In young's double-slit experiment , the spacing between the slits is 'd' and the wavelength of light used is `6000Å` If the angular width of a fringe formed on a distance screen is `1^@` then calculate 'd' .

Options

A. 1 mm
B. 0.05 mm
C. 0.03 mm
D. 0.01 mm

Answer

To solve the problem, we need to find the spacing between the slits (d) in Young's double-slit experiment given the wavelength of light (λ) and the angular width of the fringe (θ).

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Angular width of the fringe, \( \theta = 1^\circ \)

2. **Convert the angle from degrees to radians:**
\[ \theta \, \text{(in radians)} = \theta \, \text{(in degrees)} \times \frac{\pi}{180} \]
\[ \theta = 1 \times \frac{\pi}{180} \approx \frac{3.14}{180} \approx 0.01745 \, \text{radians} \]

3. **Use the formula for angular width in Young's double-slit experiment:**
The angular width \( \beta \theta \) is given by the formula:
\[ \beta \theta = \frac{\lambda}{d} \]
Rearranging this formula to solve for \( d \):
\[ d = \frac{\lambda}{\beta \theta} \]

4. **Substitute the known values into the formula:**
\[ d = \frac{6000 \times 10^{-10} \, \text{m}}{0.01745} \]

5. **Calculate \( d \):**
\[ d = \frac{6000 \times 10^{-10}}{0.01745} \approx \frac{6000 \times 10^{-10}}{0.01745} \approx 3.43 \times 10^{-6} \, \text{m} \]
To convert this to millimeters:
\[ d \approx 3.43 \times 10^{-6} \, \text{m} = 0.00343 \, \text{mm} = 0.0343 \, \text{mm} \]

6. **Final result:**
\[ d \approx 0.03 \, \text{mm} \]

### Conclusion:
The spacing between the slits \( d \) is approximately \( 0.03 \, \text{mm} \).