Light of 600.0 nm is incident upon a single slit. The resulting diffraction pattern is observed on a screen that is 0.50 m from the slit. The distance between the first and third minima of the diffraction pattern is 0.80 mm. Which range of values listed below contains the width of the slit?

Options

A. 0.1 mm to 0.4 mm
B. 0.8 mm to 1.2 mm
C. 0.4 mm to 0.8 mm
D. 1.2 mm to 1.6 mm

Answer

To solve the problem, we need to find the width of the slit (denoted as \( d \)) using the given information about the diffraction pattern.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength of light, \( \lambda = 600.0 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( L = 0.50 \, \text{m} \)
- Distance between the first and third minima, \( x_3 - x_1 = 0.80 \, \text{mm} = 0.80 \times 10^{-3} \, \text{m} \)

2. **Determine the Positions of the Minima**:
- The position of the first minimum (\( x_1 \)) is given by the formula:
\[ x_1 = \frac{\lambda L}{d} \]
- The position of the third minimum (\( x_3 \)) is given by:
\[ x_3 = \frac{3\lambda L}{d} \]

3. **Calculate the Distance Between the Minima**:
- The distance between the first and third minima can be expressed as:
\[ x_3 - x_1 = \frac{3\lambda L}{d} - \frac{\lambda L}{d} = \frac{2\lambda L}{d} \]
- Setting this equal to the given distance:
\[ \frac{2\lambda L}{d} = 0.80 \times 10^{-3} \, \text{m} \]

4. **Rearranging the Equation to Solve for \( d \)**:
- Rearranging gives:
\[ d = \frac{2\lambda L}{0.80 \times 10^{-3}} \]

5. **Substituting the Known Values**:
- Substitute \( \lambda \) and \( L \):
\[ d = \frac{2 \times (600 \times 10^{-9}) \times (0.50)}{0.80 \times 10^{-3}} \]

6. **Calculating \( d \)**:
- Calculate the numerator:
\[ 2 \times 600 \times 10^{-9} \times 0.50 = 600 \times 10^{-9} \, \text{m} = 300 \times 10^{-9} \, \text{m} \]
- Now calculate \( d \):
\[ d = \frac{300 \times 10^{-9}}{0.80 \times 10^{-3}} = \frac{300}{0.80} \times 10^{-6} = 375 \times 10^{-6} \, \text{m} = 0.375 \, \text{mm} \]

7. **Final Result**:
- The width of the slit \( d \) is approximately \( 0.375 \, \text{mm} \).

### Conclusion:
The width of the slit is \( 0.375 \, \text{mm} \).