In a Fraunhofer diffraction experiment at a single slit using a light of wavelength 400 nm, the first minimum is formed at an angle of `30^(@)`. The direction `theta` of the first secondary maximum is given by :

Options

A. `tan^(-1)(4/3)`
B. `60^@`
C. `sin^(-1)(3/4)`
D. `tan^(-1)(3/4)`

Answer

To solve the problem regarding the direction of the first secondary maximum in a Fraunhofer diffraction experiment at a single slit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Conditions for Minima and Maxima**:
- For a single slit diffraction, the condition for the minima is given by:
\[ a \sin \theta = n \lambda \]
where \( a \) is the slit width, \( \theta \) is the angle, \( n \) is the order of the minimum, and \( \lambda \) is the wavelength of light.
- The condition for the maxima is given by:
\[ a \sin \theta = \left( n - \frac{1}{2} \right) \lambda \]

2. **Given Information**:
- Wavelength \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
- The first minimum occurs at \( \theta = 30^\circ \) (i.e., \( n = 1 \)).

3. **Finding the Slit Width**:
- For the first minimum (\( n = 1 \)):
\[ a \sin 30^\circ = 1 \cdot \lambda \]
- Since \( \sin 30^\circ = \frac{1}{2} \):
\[ a \cdot \frac{1}{2} = 400 \times 10^{-9} \]
- Therefore, the slit width \( a \) can be calculated as:
\[ a = 2 \cdot 400 \times 10^{-9} = 800 \times 10^{-9} \, \text{m} \]

4. **Finding the Angle for the First Secondary Maximum**:
- For the first secondary maximum, we use \( n = 1 \) in the maxima condition:
\[ a \sin \theta = \left( 1 - \frac{1}{2} \right) \lambda = \frac{1}{2} \lambda \]
- Substituting the value of \( a \):
\[ 800 \times 10^{-9} \sin \theta = \frac{1}{2} \cdot 400 \times 10^{-9} \]
- Simplifying:
\[ 800 \times 10^{-9} \sin \theta = 200 \times 10^{-9} \]
- Dividing both sides by \( 800 \times 10^{-9} \):
\[ \sin \theta = \frac{200 \times 10^{-9}}{800 \times 10^{-9}} = \frac{1}{4} \]

5. **Calculating the Angle**:
- Now, we find \( \theta \):
\[ \theta = \sin^{-1} \left( \frac{1}{4} \right) \]

### Final Answer:
The direction \( \theta \) of the first secondary maximum is given by:
\[ \theta = \sin^{-1} \left( \frac{1}{4} \right) \]