First diffraction minima due to a single slit of width `1.0 xx 10^(-5)cm` is at `30^(@)`. The wavelength of light used is :

Options

A. `400 Å`
B. `500 Å`
C. `600 Å`
D. `700 Å`

Answer

To find the wavelength of light used in the diffraction pattern created by a single slit, we can use the formula for the position of the first diffraction minima. The condition for the first minima in single-slit diffraction is given by:

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

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Width of the slit, \( b = 1.0 \times 10^{-5} \) cm
- Angle for the first minima, \( \theta = 30^\circ \)

2. **Convert the width of the slit from cm to meters**:
\[ b = 1.0 \times 10^{-5} \text{ cm} = 1.0 \times 10^{-5} \times 10^{-2} \text{ m} = 1.0 \times 10^{-7} \text{ m} \]

3. **Use the sine function for the angle**:
\[ \sin(30^\circ) = 0.5 \]

4. **Substitute the values into the diffraction minima formula**:
\[ b \sin \theta = n \lambda \]
For the first minima (\( n = 1 \)):
\[ 1.0 \times 10^{-7} \text{ m} \times 0.5 = 1 \times \lambda \]

5. **Calculate the wavelength \( \lambda \)**:
\[ \lambda = 1.0 \times 10^{-7} \text{ m} \times 0.5 = 0.5 \times 10^{-7} \text{ m} \]

6. **Convert the wavelength to nanometers**:
\[ \lambda = 0.5 \times 10^{-7} \text{ m} = 5.0 \times 10^{-8} \text{ m} = 500 \text{ nm} \]

### Final Answer:
The wavelength of light used is \( 500 \text{ nm} \).