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

Options

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

Explanation

To find 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 Problem
We are given:
- Wavelength, \( \lambda = 628 \, \text{nm} = 628 \times 10^{-9} \, \text{m} \)
- Angular divergence, \( \theta = 2^\circ \)

We need to find the size of the aperture, denoted as \( A \).

### Step 2: Relate Angular Divergence to Aperture Size
In diffraction, the angular divergence can be related to the aperture size using the formula:
\[ A \sin(\theta) = n \lambda \]
where \( n \) is the order of the maximum (for the first minimum, \( n = 1 \)).

### Step 3: Approximate \( \sin(\theta) \)
For small angles, \( \sin(\theta) \approx \theta \) in radians. Thus, we need to convert \( 2^\circ \) to radians:
\[ \theta \, \text{(in radians)} = \frac{2 \pi}{180} \approx \frac{2 \times 3.14}{180} \approx 0.0349 \, \text{radians} \]

### Step 4: Substitute Values into the Formula
Now we can substitute \( \sin(\theta) \) with \( \theta \) in radians:
\[ A \cdot 0.0349 = 1 \cdot (628 \times 10^{-9}) \]

### Step 5: Solve for Aperture Size \( A \)
Rearranging the equation gives:
\[ A = \frac{628 \times 10^{-9}}{0.0349} \]

### Step 6: Calculate \( A \)
Calculating the above expression:
\[ A \approx \frac{628 \times 10^{-9}}{0.0349} \approx 18.0 \times 10^{-6} \, \text{m} \]
Converting to micrometers:
\[ A \approx 18.0 \, \mu m \]

### Final Answer
Thus, the size of the aperture is approximately:
\[ \boxed{18 \, \mu m} \]

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