A parallel beam of monochromatic light of wavelength 450 nm passes through a long slit of width 0.2 mm. find the angular divergence in which most of the light is diffracted.

Explanation

To solve the problem of finding the angular divergence in which most of the light is diffracted when a parallel beam of monochromatic light passes through a slit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength of light, \( \lambda = 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m} \)
- Width of the slit, \( b = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)

2. **Understand the Diffraction Condition:**
- The condition for minima in single-slit diffraction is given by:
\[ b \sin \theta = n \lambda \]
- For the first-order minima, set \( n = 1 \):
\[ b \sin \theta = \lambda \]

3. **Rearranging the Equation:**
- We can express \( \sin \theta \) as:
\[ \sin \theta = \frac{\lambda}{b} \]

4. **Substituting the Values:**
- Substitute the known values into the equation:
\[ \sin \theta = \frac{450 \times 10^{-9}}{0.2 \times 10^{-3}} \]

5. **Calculating \( \sin \theta \):**
- Perform the calculation:
\[ \sin \theta = \frac{450 \times 10^{-9}}{0.2 \times 10^{-3}} = 2.25 \times 10^{-3} \]

6. **Using Small Angle Approximation:**
- For small angles, \( \sin \theta \approx \theta \) (in radians):
\[ \theta \approx 2.25 \times 10^{-3} \, \text{radians} \]

7. **Finding Total Angular Divergence:**
- The total angular divergence is twice the angle to account for both sides of the central maximum:
\[ \text{Total Angular Divergence} = 2 \times \theta = 2 \times 2.25 \times 10^{-3} = 4.5 \times 10^{-3} \, \text{radians} \]

### Final Answer:
The angular divergence in which most of the light is diffracted is:
\[ \text{Total Angular Divergence} = 4.5 \times 10^{-3} \, \text{radians} \]

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