In a diffraction pattern due to single slit of width `'a'`, the first minimum is observed at an angle `30^(@)` when light of wavelength `5000 Å` is inclined on the slit. The first secondary maximum is observed at an angle of:

Options

A. `"sin"^(-1)(2/3)`
B. `"sin"^(-1)(1/2)`
C. `"sin"^(-1) (3/4)`
D. `"sin"^(-1) (1/4)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the condition for the first minimum in single-slit diffraction
In a single-slit diffraction pattern, the condition for the first minimum is given by the equation:
\[ a \sin \theta = n \lambda \]
where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle of the minimum,
- \( n \) is the order of the minimum (for the first minimum, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Substitute the known values into the equation
From the problem, we know:
- \( \theta = 30^\circ \)
- \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)

Substituting these values into the equation for the first minimum:
\[ a \sin(30^\circ) = 1 \cdot \lambda \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ a \cdot \frac{1}{2} = 5 \times 10^{-7} \]
This gives us:
\[ a = 2 \cdot 5 \times 10^{-7} = 1 \times 10^{-6} \, \text{m} = 1000 \, \mu m \]

### Step 3: Find the condition for the first secondary maximum
The condition for the secondary maxima in single-slit diffraction is given by:
\[ a \sin \theta_1 = \left( n + \frac{1}{2} \right) \lambda \]
For the first secondary maximum, \( n = 1 \):
\[ a \sin \theta_1 = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \]

### Step 4: Substitute the value of \( a \) into the equation for the first secondary maximum
Substituting \( a = 1 \times 10^{-6} \, \text{m} \) and \( \lambda = 5 \times 10^{-7} \, \text{m} \) into the equation:
\[ 1 \times 10^{-6} \sin \theta_1 = \frac{3}{2} \cdot 5 \times 10^{-7} \]
This simplifies to:
\[ 1 \times 10^{-6} \sin \theta_1 = \frac{15}{10} \times 10^{-7} \]
\[ \sin \theta_1 = \frac{15}{10} \times \frac{10^{-7}}{10^{-6}} = \frac{15}{10} \cdot 0.1 = \frac{3}{4} \]

### Step 5: Calculate \( \theta_1 \)
Now we find \( \theta_1 \):
\[ \theta_1 = \sin^{-1} \left( \frac{3}{4} \right) \]

### Final Answer
Thus, the angle for the first secondary maximum is:
\[ \theta_1 = \sin^{-1} \left( \frac{3}{4} \right) \]