A slit of width `d` is illuminated by a monochromatic light of wavelength `700 nm` at normal incidence. Calculate the value of `d` for position of (i) first minimum at an angle of diffraction of `30^(@)` (ii) first maximum at an angle of diffraction of `30^(@)`.

Explanation

To solve the problem, we need to calculate the width of the slit `d` for two cases: (i) the first minimum and (ii) the first maximum at an angle of diffraction of \(30^\circ\) for light of wavelength \(700 \, \text{nm}\).

### Step-by-Step Solution:

**Given:**
- Wavelength, \(\lambda = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m}\)
- Angle of diffraction, \(\theta = 30^\circ\)

**(i) Calculate `d` for the first minimum:**

1. **Understanding the condition for the first minimum:**
The condition for the first minimum in single-slit diffraction is given by:
\[ d \sin \theta = n \lambda \]
where \(n\) is the order of the minimum. For the first minimum, \(n = 1\).

2. **Substituting the values:**
\[ d \sin(30^\circ) = 1 \times 700 \times 10^{-9} \]

3. **Calculating \(\sin(30^\circ)\):**
\(\sin(30^\circ) = \frac{1}{2}\)

4. **Substituting \(\sin(30^\circ)\) into the equation:**
\[ d \cdot \frac{1}{2} = 700 \times 10^{-9} \]

5. **Solving for `d`:**
\[ d = 700 \times 10^{-9} \times 2 = 1400 \times 10^{-9} \, \text{m} = 1.4 \times 10^{-6} \, \text{m} = 14 \times 10^{-7} \, \text{m} \]

**(ii) Calculate `d` for the first maximum:**

1. **Understanding the condition for the first maximum:**
The condition for the first maximum in single-slit diffraction is given by:
\[ d \sin \theta = \left( n + \frac{1}{2} \right) \lambda \]
For the first maximum, we take \(n = 1\).

2. **Substituting the values:**
\[ d \sin(30^\circ) = \left( 1 + \frac{1}{2} \right) \times 700 \times 10^{-9} \]

3. **Calculating \(\sin(30^\circ)\):**
\(\sin(30^\circ) = \frac{1}{2}\)

4. **Substituting \(\sin(30^\circ)\) into the equation:**
\[ d \cdot \frac{1}{2} = \frac{3}{2} \times 700 \times 10^{-9} \]

5. **Solving for `d`:**
\[ d = \frac{3 \times 700 \times 10^{-9}}{1} = 2100 \times 10^{-9} \, \text{m} = 2.1 \times 10^{-6} \, \text{m} = 21 \times 10^{-7} \, \text{m} \]

### Final Answers:
- (i) Width of the slit for the first minimum: \(d = 1.4 \times 10^{-6} \, \text{m}\)
- (ii) Width of the slit for the first maximum: \(d = 2.1 \times 10^{-6} \, \text{m}\)