Two radio stations broadcast their programmes at the same amplitude `A` and at slightly different frequencies `omega_(1)` and `omega_(2)` respectively, where `omega_(1) - omega_(2) = 10^(3) Hz`. A detector receives the signals from the two stations simultaneously, it can only detect signals of intensity `ge 2A^(2)`.
(i) Find the time interval between successive maxima of the intensity of the signal received by the detector.
(ii) Find the time for which the detector remains idle in each cycle of the intensity of the signal.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (i): Finding the time interval between successive maxima of the intensity

1. **Understanding the Signals**:
We have two signals from radio stations with the same amplitude \( A \) and frequencies \( \omega_1 \) and \( \omega_2 \). The difference in frequencies is given as:
\[ \Delta \omega = \omega_1 - \omega_2 = 10^3 \text{ Hz} \]

2. **Resultant Displacement**:
The resultant displacement \( y \) due to the two signals can be expressed using the principle of superposition:
\[ y = A \sin(\omega_1 t) + A \sin(\omega_2 t) \]

3. **Using Trigonometric Identities**:
We can use the identity for the sum of sines:
\[ y = 2A \cos\left(\frac{\Delta \omega t}{2}\right) \sin\left(\frac{\omega_1 + \omega_2}{2} t\right) \]
Here, the amplitude of the resultant wave is:
\[ A' = 2A \cos\left(\frac{\Delta \omega t}{2}\right) \]

4. **Intensity Calculation**:
The intensity \( I \) is proportional to the square of the amplitude:
\[ I \propto (A')^2 = (2A \cos\left(\frac{\Delta \omega t}{2}\right))^2 = 4A^2 \cos^2\left(\frac{\Delta \omega t}{2}\right) \]

5. **Finding Maxima of Intensity**:
For the intensity to be maximum, \( \cos^2\left(\frac{\Delta \omega t}{2}\right) \) must equal 1. This occurs when:
\[ \frac{\Delta \omega t}{2} = n\pi \quad (n \in \mathbb{Z}) \]
Thus, we can write:
\[ t = \frac{2n\pi}{\Delta \omega} \]

6. **Time Interval Between Successive Maxima**:
The time interval between successive maxima (for \( n \) and \( n+1 \)):
\[ \Delta t = \frac{2(n+1)\pi}{\Delta \omega} - \frac{2n\pi}{\Delta \omega} = \frac{2\pi}{\Delta \omega} \]
Substituting \( \Delta \omega = 10^3 \text{ Hz} \):
\[ \Delta t = \frac{2\pi}{10^3} \approx 6.283 \times 10^{-3} \text{ seconds} \]

### Part (ii): Finding the time for which the detector remains idle

1. **Condition for Detection**:
The detector can only detect signals when the intensity \( I \) is greater than or equal to \( 2A^2 \):
\[ 4A^2 \cos^2\left(\frac{\Delta \omega t}{2}\right) \geq 2A^2 \]
Simplifying, we find:
\[ \cos^2\left(\frac{\Delta \omega t}{2}\right) \geq \frac{1}{2} \]

2. **Finding the Argument**:
The above condition implies:
\[ \frac{\Delta \omega t}{2} = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]
This gives us two critical points:
\[ t_1 = \frac{2(\frac{\pi}{4})}{\Delta \omega} = \frac{\pi}{2\Delta \omega}, \quad t_2 = \frac{2(\frac{3\pi}{4})}{\Delta \omega} = \frac{3\pi}{2\Delta \omega} \]

3. **Calculating Idle Time**:
The time interval during which the detector is idle is:
\[ t_{\text{idle}} = t_2 - t_1 = \frac{3\pi}{2\Delta \omega} - \frac{\pi}{2\Delta \omega} = \frac{2\pi}{2\Delta \omega} = \frac{\pi}{\Delta \omega} \]
Substituting \( \Delta \omega = 10^3 \text{ Hz} \):
\[ t_{\text{idle}} = \frac{\pi}{10^3} \approx 3.142 \times 10^{-3} \text{ seconds} \]

### Final Answers:
(i) The time interval between successive maxima of the intensity is approximately \( 6.283 \times 10^{-3} \) seconds.
(ii) The time for which the detector remains idle in each cycle of the intensity is approximately \( 3.142 \times 10^{-3} \) seconds.