima of the double-slit interference pattern within the central maximum of the single-slit diffraction pattern.

2. **Formula for Double-Slit Interference**:
- The position of the maxima in a double-slit interference pattern is given by:
\[ y_n = \frac{n \lambda D}{d} \]
where \( n \) is the order of the maxima, \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

3. **Width of Central Maximum in Single-Slit Diffraction**:
- The width of the central maximum in a single-slit diffraction pattern is given by:
\[ W = \frac{2 \lambda D}{a} \]
where \( a \) is the width of the single slit.

4. **Condition for 20 Maxima**:
- For 20 maxima to fit within the central maximum, the distance from the center to the 10th maximum (since we count from 0 to 20) must equal half the width of the central maximum:
\[ y_{10} = \frac{10 \lambda D}{d} \]
- Therefore, we can set up the equation:
\[ \frac{10 \lambda D}{d} = \frac{W}{2} = \frac{\lambda D}{a} \]

5. **Substituting the Width of Central Maximum**:
- From the above equation:
\[ \frac{10 \lambda D}{d} = \frac{\lambda D}{a} \implies a = \frac{d}{10} \]

6. **Substituting the Value of \( d \)**:
- Given \( d = 2 \, \text{mm} \):
\[ a = \frac{2 \, \text{mm}}{10} = 0.2 \, \text{mm} \]

### Final Answer:
The width of the slit \( a \) is \( 0.2 \, \text{mm} \).

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