A small aperture is illuminated with a parallel beam of `lambda= 628 nm`. The emergent beam has an angular divergence of `2^0`. The size of the aperture is

Options

A. `9 mu m`
B. `18 mu m`
C. `27 mu m`
D. `36 mu m`

Answer

To solve the problem of finding the size of the aperture given the wavelength of light and the angular divergence of the emergent beam, we can follow these steps:

### Step 1: Understand the relationship between aperture size, wavelength, and angular divergence.
The relationship can be expressed as:
\[ A \cdot \sin(\theta) = n \lambda \]
where:
- \( A \) is the size of the aperture,
- \( \theta \) is the angular divergence,
- \( n \) is an integer (for the first minimum, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Simplify the equation for small angles.
Since the angle \( \theta \) is small, we can use the small angle approximation:
\[ \sin(\theta) \approx \theta \]
Thus, the equation simplifies to:
\[ A \cdot \theta = n \lambda \]

### Step 3: Set \( n = 1 \) for the first minimum.
For the first minimum, we have:
\[ A \cdot \theta = \lambda \]
This implies:
\[ \theta = \frac{\lambda}{A} \]

### Step 4: Convert the angular divergence from degrees to radians.
Given that \( \theta = 2^\circ \), we need to convert this to radians:
\[ \theta \text{ (in radians)} = \theta \text{ (in degrees)} \cdot \frac{\pi}{180} \]
Calculating this gives:
\[ \theta = 2 \cdot \frac{\pi}{180} = \frac{2\pi}{180} = \frac{\pi}{90} \approx 0.0349 \text{ radians} \]

### Step 5: Substitute the values into the equation.
Now we can substitute \( \theta \) and \( \lambda \) into the equation:
\[ \frac{\lambda}{A} = \theta \]
Rearranging gives:
\[ A = \frac{\lambda}{\theta} \]

### Step 6: Substitute the values of \( \lambda \) and \( \theta \).
Given \( \lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m} \) and \( \theta \approx 0.0349 \text{ radians} \):
\[ A = \frac{628 \times 10^{-9}}{0.0349} \]

### Step 7: Calculate the size of the aperture.
Calculating this gives:
\[ A \approx \frac{628 \times 10^{-9}}{0.0349} \approx 18 \times 10^{-6} \text{ m} = 18 \text{ micrometers} \]

### Final Answer:
The size of the aperture is approximately **18 micrometers**.

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