A silt of width `'a'` is illuminated by light of wavelength `6000 A^(@) FOR` What value of `'a` will the :-
(i) First maximum fall at an angle of diffraction of `30^(@)` ?
(ii) First minimum fall at an angle of diffracton `30^(@)` ?

Explanation

To solve the problem, we need to find the slit width 'a' for two cases:

1. When the first maximum falls at an angle of diffraction of \(30^\circ\).
2. When the first minimum falls at an angle of diffraction of \(30^\circ\).

Given:
- Wavelength, \(\lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m}\)
- Angle of diffraction, \(\theta = 30^\circ\)

### Part (i): First Maximum at \(30^\circ\)

For the first maximum, the condition is given by the formula:
\[ \sin \theta = \frac{(m + 1) \lambda}{2a} \]
where \(m = 1\) for the first maximum.

Substituting the values:
\[ \sin 30^\circ = \frac{(1 + 1) \lambda}{2a} \]
\[ \frac{1}{2} = \frac{2 \lambda}{2a} \]
This simplifies to:
\[ \frac{1}{2} = \frac{\lambda}{a} \]
Rearranging gives:
\[ a = 2\lambda \]
Substituting \(\lambda = 6 \times 10^{-7} \, \text{m}\):
\[ a = 2 \times (6 \times 10^{-7}) = 12 \times 10^{-7} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \]

### Part (ii): First Minimum at \(30^\circ\)

For the first minimum, the condition is given by:
\[ \sin \theta = \frac{m \lambda}{a} \]
where \(m = 1\) for the first minimum.

Substituting the values:
\[ \sin 30^\circ = \frac{1 \cdot \lambda}{a} \]
\[ \frac{1}{2} = \frac{\lambda}{a} \]
Rearranging gives:
\[ a = 2\lambda \]
Substituting \(\lambda = 6 \times 10^{-7} \, \text{m}\):
\[ a = 2 \times (6 \times 10^{-7}) = 12 \times 10^{-7} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \]

### Final Answers
- (i) For the first maximum at \(30^\circ\), \(a = 1.2 \times 10^{-6} \, \text{m}\)
- (ii) For the first minimum at \(30^\circ\), \(a = 1.2 \times 10^{-6} \, \text{m}\)