A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000Å 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 _______ 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 mm = \(0.1 \times 10^{-3}\) m
- Wavelength of light (\(\lambda\)) = 6000 Å = \(6000 \times 10^{-10}\) m = \(6 \times 10^{-7}\) m
- Distance from the slit to the screen (d) = 0.5 m

### Step 2: Write the condition for the dark bands
The condition for the dark bands in a single slit diffraction pattern is given by:
\[ a \sin \theta = n \lambda \]
For the third dark band, \(n = 3\):
\[ a \sin \theta = 3 \lambda \]

### Step 3: Express \(\sin \theta\)
From the equation above, we can express \(\sin \theta\):
\[ \sin \theta = \frac{3 \lambda}{a} \]

### Step 4: Relate \(\sin \theta\) to the position on the screen
In the small angle approximation (which is valid for diffraction patterns), we can relate \(\sin \theta\) to the position \(y\) of the dark band on the screen:
\[ \sin \theta \approx \frac{y}{d} \]
where \(y\) is the distance of the dark band from the central bright band.

### Step 5: Set the two expressions for \(\sin \theta\) equal
Equating the two expressions for \(\sin \theta\):
\[ \frac{y}{d} = \frac{3 \lambda}{a} \]

### Step 6: Solve for \(y\)
Rearranging gives:
\[ y = \frac{3 \lambda d}{a} \]

### Step 7: Substitute the values
Substituting the known values:
\[ y = \frac{3 \times (6 \times 10^{-7} \text{ m}) \times (0.5 \text{ m})}{0.1 \times 10^{-3} \text{ m}} \]

### Step 8: Calculate \(y\)
Calculating the above expression:
\[ y = \frac{3 \times 6 \times 0.5}{0.1} \times 10^{-7 + 3} = \frac{9}{0.1} \times 10^{-4} = 90 \times 10^{-4} \text{ m} = 9 \times 10^{-3} \text{ m} \]
Converting to millimeters:
\[ y = 9 \text{ mm} \]

### Final Answer
The distance of the third dark band from the central bright band is **9 mm**.