In interference, `I_(max)/I_(min) = alpha` , find
(a) ` A_(max)/A_(min)` (b)` A_1/A_2` (c)`I_1/I_2` .

Explanation

To solve the problem step by step, we will derive the required quantities from the given relationship \( \frac{I_{max}}{I_{min}} = \alpha \).

### Given:
\[ \frac{I_{max}}{I_{min}} = \alpha \]

### Step 1: Relate Intensities to Amplitudes
The intensity \( I \) of a wave is proportional to the square of its amplitude \( A \):
\[ I \propto A^2 \]
Thus, we can express the maximum and minimum intensities in terms of amplitudes:
\[ I_{max} = k A_{max}^2 \quad \text{and} \quad I_{min} = k A_{min}^2 \]
where \( k \) is a proportionality constant.

### Step 2: Substitute into the Given Relationship
Substituting the expressions for \( I_{max} \) and \( I_{min} \) into the given equation:
\[ \frac{k A_{max}^2}{k A_{min}^2} = \alpha \]
This simplifies to:
\[ \frac{A_{max}^2}{A_{min}^2} = \alpha \]

### Step 3: Find \( \frac{A_{max}}{A_{min}} \)
Taking the square root of both sides gives:
\[ \frac{A_{max}}{A_{min}} = \sqrt{\alpha} \]
This is the answer for part (a).

### Step 4: Find the Ratio of Amplitudes \( \frac{A_1}{A_2} \)
Using the relationship for maximum and minimum amplitudes in interference:
\[ \frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} \]
This implies:
\[ \alpha = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} \]
Taking the square root:
\[ \sqrt{\alpha} = \frac{A_1 + A_2}{A_1 - A_2} \]

### Step 5: Rearranging to Find \( \frac{A_1}{A_2} \)
Cross-multiplying gives:
\[ \sqrt{\alpha}(A_1 - A_2) = A_1 + A_2 \]
This can be rearranged to:
\[ A_1(\sqrt{\alpha} - 1) = A_2(\sqrt{\alpha} + 1) \]
Thus, we find:
\[ \frac{A_1}{A_2} = \frac{\sqrt{\alpha} + 1}{\sqrt{\alpha} - 1} \]
This is the answer for part (b).

### Step 6: Find the Ratio of Intensities \( \frac{I_1}{I_2} \)
Using the relationship between intensity and amplitude:
\[ \frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \]
Substituting the expression found for \( \frac{A_1}{A_2} \):
\[ \frac{I_1}{I_2} = \left(\frac{\sqrt{\alpha} + 1}{\sqrt{\alpha} - 1}\right)^2 \]
This is the answer for part (c).

### Final Answers:
(a) \( \frac{A_{max}}{A_{min}} = \sqrt{\alpha} \)
(b) \( \frac{A_1}{A_2} = \frac{\sqrt{\alpha} + 1}{\sqrt{\alpha} - 1} \)
(c) \( \frac{I_1}{I_2} = \left(\frac{\sqrt{\alpha} + 1}{\sqrt{\alpha} - 1}\right)^2 \)