First diffraction minima due to a single slit of width `10^-4`cm is at `theta=30^@`. The wavelength of the light used is –

Options

A. `4000Å`
B. `5000Å`
C. `6000Å`
D. `6250Å`

Answer

To find the wavelength of light used in a single slit diffraction experiment, we can use the formula for the position of the minima:

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

where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle at which the minima occurs,
- \( n \) is the order of the minima (for the first minima, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Identify the given values:**
- Slit width \( a = 10^{-4} \) cm
- Angle \( \theta = 30^\circ \)
- Order of the minima \( n = 1 \)

2. **Convert the slit width from centimeters to meters:**
\[ a = 10^{-4} \text{ cm} = 10^{-4} \times 10^{-2} \text{ m} = 10^{-6} \text{ m} \]

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

4. **Substitute the values into the minima condition formula:**
\[ a \sin \theta = n \lambda \]
\[ 10^{-6} \sin 30^\circ = 1 \cdot \lambda \]
\[ 10^{-6} \cdot \frac{1}{2} = \lambda \]

5. **Calculate the wavelength \( \lambda \):**
\[ \lambda = \frac{10^{-6}}{2} = 5 \times 10^{-7} \text{ m} \]

6. **Convert the wavelength from meters to angstroms:**
\[ 1 \text{ m} = 10^{10} \text{ angstroms} \]
\[ \lambda = 5 \times 10^{-7} \text{ m} \times 10^{10} \text{ angstroms/m} = 5000 \text{ angstroms} \]

### Final Answer:
The wavelength of the light used is \( 5000 \) angstroms.