A single slit of width 0.1mm is illuminated by parallel light of wavelength 6000Å, and diffraction bands are observed on a screen 40cm from the slit. The distance of third dark band from the central bright band is:

Options

A. 7.2 mm
B. 3.6 mm
C. 2.4 mm
D. 0.6 mm

Explanation

To solve the problem of finding the distance of the third dark band from the central bright band in a single slit diffraction experiment, we will follow these steps:

### Step-by-Step Solution:

1. **Convert the Given Values to SI Units**:
- Width of the slit \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1.0 \times 10^{-4} \, \text{m} \)
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6.0 \times 10^{-7} \, \text{m} \)
- Distance from the slit to the screen \( D = 40 \, \text{cm} = 0.4 \, \text{m} \)

2. **Use the Formula for Dark Bands in Single Slit Diffraction**:
- The condition for dark bands in single slit diffraction is given by:
\[ a \sin \theta = m \lambda \]
where \( a \) is the slit width, \( \lambda \) is the wavelength, and \( m \) is the order of the dark band (for the third dark band, \( m = 3 \)).

3. **Relate \( \sin \theta \) to \( x \)**:
- For small angles, \( \sin \theta \approx \tan \theta \approx \frac{x}{D} \), where \( x \) is the distance of the dark band from the central maximum.
- Thus, we can rewrite the equation as:
\[ \sin \theta = \frac{x}{D} \]

4. **Substituting into the Dark Band Condition**:
- Substitute \( \sin \theta \) into the dark band condition:
\[ \frac{x}{D} = \frac{m \lambda}{a} \]
- Rearranging gives:
\[ x = \frac{m \lambda D}{a} \]

5. **Plugging in the Values**:
- For the third dark band (\( m = 3 \)):
\[ x = \frac{3 \times (6.0 \times 10^{-7} \, \text{m}) \times (0.4 \, \text{m})}{1.0 \times 10^{-4} \, \text{m}} \]

6. **Calculating \( x \)**:
- Calculate:
\[ x = \frac{3 \times 6.0 \times 0.4}{1.0} \times 10^{-3} \, \text{m} \]
\[ x = \frac{7.2}{1.0} \times 10^{-3} \, \text{m} = 7.2 \times 10^{-3} \, \text{m} = 7.2 \, \text{mm} \]

### Final Answer:
The distance of the third dark band from the central bright band is **7.2 mm**.