A light wave is incident normally over a slit of width `24xx10^-5cm`. The angular position of second dark fringe from the central maxima is `30^@`. What is the wavelength of light?

Options

A. `6000 Å`
B. `5000 Å`
C. `3000 Å`
D. `1500 Å`

Explanation

To solve the problem, we will use the formula for the position of dark fringes in single-slit diffraction. The formula for the angular position of the dark fringes is given by:

\[ a \sin \theta = n \lambda \]

where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle of the dark fringe,
- \( n \) is the order of the dark fringe (for the second dark fringe, \( n = 2 \)),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Convert the slit width to meters:**
\[ a = 24 \times 10^{-5} \text{ cm} = 24 \times 10^{-7} \text{ m} \]

2. **Identify the angle and order of the dark fringe:**
\[ \theta = 30^\circ, \quad n = 2 \]

3. **Use the sine function for the angle:**
\[ \sin(30^\circ) = \frac{1}{2} \]

4. **Substitute the values into the diffraction formula:**
\[ a \sin \theta = n \lambda \]
\[ 24 \times 10^{-7} \cdot \frac{1}{2} = 2 \lambda \]

5. **Simplify the equation:**
\[ 12 \times 10^{-7} = 2 \lambda \]

6. **Solve for \( \lambda \):**
\[ \lambda = \frac{12 \times 10^{-7}}{2} = 6 \times 10^{-7} \text{ m} \]

7. **Convert the wavelength to angstroms:**
\[ 1 \text{ m} = 10^{10} \text{ angstroms} \implies \lambda = 6 \times 10^{-7} \text{ m} = 6000 \text{ angstroms} \]

### Final Answer:
The wavelength of light is \( 6 \times 10^{-7} \text{ m} \) or \( 6000 \text{ angstroms} \).