In Young's double-slit experiment the angular width of a fringe formed on a distant screen is `1^(@)`. The wavelength of light used is `6000 Å`. What is the spacing between the slits?

Options

A. 344 mm
B. 0.1344mm
C. 0.0344 mm
D. 0.034 mm

Explanation

To find the spacing between the slits in Young's double-slit experiment given the angular width of a fringe and the wavelength of light, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between angular fringe width and fringe width**:
The angular width of the fringe (θ) is related to the distance between the slits (d) and the distance to the screen (D) as follows:
\[ \theta = \frac{y}{D} \]
where \(y\) is the fringe width.

2. **Express fringe width in terms of wavelength and slit separation**:
The fringe width (β) can be expressed as:
\[ \beta = \frac{\lambda D}{d} \]
where \(λ\) is the wavelength of light.

3. **Relate angular fringe width to fringe width**:
Since the angular width is given as \(1^\circ\), we can convert this to radians:
\[ \theta_0 = 1^\circ = \frac{\pi}{180} \text{ radians} \]

4. **Substituting values**:
We can substitute \(y\) with the fringe width (β) and rewrite the equation:
\[ \theta_0 = \frac{\beta}{D} \]
Therefore, we can express \(d\) in terms of \(λ\), \(D\), and θ:
\[ d = \frac{\lambda D}{\beta} \]

5. **Substituting the expression for β**:
From the previous expression for β, we can substitute it back into the equation:
\[ \beta = \theta_0 D \]
Thus, we have:
\[ d = \frac{\lambda D}{\theta_0 D} = \frac{\lambda}{\theta_0} \]

6. **Plugging in the values**:
Given:
- Wavelength \(λ = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}\)
- Angular width \(θ_0 = \frac{\pi}{180} \text{ radians}\)

Now, substituting these values:
\[ d = \frac{6 \times 10^{-7}}{\frac{\pi}{180}} = 6 \times 10^{-7} \times \frac{180}{\pi} \]

7. **Calculating the final value**:
\[ d \approx 6 \times 10^{-7} \times 57.2958 \approx 3.44 \times 10^{-5} \text{ m} \]

8. **Converting to micrometers**:
\[ d \approx 0.0344 \text{ mm} = 34.4 \text{ micrometers} \]

### Final Answer:
The spacing between the slits is approximately **34.4 micrometers**.