In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:

Options

A. `0.17 ^@`
B. `0.07^@`
C. `0.57 ^@`
D. `1.7^@`

Explanation

To solve the problem of finding the angular width of the fringes in a Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given parameters**:
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Distance between the slits, \( d = 0.05 \, \text{mm} = 0.05 \times 10^{-3} \, \text{m} \)

2. **Formula for angular width**:
The angular width \( \theta \) of the fringe can be approximated for small angles using the formula:
\[ \theta \approx \frac{\lambda}{d} \]

3. **Substitute the values**:
\[ \theta \approx \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} \]

4. **Calculate the value**:
- First, simplify the fraction:
\[ \theta \approx \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} = \frac{500 \times 10^{-9}}{5 \times 10^{-5}} = \frac{500}{5} \times 10^{-9 + 5} = 100 \times 10^{-4} = 10^{-2} \, \text{radians} \]

5. **Convert radians to degrees**:
To convert radians to degrees, use the conversion factor \( \frac{180}{\pi} \):
\[ \theta \text{ (in degrees)} = 10^{-2} \times \frac{180}{\pi} \]
- Using \( \pi \approx 3.14 \):
\[ \theta \approx 10^{-2} \times \frac{180}{3.14} \approx 10^{-2} \times 57.3 \approx 0.573 \, \text{degrees} \]

6. **Final Result**:
The angular width of the fringes is approximately \( 0.57 \, \text{degrees} \).

### Conclusion:
The angular width of the fringes formed on the distant screen is close to \( 0.57 \, \text{degrees} \).