Radiowaves of wavelength `3m` are incident on a rectangular hole of width `5m`. A screen is placed at a distance `200m` from the hole. The width of the central is `100bm`. Then find `b`.

Explanation

To solve the problem, we need to find the value of \( b \) given the parameters of the setup involving radio waves and a rectangular hole.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Wavelength of the radio waves, \( \lambda = 3 \, \text{m} \)
- Width of the rectangular hole, \( d = 5 \, \text{m} \)
- Distance from the hole to the screen, \( L = 200 \, \text{m} \)
- Width of the central maximum, \( W = 100 \, b \, \text{m} \)

2. **Understanding the formation of minima:**
- The positions of the minima in a single slit diffraction pattern are given by the formula:
\[ d \sin \theta = n \lambda \]
- For small angles, \( \sin \theta \approx \theta \) (in radians), hence:
\[ d \theta = n \lambda \]

3. **Determine the angular spread for the first minima:**
- The first minima occur at \( n = 1 \) on either side of the central maximum. Thus, the angular spread for the first minima on either side is:
\[ \theta = \frac{\lambda}{d} \]

4. **Calculate the angular spread:**
- Substituting the values:
\[ \theta = \frac{3 \, \text{m}}{5 \, \text{m}} = 0.6 \, \text{radians} \]

5. **Calculate the linear width of the central maximum:**
- The width of the central maximum is the distance between the first minima on either side, which is:
\[ W = 2 \times L \times \theta \]
- Substituting the values:
\[ W = 2 \times 200 \, \text{m} \times 0.6 = 240 \, \text{m} \]

6. **Relate the width of the central maximum to \( b \):**
- We know that the width of the central maximum is also given by:
\[ W = 100b \]
- Setting the two expressions for \( W \) equal gives:
\[ 100b = 240 \]

7. **Solve for \( b \):**
- Dividing both sides by 100:
\[ b = \frac{240}{100} = 2.4 \]

### Final Answer:
The value of \( b \) is \( 2.4 \).

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