Light of wavelength `5900 xx 10^(-10)m` falls normally on a slit of width `11.8 xx 10^(-7)m`. The resulting diffraction pattern is received on a screen. The angular position of the first minimum is.

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

To find the angular position of the first minimum in a single-slit diffraction pattern, we can use the formula for the position of minima given by:

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

where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle of the minimum,
- \( m \) is the order of the minimum (for the first minimum, \( m = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Identify the values**:
- Wavelength \( \lambda = 5900 \times 10^{-10} \, \text{m} \)
- Width of the slit \( a = 11.8 \times 10^{-7} \, \text{m} \)

2. **Set up the equation for the first minimum**:
- For the first minimum (\( m = 1 \)):
\[ a \sin \theta = \lambda \]

3. **Substitute the known values into the equation**:
\[ 11.8 \times 10^{-7} \sin \theta = 5900 \times 10^{-10} \]

4. **Solve for \( \sin \theta \)**:
\[ \sin \theta = \frac{5900 \times 10^{-10}}{11.8 \times 10^{-7}} \]
\[ \sin \theta = \frac{5900}{11.8} \times 10^{-3} \]
\[ \sin \theta = 0.5 \]

5. **Find \( \theta \) using the inverse sine function**:
\[ \theta = \sin^{-1}(0.5) \]
\[ \theta = 30^\circ \]

### Final Answer:
The angular position of the first minimum is \( 30^\circ \).