A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength `6000 Å` and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is:

Options

A. 3 mm
B. 9 mm
C. 4.5 mm
D. 1.5 mm

Explanation

To find the distance of the third dark band from the central bright band in a single-slit diffraction pattern, we can follow these steps:

### Step 1: Identify the given values
- Width of the slit, \( a = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \)
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Distance from the slit to the screen, \( D = 0.5 \, \text{m} \)

### Step 2: Use the formula for dark bands in single-slit diffraction
The condition for the dark bands (minima) in a single-slit diffraction pattern is given by:
\[ a \sin \theta = n \lambda \]
where \( n \) is the order of the dark band (for the third dark band, \( n = 3 \)).

### Step 3: Substitute for the third dark band
For the third dark band:
\[ a \sin \theta = 3 \lambda \]
Thus,
\[ \sin \theta = \frac{3 \lambda}{a} \]

### Step 4: Calculate \( \sin \theta \)
Substituting the values:
\[ \sin \theta = \frac{3 \times 6 \times 10^{-7}}{1 \times 10^{-4}} = \frac{18 \times 10^{-7}}{1 \times 10^{-4}} = 1.8 \times 10^{-3} \]

### Step 5: Use the small angle approximation
For small angles, \( \sin \theta \approx \theta \) (in radians). Therefore:
\[ \theta \approx 1.8 \times 10^{-3} \]

### Step 6: Relate \( x \) (distance of the dark band) to \( \theta \)
The distance \( x \) of the dark band from the central maximum can be related to \( \theta \) by:
\[ x = D \tan \theta \approx D \sin \theta \quad (\text{for small } \theta) \]
Thus,
\[ x \approx D \cdot \theta \]

### Step 7: Substitute \( D \) and \( \theta \)
Substituting the values:
\[ x \approx 0.5 \cdot 1.8 \times 10^{-3} = 0.0009 \, \text{m} = 9 \, \text{mm} \]

### Final Answer
The distance of the third dark band from the central bright band is:
\[ \boxed{9 \, \text{mm}} \]