Light of wavelength `600 nm` is incident on an aperture of size `2 mm`. Calculate the distance light can travel before its spread is more than the size of aperture.

Options

A. 12.13 m
B. 6.67 m
C. 3.33 m
D. 2.19 m

Explanation

To solve the problem, we will use the concept of Fresnel's distance, which is a measure of how far light can travel before its spreading due to diffraction becomes significant compared to the size of the aperture.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Size of the aperture, \( a = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

2. **Fresnel's Distance Formula:**
The formula for Fresnel's distance \( Z_f \) is given by:
\[ Z_f = \frac{a^2}{\lambda} \]

3. **Substitute the Values into the Formula:**
\[ Z_f = \frac{(2 \times 10^{-3})^2}{600 \times 10^{-9}} \]

4. **Calculate \( a^2 \):**
\[ a^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{m}^2 \]

5. **Substitute \( a^2 \) into the Fresnel's Distance Formula:**
\[ Z_f = \frac{4 \times 10^{-6}}{600 \times 10^{-9}} \]

6. **Perform the Division:**
\[ Z_f = \frac{4 \times 10^{-6}}{600 \times 10^{-9}} = \frac{4}{600} \times 10^{3} = \frac{1}{150} \times 10^{3} = \frac{1000}{150} \approx 6.67 \, \text{m} \]

7. **Final Result:**
The distance light can travel before its spread is more than the size of the aperture is approximately:
\[ Z_f \approx 6.67 \, \text{m} \]

### Conclusion:
The light can travel approximately **6.67 meters** before its spread exceeds the size of the aperture.