A parallel beam of light of wavelength `lambda` is incident normally on a single slit of width d. Diffraction bands are obtained on a screen placed at a distance D from the slit. The second dark band from the central bright band will be at a distance gives by :

Options

A. `(2lambda D)/(d)`
B. `lambda d D`
C. `(lambda D)/(2 d)`
D. `(2 lambda D)/(D)`

Explanation

To find the position of the second dark band in a single-slit diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A parallel beam of light with wavelength \( \lambda \) is incident normally on a single slit of width \( d \).
- The diffraction pattern is observed on a screen placed at a distance \( D \) from the slit.

2. **Identifying the Condition for Dark Bands**:
- The condition for dark bands (minima) 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,...).

3. **Using Small Angle Approximation**:
- For small angles, \( \sin \theta \approx \tan \theta \approx \theta \).
- Therefore, we can rewrite the equation as:
\[ d \tan \theta = n \lambda \]

4. **Relating \( \tan \theta \) to the Position on the Screen**:
- From the geometry of the setup, we have:
\[ \tan \theta = \frac{x}{D} \]
where \( x \) is the distance from the central maximum to the dark band on the screen.

5. **Substituting \( \tan \theta \) into the Dark Band Condition**:
- Substituting \( \tan \theta \) into the dark band condition gives:
\[ d \frac{x}{D} = n \lambda \]
- Rearranging this, we find:
\[ x = \frac{n \lambda D}{d} \]

6. **Finding the Position of the Second Dark Band**:
- For the second dark band, we set \( n = 2 \):
\[ x = \frac{2 \lambda D}{d} \]

### Final Answer:
The position of the second dark band from the central bright band is given by:
\[ x = \frac{2 \lambda D}{d} \]