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.

Options

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

Answer

To solve the problem, we need to find the distance of the third dark band from the central bright band in a single-slit diffraction pattern. Here are the steps to derive the solution:

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

### Step 2: Use the formula for dark bands in single-slit diffraction
The condition for dark bands in single-slit diffraction is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the dark band (n = 1, 2, 3,...).

### Step 3: Express sin θ in terms of Y and D
From geometry, we know:
\[ \sin \theta \approx \frac{Y}{D} \]
where \( Y \) is the distance of the dark band from the central maximum.

### Step 4: Substitute sin θ into the dark band formula
Substituting this into the dark band condition gives:
\[ d \frac{Y}{D} = n \lambda \]
Rearranging this, we find:
\[ Y = \frac{n \lambda D}{d} \]

### Step 5: Calculate Y for the third dark band (n = 3)
Now, substituting \( n = 3 \):
\[ Y = \frac{3 \lambda D}{d} \]
Substituting the known values:
\[ Y = \frac{3 \times (6 \times 10^{-7} \text{ m}) \times (0.5 \text{ m})}{1 \times 10^{-4} \text{ m}} \]

### Step 6: Perform the calculation
Calculating this gives:
\[ Y = \frac{3 \times 6 \times 0.5}{1} \times 10^{-3} \]
\[ Y = \frac{9}{1} \times 10^{-3} \text{ m} \]
\[ Y = 9 \times 10^{-3} \text{ m} \]

### Step 7: Convert to mm
Since \( 1 \text{ m} = 1000 \text{ mm} \):
\[ Y = 9 \text{ mm} \]

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

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